Lecture on infinite series and differential equations120

INFINITE SERIES

Definitions of Infinite Series and Fundamental Facts

1.1 Deﬁnitions Let {un} be a sequence of real numbers Then, the formal sum

Its value, if one exists, is the limit of the sequence of partial sums {Sn= ∞ =

If the limit exists, the series is said to converge to that sum If the limit does not exist, the , S series is said to diverge

Sometimes the character of a series is obvious For example, the series generated by the frog on the log surely converges, while ∑ ∞

=1 n n diverges On the other hand, the variable series raises questions

The series can be derived from the division 1/(1-x), valid for -1 < x < 1, where the sums Sn provide approximations to 1/(1-x) that converge to the exact value in the limit However, an indeterminate form occurs at x = -1 Notably, prominent mathematicians like Leonard Euler proposed that S should equal 1/2, based on substituting -1 into the equation 1/(1-x) This conclusion presents challenges upon further investigation.

The alternating series of 1 and -1 demonstrates that specific associations can yield values of either 1 or 0 However, the requirement for uniqueness in convergence classifies this series as divergent, thereby removing any potential ambiguity in other scenarios.

∞ lim =0 The converse, however, is not necessarily true, i.e., if → n un

= 1 n un may or may not converge It follows that if the n th term of a series does not approach zero, the series is divergent

2 Multiplication of each term of a series by a constant diﬀerent from zero does not aﬀect the convergence or divergence

3 Removal (or addition) of a ﬁnite number of terms from (or to) a series does not aﬀect the convergence or divergence

=1 n vn converges, limn-> ∞Tn exists and equals T, say Also, since vn ≥ 0, Tn ≤T

Then Sn =u1+ u2 + •••+un ≤ v1+ v2 + ••• + vn ≤ T or 0 ≤ Sn ≤ T

Thus {Sn}is a bounded monotonic increasing sequence and must have a limit, i.e., ∑ ∞

(b) The proof of (b) is left for the reader as an exercise

2.2 The Limit-Comparison or Quotient Test for series of non-negative terms

A=0 or A=∞, it is easy to prove the assertions (b) and (c)

This test is related to the comparison test and is often a very useful alternative to it In particular, taking vn = l/n p , we have the following theorem

2.3 Integral test for series of non-negative terms

PROOF of Integral test: conditions (a) and (b)

= 1 n un is called absolutely convergent if ∑ ∞

= 1 n un is called conditionally convergent.

Lemma: The absolutely convergent series is convergent

Proof: a) Since L 0 such that 00 such that M

|f (n) (x)| ≤ M for all x∈(-R,R) and all , n then the series ∑ ∞

!(0) n n n n x f is convergent to f(x) on (-R,R) In other words: f(x)=∑ ∞

PROOF This is direct consequence of the Taylor’s formula with Lagrange’s Remainder.

The following series, convergent to the given function in the indicated intervals, are frequently employed in practice:

Fourier Series

In the eighteenth century, mathematicians like Daniel Bernoulli and Leonard Euler tackled the vibratory motion of a stretched string using partial differential equations that lacked solutions in elementary functions To overcome this challenge, they introduced infinite series of sine and cosine functions that satisfied these equations This foundational work paved the way for Joseph Fourier in the early nineteenth century, who further explored these concepts while investigating heat transfer.

Consequently, they were named after him Fourier series are investigated in this section As you explore the ideas, notice the similarities and diﬀerences with the inﬁnite series

4.1 Periodic functions: A function f(x) is said to have a period T or to be periodic with period T if for all x, f{x + T) = f(x), where T is a positive constant The least value of T > 0 is called the least period or simply the period of f(x)

EXAMPLE 1 The function sinx has periods 2 , 4 , 6 , , since sin(x + 2 ), sin(x + 4 ), sin π π π π π (x +6 ), all equal sinx However, 2 is the least period or the period of sinx π π

EXAMPLE 2 The period of sinn πx or cosnπx, where n is a positive integer, is 2 /n π

EXAMPLE 3 The period of tanx is π

EXAMPLE 4 A constant has any positive number as period

Other examples of periodic functions are shown in the graphs of Figures 13-1 (a), (b), and (c) below

A function f(x) is called odd if f(-x) =-f(x) Thus, x 3 + x 5 - 3x 3 + 2x, sin x, tan 3x are odd functions

A function f(x) is called even if f(-x)=f(x) Thus, x 2 , 2x 4 -4x 2 +5, cos x, e x + e -x are even functions

The functions portrayed graphically in Figures 13-1 (a) and 13-1 (b) are odd and even respectively, but that of Fig 13-l(c) is neither odd nor even

In Fourier series, odd functions exclusively contain sine terms, while even functions consist solely of cosine terms, which may include a constant term treated as a cosine.

4.5 Half Range Fourier Sine or Cosine Series

A half range Fourier sine or cosine series consists solely of sine or cosine terms, respectively To construct a half range series for a specific function, the function is typically defined over the interval (0, L), which represents half of the full interval (-L, L) This definition allows the function to be categorized as either odd or even, thereby extending its definition to the other half of the interval, (-L, 0).

If an and bn are the Fourier coeﬀicients corresponding to f(x) and if f(x) satisﬁes the Dirichlet conditions Then

4.7 Diﬀerentiation and Integration of Fourier Series

Differentiation and integration of Fourier series can be justified through established theorems applicable to series in general It is important to note that these theorems offer sufficient but not necessary conditions A particularly useful theorem for integration will be discussed.

4.8 Complex Notation for Fourier Series

BASIC CONCEPT OF DIFFERENTIAL EQUATIONS

Examples of Differential Equations

In the study of population dynamics, let N(t) represent the quantity of a substance that is experiencing growth or decay If we assume that the rate of change of this substance over time, denoted as dN/dt, is directly proportional to the current amount present, we can express this relationship mathematically as dN/dt = kN, or alternatively, dN/dt - kN = 0 (1.1), where k signifies the constant of proportionality.

Assuming N(t) is a differentiable and continuous function of time is a common approach; however, this assumption is not accurate for population problems where N(t) is discrete and integer-valued Despite this limitation, the equation (1.1) still offers a valuable approximation of the physical laws that govern such systems.

Newton's law of cooling, which also applies to heating, indicates that the rate of temperature change of an object is proportional to the temperature difference between the object and its environment Denoting the object's temperature as T and the surrounding medium's temperature as Tm, the rate of temperature change can be expressed as dT/dt This relationship can be mathematically formulated as dT/dt = -k(T - Tm), highlighting the inverse correlation between the object's temperature and the surrounding temperature.

In Newton's law of cooling, a positive constant of proportionality, denoted as k, is essential to ensure that the rate of temperature change (dT/dt) is negative during the cooling process when the object's temperature (T) exceeds the surrounding temperature.

Tm, and positive in a heating process, when T is less than Tm

The basic equation governing the amount of current / (in amperes) in a simple RL circuit (see

Figure 1-2) consisting of a resistance R (in ohms), an inductor L (in henries), and an electromotive force (abbreviated emf) E (in volts) is

In an RC circuit that includes resistance, capacitance (C in farads), and an electromotive force (emf) but lacks inductance, the relationship governing the electrical charge (q in coulombs) on the capacitor is defined by a specific equation.

Definitions and Related Concepts

2.1 Definition diﬀerential equationA is an equation involving an unknown function and its derivatives

The following are diﬀerential equations involving the unknown function y

A differential equation is classified as an ordinary differential equation when the unknown function relies on a single independent variable Conversely, if the unknown function depends on two or more independent variables, it is categorized as a partial differential equation.

The order of a diﬀerential equation is the order of the highest derivative appearing in the equation

2.2 Solution A solution of a differential equation in the unknown function y and the independent variable x on the interval is a function y(x) that satisfies the differential J equation identically for all x in J

Example: The function y(x) = c1sin2x + c 2 cos2x, where c 1 and c 2 are arbitrary constants, is a solution of y" + 4y = 0 in the interval (- , ) ∞ ∞

A particular solution of a diﬀerential equation is any one solution The general solution of a diﬀerential equation is the set of all solutions

2.4 Initial-Value and Boundary-Value Problems always be written as a quotient of two other functions -M(x,y) and N(x,y) Then (2.1) becomes dy/dx = -M(x,y)/N(x, y), which is equivalent to the diﬀerential form

SOLUTIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS

Separable Equations

1.1 Definition: Consider a differential equation in differential form (1.4) If M(x,y) =A(x)

(a function only of x) and N(x,y) = B(y) (a function only of y), diﬀerential equation is separable, or has its variables separated

1.2 General Solution:The solution to the ﬁrst-order separable diﬀerential equation

(1.2) where c represents an arbitrary constant

Example Solve the equation: dx dy y x 2 +2

This equation may be rewritten in the diﬀerential form

(x 2 +2)dx-ydy = 0 which is separable with A(x) =x 2 + 2 and B(y) = -y Its solution is

The integrals presented in Equation (1.2) can often be challenging or impossible to evaluate In these situations, numerical methods are employed to find an approximate solution Additionally, even if the integrations can be completed, it may not be feasible to express y explicitly in terms of x, leading to solutions that remain in implicit form.

1.3 Solutions to the Initial-Value Problem:

The solution to the initial-value problem

A(x)dx + B(y)dy = 0; y(x) = y (1.3) y = xv (2.6) along with its corresponding derivative: dx dy = v+ dx xdv (2.7)

Then we obtain v+ dx xdv=g(v) This can be rewritten as x dx v v g dv )−

The resulting equation in the variables v and x is solved as a separable diﬀerential equation; the required solution to Equation (2.5) is obtained by back substitution

The case g(v) = v yields another solution of the form y = kx for any constant k

To solve the differential equation y' = x^2 y + for x ≠ 0, we recognize that it is not separable but takes the form y' = f(x, y), where f(x, y) = x^2 y + This function is homogeneous, as it satisfies the condition f(tx, ty) = t^2 f(x, y) By substituting the appropriate transformations into the equation, we arrive at v + x^2 dv = x^2 v +, which can be further simplified algebraically.

This last equation is separable; its solution is which, when evaluated, yields v = ln |x | - c, or v = ln|kx| (26) where we have set c = -ln|k|; and have noted that ln|x| +ln|k| = ln|xk|

Finally, substituting v = y/x back into (26), we obtain the solution to the given diﬀerential equation as y = xln|kx|

3.1 Definition: A differential equation in differential form

M(x, y)dx + N(x, y)dy = 0 (27) is exact if there exists a function g(x, y) such that dg(x, y) = M(x, y)dx + N(x, y)dy (28)

3.2 Test for exactness: If M(x,y) and N(x,y) are continuous functions and have continuous ﬁrst partial derivatives on some rectangle of the xy-plane, then Equation (27) is exact if and only if x y x

3.3 Solution: To solve Equation (27), assuming that it is exact, ﬁrst solve the equations

∂ (29) for g(x, y) The solution to (27) is then given implicitly by g(x,y) = c (30) where c represents an arbitrary constant

Equation (30) is immediate from Equations (26) and (27) If (27) is substituted into (26), we obtain dg(x, y(x)) = 0 Integrating this equation (note that we can write 0 as 0dx), we have

∫ dg ( x , y ( x )) = 0 dx, which, in turn, implies (30)

This equation has the form of Equation (26) with M(x, y) = 2xy and N(x, y) = 1 + x 2 Since x y x

∂ ( , ) ( , ) = 2x, the diﬀerential equation is exact Because this equation is exact, we now determine a function g(x, y) that satisﬁes Equations (2.28) and (2.29) Substituting M(x, y) = 2xy into (2.28), we obtain x y x g

The equation ∂(x, y) = 2xy can be integrated with respect to x, yielding g(x, y) = x²y + h(y) It is important to note that the integration constant can vary with y To find h(y), we differentiate the resulting equation with respect to y, which leads to the expression y y x g.

Substituting this equation along with N(x, y) =1 + x 2 into (29), we have x 2 +h'(y) = 1 + x 2 or h'(y) = 1

Integrating this last equation with respect to y, we obtain h(y) = y + c 1 (c 1 = constant)

Substituting this expression into (31) yields g(x, y) = x 2 y + y + c 1 The solution to the diﬀerential equation, which is given implicitly by (30) as g(x, y) = c, is x 2 y + y = c 2

(c 2 = c-c 1 ) Solving for y explicitly, we obtain the solution as y = c2/(x 2 +1)

In general, Equation (27) is not exact Occasionally, it is possible to transform (27) into an exact diﬀerential equation by a judicious multiplication A function I(x, y) is an integrating

Integrating factors can be challenging to identify, and if a differential equation does not fit the specified forms, the search for an integrating factor is unlikely to succeed In such cases, it is advisable to explore alternative solution methods.

This equation is not exact It is easy to see that an integrating factor is I(x)=1/x 2 Therefore, we can rewrite the given diﬀerential equation as

− x ydx xdy which is exact This equation can be solved using the steps described in equations (28) through (30)

Alternatively, we can rewrite the above equation as d (y/x) = 0 Hence, by direct integration, we have y / x = c, or y = cx, as the solution

4.1 Definition: A first-order linear differential equation has the form y′ + p(x)y = q(x) (33)

4.2 Method of Solutions: An integrating factor for Equation (2.33) is

I(x)=e∫ p ( x ) dx (34) which depends only on x and is independent of y When both sides of (33) are multiplied by

I(x) y′+ I(x)p(x)y = q(x)I(x) (35) is exact This equation can be solved by the method described previously

A simpler procedure is to rewrite (23) as dx Iq

Iy d( )= , and integrate both sides of this last equation with respect to x, then solve the resulting equation for y The general solution for Equation (33) is y ( x ) = e − ∫ p ( x ) dx ( ∫ e ∫ p ( x ) dx q ( x ) dx + C ) (36)

Using (36) for p(x)=4/x and q(x)=x 4 , we obtain the general solution of the given equation as

A Bernoulli differential equation is expressed as y′ + p(x)y = q(x)y^α, where α is a real number not equal to 0 or 1 If α > 0, then y ≡ 0 is a solution; if α < 0, the condition requires y > 0 To find solutions for y > 0, we manipulate the equation by dividing both sides by y^α, leading to y^(1-α)y′ + p(x)y^(1-α) = q(x) By substituting z = y^(1-α), the equation transforms into a linear differential equation in terms of z(x) This mathematical framework is significant in various fields, including social sciences, as it establishes relationships between variables and parameters in complex problems.

Mathematical equations play a crucial role in predicting the behavior of natural processes under various conditions, allowing for easy manipulation of parameters compared to the time-consuming or costly nature of experimental methods Both mathematical modeling and experimental observation are essential and complement each other in scientific research Validation of mathematical models occurs through the comparison of their predictions with experimental data, while mathematical analyses can guide experimental directions and highlight the most valuable data to collect In Section 1.1, we explored several simple mathematical models and will now expand on the key conclusions drawn Regardless of the application field, three fundamental steps are consistently present in the mathematical modeling process.

6.1 Construction of the Model This involves a translation of the physical situation into mathematical terms, often using the steps listed at the end of Section 1.1 Perhaps most critical at this stage is to state clearly the physical principle(s) that are believed to govern the process For example, it has been observed that in some circumstances heat passes from a warmer to a cooler body at a rate proportional to the temperature difference, that objects move about in accordance with Newton’s laws of motion, and that isolated insect populations grow at a rate proportional to the current population Each of these statements involves a rate of change (derivative) and consequently, when expressed mathematically, leads to a differential equation The differential equation is a mathematical model of the process It is important to realize that the mathematical equations are almost always only an approximate description of the actual process For example, bodies moving at speeds comparable to the speed of light are not governed by Newton’s laws, insect populations do not grow indefinitely as stated because of eventual limitations on their food supply, and heat transfer is affected by factors other than the temperature difference Alternatively, one can adopt the point of view that the mathematical equations exactly describe the operation of a simplified physical model, which has been constructed (or conceived of) so as to embody the most important features of the actual process Sometimes, the process of mathematical modelling involves the conceptual replacement of a discrete process by a continuous one For instance, the number of members in an insect population changes by discrete amounts; however, if the population is large, it seems reasonable to consider it as a continuous variable and even to speak of its derivative

6.2 Analysis of the Model Once the problem has been formulated mathematically, one is often faced with the problem of solving one or more differential equations or, failing that, of finding out as much as possible about the properties of the solution It may happen that this mathematical problem is quite difficult and, if so, further approximations may be indicated at this stage to make the problem mathematically tractable For example, a nonlinear equation may be approximated by a linear one, or a slowly varying coefficient may be replaced by a constant Naturally, any such approximations must also be examined from the physical point of view to make sure that the simplified mathematical problem still reflects the essential features of the physical process under investigation At the same time, an intimate knowledge of the physics of the problem may suggest reasonable mathematical approximations that will make the mathematical problem more amenable to analysis This interplay of understanding of physical phenomena and knowledge of mathematical techniques and their limitations is characteristic of applied mathematics at its best, and is indispensable in successfully constructing useful mathematical models of intricate physical processes

6.3 Comparison with Experiment or Observation Finally, having obtained the solution (or at least some information about it), you must interpret this information in the context in which the problem arose In particular, you should always check that the mathematical solution appears physically reasonable If possible, calculate the values of the solution at selected points and compare them with experimentally observed values Or, ask whether the behavior of the solution after a long time is consistent with observations Or, examine the solutions corresponding to certain special values of parameters in the problem Of course, the fact that the mathematical solution appears to be reasonable does not guarantee it is correct However, if the predictions of the mathematical model are seriously inconsistent with observations of the physical system it purports to describe, this suggests that either errors have been made in solving the mathematical problem, or the mathematical model itself needs refinement, or observations must be made with greater care In Chapter 1 we have given some examples which are typical of applications in which first-order differential equations arise In this section we pay our attention to a concrete model, that is a mathematical model of electric circuits We start with some important facts from electric circuits

6.4 Electric circuits The simplest electric circuit is a series circuit in which we have a source of electric energy (electromotive force) such as a generator or a battery, and a resistor, which uses the energy Experiments show that the following law holds

The voltage drop ER across a resistor is proportional to the instantaneous the current I, say,

Ohm's Law states that the voltage (E) across a resistor is equal to the product of the current (I) flowing through it and its resistance (R), expressed as E = RI In this equation, resistance is measured in Ohms, current in amperes, and voltage in volts.

Inductors and capacitors are crucial components in complex circuits An inductor resists changes in current, exhibiting an inertia effect in electricity akin to mass in mechanics This analogy will be explored further in subsequent discussions.

The voltage drop ER across an inductor is proportional to the instantaneous time rate of change of the current I, say, dI

In any closed loop, the algebraic sum of all instantaneous voltage drops equals zero, meaning that the total voltage applied to the loop is balanced by the sum of the voltage drops within that loop.

Model the “RL-circuit” in ﬁg 2.2 and solve the resulting equation for: (A) a constant electromotive force; (B) a period electromotive force

In the first step of the solution, we model the circuit by applying Kirchhoff's Voltage Law (KVL) According to the first law, the voltage drop across the resistor is represented as RI, while the voltage drop across the inductor is expressed as LdI/dt The total of these voltage drops must equal the electromotive force E(t).

LdI + 2 nd Step Solution of the equation In order to use the formula (2.36) we transform the above equation to the standard form by deviding both side to L and obtain

Using now formula (2.36) with x=t, y=I, p=R/L, and q=E/L we get

3 rd Step Case A: Constant electromotive force E=E0 The above equality for I(t) yields t t t E ce c

As time approaches infinity, the current I(t) stabilizes to a constant value of E0/R, which reflects the immediate current value predicted by Ohm's law in the absence of an inductor in the circuit This steady-state current is unaffected by the initial current value I(0).

Case B: Periodic electromotive force E=E0 Sin ωt For this E(t) we have that

I( ) α t 0 α t sinω for = R/L α Integration by part yields

The exponential term will approach zero as t tends to inﬁnity This mean that after some time the current I(t) will execute practically harmonic oscillations

We now ﬁnish this chapter by stating the theorem on existence and uniqueness of the solution of an initial-value problem for a ﬁrst-order differential equation

Let the functions f(t,y) and ∂f/∂y be continuous in some rectangle α < t < β γ, < y < δ containing the point (t 0 ,y 0 ) Then, in some interval t 0 − h < t < t 0 + h contained in α < t < β, there is a unique solution y = φ(t) of the initial value problem y’ = f (t, y), y(t 0 ) = y 0

I In each of Problems 1 through 8 solve the given diﬀerential equation

II In each of Problems 9 through 20 ﬁnd the solution of the given initial value problem in explicit form

V Show that the equations in Problems 1 through 2 are not exact, but become exact when multiplied by the given integrating factor Then solve the equations

Exact equations

3.1 Definition: A differential equation in differential form

M(x, y)dx + N(x, y)dy = 0 (27) is exact if there exists a function g(x, y) such that dg(x, y) = M(x, y)dx + N(x, y)dy (28)

3.2 Test for exactness: If M(x,y) and N(x,y) are continuous functions and have continuous ﬁrst partial derivatives on some rectangle of the xy-plane, then Equation (27) is exact if and only if x y x

3.3 Solution: To solve Equation (27), assuming that it is exact, ﬁrst solve the equations

∂ (29) for g(x, y) The solution to (27) is then given implicitly by g(x,y) = c (30) where c represents an arbitrary constant

Equation (30) is immediate from Equations (26) and (27) If (27) is substituted into (26), we obtain dg(x, y(x)) = 0 Integrating this equation (note that we can write 0 as 0dx), we have

∫ dg ( x , y ( x )) = 0 dx, which, in turn, implies (30)

This equation has the form of Equation (26) with M(x, y) = 2xy and N(x, y) = 1 + x 2 Since x y x

∂ ( , ) ( , ) = 2x, the diﬀerential equation is exact Because this equation is exact, we now determine a function g(x, y) that satisﬁes Equations (2.28) and (2.29) Substituting M(x, y) = 2xy into (2.28), we obtain x y x g

The equation ∂(x, y) = 2xy can be integrated with respect to x, leading to the expression g(x, y) = x²y + h(y), where h(y) is an arbitrary function dependent on y It's important to note that when integrating with respect to x, the constant of integration may vary with y To find the function h(y), we differentiate the resulting equation with respect to y, resulting in the expression y * ∂g/∂y = x.

Substituting this equation along with N(x, y) =1 + x 2 into (29), we have x 2 +h'(y) = 1 + x 2 or h'(y) = 1

Integrating this last equation with respect to y, we obtain h(y) = y + c 1 (c 1 = constant)

Substituting this expression into (31) yields g(x, y) = x 2 y + y + c 1 The solution to the diﬀerential equation, which is given implicitly by (30) as g(x, y) = c, is x 2 y + y = c 2

(c 2 = c-c 1 ) Solving for y explicitly, we obtain the solution as y = c2/(x 2 +1)

In general, Equation (27) is not exact Occasionally, it is possible to transform (27) into an exact diﬀerential equation by a judicious multiplication A function I(x, y) is an integrating

Integrating factors can be challenging to identify, and if a differential equation does not match the specified forms, it is unlikely that an integrating factor will be found In such cases, it is advisable to explore alternative solution methods.

This equation is not exact It is easy to see that an integrating factor is I(x)=1/x 2 Therefore, we can rewrite the given diﬀerential equation as

− x ydx xdy which is exact This equation can be solved using the steps described in equations (28) through (30)

Alternatively, we can rewrite the above equation as d (y/x) = 0 Hence, by direct integration, we have y / x = c, or y = cx, as the solution.

Linear Equations

4.1 Definition: A first-order linear differential equation has the form y′ + p(x)y = q(x) (33)

4.2 Method of Solutions: An integrating factor for Equation (2.33) is

I(x)=e∫ p ( x ) dx (34) which depends only on x and is independent of y When both sides of (33) are multiplied by

I(x) y′+ I(x)p(x)y = q(x)I(x) (35) is exact This equation can be solved by the method described previously

A simpler procedure is to rewrite (23) as dx Iq

Iy d( )= , and integrate both sides of this last equation with respect to x, then solve the resulting equation for y The general solution for Equation (33) is y ( x ) = e − ∫ p ( x ) dx ( ∫ e ∫ p ( x ) dx q ( x ) dx + C ) (36)

Using (36) for p(x)=4/x and q(x)=x 4 , we obtain the general solution of the given equation as

Bernoulli Equations

A Bernoulli differential equation is expressed as y′ + p(x)y = q(x)y^α, where α is a real number not equal to 0 or 1 If α > 0, then y ≡ 0 serves as a solution; if α < 0, the condition y > 0 must be satisfied To find solutions when y > 0, we divide both sides by y^α, resulting in the equation y^(1-α)y′ + p(x)y^(1-α) = q(x) By substituting z = y^(1-α), the original equation transforms into a linear differential equation in terms of the function z(x), facilitating the solution process This approach highlights the significance of mathematical models in connecting variables and parameters within various fields, including social sciences.

Mathematical equations play a crucial role in predicting natural processes across various scenarios, allowing for easy parameter adjustments that would be costly and time-consuming in experimental settings Both mathematical modeling and experimental observation are essential and complement each other in scientific research; models are validated by comparing predictions with experimental data, while mathematical analyses can guide experimental directions and identify useful data In Section 1.1, we explored several simple mathematical models, and we will now expand on the conclusions drawn from that section Regardless of the specific application field, the mathematical modeling process consistently involves three identifiable steps.

6.1 Construction of the Model This involves a translation of the physical situation into mathematical terms, often using the steps listed at the end of Section 1.1 Perhaps most critical at this stage is to state clearly the physical principle(s) that are believed to govern the process For example, it has been observed that in some circumstances heat passes from a warmer to a cooler body at a rate proportional to the temperature difference, that objects move about in accordance with Newton’s laws of motion, and that isolated insect populations grow at a rate proportional to the current population Each of these statements involves a rate of change (derivative) and consequently, when expressed mathematically, leads to a differential equation The differential equation is a mathematical model of the process It is important to realize that the mathematical equations are almost always only an approximate description of the actual process For example, bodies moving at speeds comparable to the speed of light are not governed by Newton’s laws, insect populations do not grow indefinitely as stated because of eventual limitations on their food supply, and heat transfer is affected by factors other than the temperature difference Alternatively, one can adopt the point of view that the mathematical equations exactly describe the operation of a simplified physical model, which has been constructed (or conceived of) so as to embody the most important features of the actual process Sometimes, the process of mathematical modelling involves the conceptual replacement of a discrete process by a continuous one For instance, the number of members in an insect population changes by discrete amounts; however, if the population is large, it seems reasonable to consider it as a continuous variable and even to speak of its derivative

6.2 Analysis of the Model Once the problem has been formulated mathematically, one is often faced with the problem of solving one or more differential equations or, failing that, of finding out as much as possible about the properties of the solution It may happen that this mathematical problem is quite difficult and, if so, further approximations may be indicated at this stage to make the problem mathematically tractable For example, a nonlinear equation may be approximated by a linear one, or a slowly varying coefficient may be replaced by a constant Naturally, any such approximations must also be examined from the physical point of view to make sure that the simplified mathematical problem still reflects the essential features of the physical process under investigation At the same time, an intimate knowledge of the physics of the problem may suggest reasonable mathematical approximations that will make the mathematical problem more amenable to analysis This interplay of understanding of physical phenomena and knowledge of mathematical techniques and their limitations is characteristic of applied mathematics at its best, and is indispensable in successfully constructing useful mathematical models of intricate physical processes

6.3 Comparison with Experiment or Observation Finally, having obtained the solution (or at least some information about it), you must interpret this information in the context in which the problem arose In particular, you should always check that the mathematical solution appears physically reasonable If possible, calculate the values of the solution at selected points and compare them with experimentally observed values Or, ask whether the behavior of the solution after a long time is consistent with observations Or, examine the solutions corresponding to certain special values of parameters in the problem Of course, the fact that the mathematical solution appears to be reasonable does not guarantee it is correct However, if the predictions of the mathematical model are seriously inconsistent with observations of the physical system it purports to describe, this suggests that either errors have been made in solving the mathematical problem, or the mathematical model itself needs refinement, or observations must be made with greater care In Chapter 1 we have given some examples which are typical of applications in which first-order differential equations arise In this section we pay our attention to a concrete model, that is a mathematical model of electric circuits We start with some important facts from electric circuits

6.4 Electric circuits The simplest electric circuit is a series circuit in which we have a source of electric energy (electromotive force) such as a generator or a battery, and a resistor, which uses the energy Experiments show that the following law holds

The voltage drop ER across a resistor is proportional to the instantaneous the current I, say,

Ohm's Law states that the voltage (E) across a resistor is equal to the product of the current (I) flowing through it and its resistance (R), expressed as E = RI In this equation, current is measured in amperes, resistance in ohms, and voltage in volts, with resistance being a constant that characterizes the resistor.

Inductors and capacitors are crucial components in complex circuits An inductor resists changes in current, exhibiting an inertia effect in electricity akin to mass in mechanics This analogy will be explored further in subsequent discussions.

The voltage drop ER across an inductor is proportional to the instantaneous time rate of change of the current I, say, dI

The algebraic sum of all instantaneous voltage drops in a closed loop equals zero, indicating that the total voltage applied to the loop matches the sum of the voltage drops throughout the loop.

Model the “RL-circuit” in ﬁg 2.2 and solve the resulting equation for: (A) a constant electromotive force; (B) a period electromotive force

In the first step of the solution, we model the circuit by applying Kirchhoff's Voltage Law (KVL) According to the first law, the voltage drop across the resistor is represented as RI, while the voltage drop across the inductor is expressed as LdI/dt The sum of these voltage drops must equal the electromotive force E(t).

LdI + 2 nd Step Solution of the equation In order to use the formula (2.36) we transform the above equation to the standard form by deviding both side to L and obtain

Using now formula (2.36) with x=t, y=I, p=R/L, and q=E/L we get

3 rd Step Case A: Constant electromotive force E=E0 The above equality for I(t) yields t t t E ce c

As time approaches infinity, the current I(t) approaches a constant value of E0/R, which reflects the immediate current determined by Ohm's law in the absence of an inductor in the circuit Notably, this limiting value remains unaffected by the initial current I(0).

Case B: Periodic electromotive force E=E0 Sin ωt For this E(t) we have that

I( ) α t 0 α t sinω for = R/L α Integration by part yields

The exponential term will approach zero as t tends to inﬁnity This mean that after some time the current I(t) will execute practically harmonic oscillations

VI Show that if (N'x − M' y)/( xM − yN) = G, where G depends on the quantity xy only, then the diﬀerential equation Mdx + Ndy = 0 has an integrating factor of the form μ( xy) Find a general formula for this integrating factor

VII In each of Problems 1 through 5 ﬁnd an integrating factor and solve the given equation

VIII In each of Problems 1 through 12 ﬁnd the general solution of the given diﬀerential equation and use it to determine howsolutions behave as t →∞.

IX In each of Problems 28 through 31 solve the given Bernoulli equation:

(a) Determine the diﬀerential equation governing the current I (in amperes) on the circuit

(b)Solve the equation to ﬁnd the current in the case of constant electromotive force

Existence and Uniqueness Theorem

VI Show that if (N'x − M' y)/( xM − yN) = G, where G depends on the quantity xy only, then the diﬀerential equation Mdx + Ndy = 0 has an integrating factor of the form μ( xy) Find a general formula for this integrating factor

VII In each of Problems 1 through 5 ﬁnd an integrating factor and solve the given equation

VIII In each of Problems 1 through 12 ﬁnd the general solution of the given diﬀerential equation and use it to determine howsolutions behave as t →∞.

IX In each of Problems 28 through 31 solve the given Bernoulli equation:

(a) Determine the diﬀerential equation governing the current I (in amperes) on the circuit

(b)Solve the equation to ﬁnd the current in the case of constant electromotive force

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

Definitions and Notations

A second-order ordinary differential equation (ODE) is represented as y’’(t) = f(t, y, y’), where f is a specified function While the independent variable is typically denoted by t, it can also be represented by x in certain contexts The equation is classified as linear if f takes the form f(t, y, y’) = r(t) - q(t)y - p(t)y’, indicating that f is linear in both y and y’ Here, r, p, and q are functions of the independent variable t and do not depend on y Consequently, the equation can be rewritten as y’’ + p(t)y’ + q(t)y = r(t), leading to a precise definition of linear second-order ODEs.

1.1 Deﬁnition A second-order diﬀerential equation is called linear if it can be written in the form y’’ + p(t)y’ + q(t)y = r(t) (1)

Instead of Eq (1), we often see the equation

Of course, if P(t) ≠ 0, we can divide this Eq by P(t) and thereby obtain Eq (1) with p(t) =Q(t)/P(t), q(t) =G(t)/P(t), r(t) =R(t)/P(t).

In discussing Eq (1) and in trying to solve it, we will restrict ourselves to intervals in which with continuous coeﬀicients p, q and initial conditions y(t0)=y0; y’(t0)=y1 for given y 0 , y 1 , and some ﬁxed t 0 ∈I (3)

We accept the following theorem on existence and uniqueness theorem of the solution to the initial-value problem (2), (3)

2.1 Existence and Uniqueness Theorem:Consider the initial-value problem y’’ + p(t)y’ + q(t)y = r(t), y(t0) = y0, y’(t0) = y1 (IVP) where p, q, and r are continuous on an open interval I Then there is exactly one solution y = φ(t) of this problem, and the solution exists throughout the interval I

2.3 Linearity Principle If y 1 and y 2 are two solutions of the diﬀerential equation (2), then the linear combination c 1 y 1 + c 2 y 2 is also a solution for any values of the constants c 1 and c 2

PROOF The assertion follows from the following direct substitution:

Two solutions, y₁ and y₂, are considered linearly independent on an interval I if the equation k₁y₁(t) + k₂y₂(t) = 0 holds for all t in I only when k₁ = k₂ = 0 Conversely, they are linearly dependent if there exist constants k₁ and k₂, not both zero, such that the same equation holds In the case of linear dependence, y₁ and y₂ are proportional, meaning y₁ = ky₂ for some non-zero k, or y₂ = ly₁ for some non-zero l Therefore, y₁ and y₂ are linearly independent if and only if they are not proportional to each other.

Example y 1 (t) =e t and y 2 (t)=e -2 t are linearly independent, because they are not proportional

The following notion of Wronski determinant is very helpful in characterizing the linear independence of solutions

2.5 Deﬁnition The Wronski Determinant (or Wronskian) of the two solutions y 1 , y2 of the equation (2) is deﬁned by

The following theorems connect the linear dependence and independence of the two solutions of Eq (2) with the properties of their Wronskian

2.6 Theorem Two solutions y1 and y2 of Eq (2) are linearly dependent on I if and only if their Wronskian W(y1, y2) is zero at some point t0∈I

PROOF If y 1 and y 2 arelinearly dependent on I, then they are proportional, say, y 1 =k y 2 on I. This follows that W(y 1 , y 2 )(t)or all t∈I

If there exists a point $ t_0 \in I $ such that the Wronskian $ W(f,g)(t_0) = 0 $, it can be demonstrated that the functions $ f $ and $ g $ are linearly dependent on the interval $ I $ Specifically, the equations $ c_1 y_1(t_0) + c_2 y_2(t_0) = 0 $ and $ c_1 y_1'(t_0) + c_2 y_2'(t_0) = 0 $ imply that there is a nontrivial solution for the coefficients $ c_1 $ and $ c_2 $ Defining $ \phi(t) = c_1 y_1(t) + c_2 y_2(t) $, we find that $ \phi $ satisfies the initial conditions $ \phi(t_0) = 0 $ and $ \phi'(t_0) = 0 $ By applying the existence and uniqueness theorem, it follows that $ \phi(t) = 0 $ for all $ t $ in $ I $ Since $ \phi(t) $ is expressed as a linear combination of $ y_1(t) $ and $ y_2(t) $ with at least one non-zero coefficient, this confirms that $ y_1 $ and $ y_2 $ are indeed linearly dependent.

Remark: The above proof also shows that two solutionsy 1 and y 2 of Eq (2) are linearly dependent on I if and only if their Wronskian W(y 1 , y 2 ) is zero for all t∈I.

2.7 Theorem Let y1 and y2 be two solutions of the equation (2) on an interval I Then, the following assertions are equivalent

(ii) W(y1, y2)(t0) ≠0 for some point t 0 in I

PROOF “(i)⇒ (ii)”: For the purpose of contradiction let W(y 1 , y 2 )(t) =0 for all t in I Then, by Theorem 2.6, y 1 and y 2 arelinearly dependent This contradicts to (i)

Assuming for contradiction that W(y1, y2)(t1) = 0 for some t1 in I leads to the conclusion that y1 and y2 are linearly dependent, as established by theorem 2.6 Consequently, this implies that W(y1, y2) is zero for all t in I, which contradicts the initial assumption (ii).

“(iii) ⇒ (i)”: If y 1 and y 2 are linearly dependent, then, by theorem 2.6, there exists t0 such thatW(y 1 , y 2 )(t0) =0 This contradicts to (iii)

2.8 Theorem (Existence of Linearly Independent Solutions)

Consider Eq (2) with continuous coefficients p, q on I Then there exists two linearly independent solutions y1, y2 on I of Eq (2)

PROOF By theorem 2.1, there exists solution y 1 of Eq (2) satisfying y 1 (t 0 )=1, y 1 ’(t 0 )=0 for some t 0 in I Also, there exists solution y 2 of Eq (2) satisfying y 2 (t 0 )=0, y 2 ’(t 0 )=1 Therefore, W(y 1 , y 2 )(t0)=1 ≠0 Hence, y 1 and y 2 are linearly independent

(i) To solve the homogeneous equation (2) is to ﬁnd a basic of solutions y 1 ,y 2 of Eq (2)

Then, the general solution is y=c 1 y 1 +c 2 y 2

Most problems of the form (2) do not allow for a useful expression of the solution, highlighting a significant distinction between first-order and second-order linear equations.

To solve the first-order linear differential equation y’ + ky = 0 with a constant coefficient k, we recall that its solution is an exponential function, specifically y = e^(-kt) This insight leads us to propose a solution of the form y = e^(ct) By substituting this into the original equation, we derive the characteristic equation (c² - 3c + 2) e^(ct) = 0, which yields the roots c = 1 and c = 2 Consequently, we identify two linearly independent solutions: y₁ = e^t and y₂ = e^(2t) Thus, the general solution of the differential equation is expressed as y = c₁ e^t + c₂ e^(2t).

If a nontrivial solution y1 is known, then we can ﬁnd the solution y2 linearly independent with y1 by the following procedure which is called the method of reduction of order

Putting y 2 =u y1 and substituting it to Eq (2), we obtain u’’ y1 +u’(2 y 1 ’+p y 1 )=0

Setting U=u’, it follows that U’ y 1 +U(2 y 1 ’+p y 1 )=0, this yields U= 2 y1 e − ∫ pdt Returning to u we have that 2

Example: Solve (t 2 -1)y’’-2ty’+2y=0 given a solution y1=t

To use the formula (9) we write the equation in the standard form y’’-2ty’/(t 2 -1)+2y/(t 2 -1)=0

Applying (9) we have dt t t e y t dt t

=t 2 +1 Therefore, the general solution of given equation is y=c t+c (t 2 +1)

3 Homogeneous Equations with Constant Coefficients

This section focuses on homogeneous equations with constant coefficients, specifically of the form y’’ + ay’ + by = 0, where a and b are arbitrary real constants Drawing from our previous experience with similar equations, we will explore exponential solutions for this equation We assume a solution of the form y = e^(kt), where k is a parameter to be determined By differentiating this expression, we find that y’ = ke^(kt) and y’’ = k²e^(kt) Substituting these derivatives back into the original equation allows us to analyze the resulting expressions.

(k 2 + + ak b)e kt = 0 or, since e kt is never zero, k 2 + + = 0ak b (11)

The characteristic equation for the differential equation is represented as Equation (11), which is significant because if a root of this polynomial equation is found, then the solution to the differential equation can be expressed as $ y = k e^{kt} $ As a quadratic equation with real coefficients, Equation (11) can yield two types of roots: real and distinct, real and repeated, or complex conjugates Each of these cases will be examined in detail.

1 st Case: Distinct real roots

Assuming the roots of the characteristic equation (11) are real and distinct, we denote them as k1 and k2, with k1 ≠ k2 The solutions y1(t) = e^(k1 t) and y2(t) = e^(k2 t) are linearly independent Consequently, according to Theorem 2.9 from the previous section, the general solution of equation (11) can be expressed as y = c1 e^(k1 t) + c2 e^(k2 t).

Example Find the general solution of y’’ + 5y’ + 6y = 0 (13) The characteristic equation is k 2 + 5k + 6 = 0

It has two distinct real roots: k 1 = −2 and k 2 = −3; then the general solution of Eq (13) is y = c 1 e −2 t + c 2 e −3 t

2 nd Case: Double real root

In the scenario where the two real roots k₁ and k₂ are equal, the discriminant (∆ = a² - 4b) is zero, leading to the conclusion that k₁ = k₂ = -a/2 This presents a challenge, as both roots produce the same solution, y₁(t) = e^(k₁t) = e^(-a/2 t) for the given differential equation To find a second solution y₂ that is linearly independent from y₁, we utilize the appropriate formula, resulting in a new expression involving the exponential function.

2 =t 2 at e − Therefore, the general solution of Eq (11) in this case is

(e λt (cosàt + i sinàt) - e λt (cosàt − i sinàt))= e λt sinàt

The linear combination of two solutions to Eq (2) yields additional solutions, leading to the identification of two linearly independent real-valued solutions: y1 = e^(λt) cos(αt) and y2 = e^(λt) sin(αt) Consequently, the general solution of Eq (2) can be expressed as y = c1 e^(λt) cos(αt) + c2 e^(λt) sin(αt), where c1 and c2 are constants.

Example 1 Find the general solution of y’’+ y’ + y = 0 (17) The characteristic equation is k 2 + k + 1 = 0, and its roots are k = −

23 Thus λ = −1/2 and à = 3/2, so the general solution of Eq (17) is y = c 1 e − 2 t/ cos( 3t/2) + c 2 e − 2 t/ sin( 3t/2).

Example 2 Find the general solution of y’’+ 9y = 0

The characteristic equation is k 2 + 9 = 0 with the roots k = ±3i ; thus λ = 0 and à = 3 The general solution is y = c 1 cos3t + c 2 sin3t

Note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution

4 Modelling: Free Oscillation (Mass-spring problem)

The study of a mass on a spring is essential for understanding the dynamics of more complex vibrating systems, as it introduces fundamental principles applicable to various problems When a mass $ m $ is suspended from a vertical spring of original length $ l $, it causes an elongation $ L $ in the spring's length At the attachment point, two forces are at play: the gravitational force, or weight, acting downward with a magnitude of $ mg $, and the spring force $ F_s $, which acts upward Assuming the elongation $ L $ is small, the spring force can be described by Hooke's law, expressed as $ F_s = -kL $, where $ k $ is the spring constant, and the negative sign indicates that the spring force opposes the direction of elongation.

Fig 4.2 Force diagram for a spring–mass system.

Since the mass is in equilibrium, the two forces balance each other, which means that mg − kL = 0 (1)

To determine the spring constant k for a weight w = mg, one can measure the length L and apply the relevant equation, noting that k is expressed in units of force per length In dynamic scenarios, we analyze the mass's motion when influenced by an external force or after being displaced The downward displacement of the mass from its equilibrium position at time t is denoted as u(t), which is related to the forces acting on it via Newton’s second law of motion, expressed as mu’’(t) = f(t), where u’’ represents the mass's acceleration and f signifies the net force acting on it It is important to recognize that both u and f are time-dependent functions, and various cases will be considered to determine the net force f.

4.1 Undamped Systems In this case there are two separate forces that must be considered :

1 The weight w = mg of the mass always acts downward

2 The spring force FS is assumed to be proportional to the total elongation L + u of the spring and always acts to restore the spring to its natural position If L + u > 0, then the spring is extended, and the spring force is directed upward In this case

On the other hand, if L + u < 0, then the spring is compressed a distance |L + u|, and the spring force, which is now directed downward, is given by Fs = k|L + u| However, when

L + u < 0, it follows that |L + u| = −(L + u), so Fsis again given by Eq (3) Thus, regardless of where

In calculating δ care must be taken to choose the correct quadrant; this can be done by checking the signs of cos δ and sin δ in Eqs (5)

Homogeneous Equations with Constant Coefficients

In this section, we analyze homogeneous equations with constant coefficients, specifically the form y’’ + ay’ + by = 0, where a and b are arbitrary real constants Drawing from our earlier work with related equations, we will explore exponential solutions for this equation We assume a solution of the form y = e^(kt), where k is a parameter to be determined Consequently, the first derivative is y’ = ke^(kt) and the second derivative is y’’ = k²e^(kt) By substituting these expressions into the original equation, we can derive further insights into the behavior of the solutions.

(k 2 + + ak b)e kt = 0 or, since e kt is never zero, k 2 + + = 0ak b (11)

The characteristic equation for the differential equation is given by Equation (11), which plays a crucial role in determining solutions If a root of this polynomial equation is identified, then the solution can be expressed as $y = k e^{kt}$ Since Equation (11) is a quadratic equation with real coefficients, it can yield two roots that may be real and distinct, real and repeated, or complex conjugates Each of these scenarios will be examined in detail to understand their implications for the solutions of the differential equation.

1 st Case: Distinct real roots

Assuming the roots of the characteristic equation are real and distinct, we denote them as k1 and k2, with k1 ≠ k2 The functions y1(t) = e^(k1 t) and y2(t) = e^(k2 t) represent two linearly independent solutions to the equation Consequently, applying Theorem 2.9 from the previous section, we derive the general solution of the equation as y = c1 e^(k1 t) + c2 e^(k2 t).

Example Find the general solution of y’’ + 5y’ + 6y = 0 (13) The characteristic equation is k 2 + 5k + 6 = 0

It has two distinct real roots: k 1 = −2 and k 2 = −3; then the general solution of Eq (13) is y = c 1 e −2 t + c 2 e −3 t

2 nd Case: Double real root

In the scenario where the two real roots $ k_1 $ and $ k_2 $ are equal, the discriminant $ \Delta = a^2 - 4b $ becomes zero, leading to the conclusion that $ k_1 = k_2 = -\frac{a}{2} $ This results in both roots providing the same solution, $ y_1(t) = e^{k_1 t} = e^{-\frac{a}{2} t} $, for the differential equation To find a second solution $ y_2 $ that is linearly independent from $ y_1 $, we apply the appropriate formula, ultimately deriving a new solution that complements the existing one.

2 =t 2 at e − Therefore, the general solution of Eq (11) in this case is

(e λt (cosàt + i sinàt) - e λt (cosàt − i sinàt))= e λt sinàt

The linear combination of two solutions to Eq (2) yields two linearly independent real-valued solutions: y1 = e^(λt) cos(αt) and y2 = e^(λt) sin(αt) Consequently, the general solution of Eq (2) can be expressed as y = c1 e^(λt) cos(αt) + c2 e^(λt) sin(αt), where c1 and c2 are constants.

Example 1 Find the general solution of y’’+ y’ + y = 0 (17) The characteristic equation is k 2 + k + 1 = 0, and its roots are k = −

23 Thus λ = −1/2 and à = 3/2, so the general solution of Eq (17) is y = c 1 e − 2 t/ cos( 3t/2) + c 2 e − 2 t/ sin( 3t/2).

Example 2 Find the general solution of y’’+ 9y = 0

The characteristic equation is k 2 + 9 = 0 with the roots k = ±3i ; thus λ = 0 and à = 3 The general solution is y = c 1 cos3t + c 2 sin3t

Note that if the real part of the roots is zero, as in this example, then there is no exponential factor in the solution.

Modelling: Free Oscillation (Mass-spring problem)

Understanding the motion of a mass on a spring is crucial for exploring more complex vibrating systems, as the principles involved are applicable to various problems Consider a mass $ m $ suspended from a vertical spring of original length $ l $, which experiences an elongation $ L $ in the downward direction At the attachment point, two forces are present: the gravitational force $ mg $ acting downward and the spring force $ F_s $ acting upward Assuming a small elongation $ L $, the spring force can be described by Hooke's law, expressed as $ F_s = -kL $, where $ k $ is the spring constant, and the negative sign indicates that the spring force opposes the direction of elongation.

Fig 4.2 Force diagram for a spring–mass system.

Since the mass is in equilibrium, the two forces balance each other, which means that mg − kL = 0 (1)

To determine the spring constant k for a weight w = mg, one can measure the length L and apply the relevant equation, noting that k is expressed in units of force per length In dynamic scenarios, we analyze the motion of the mass when subjected to an external force or initial displacement Let u(t) represent the downward displacement of the mass from its equilibrium position at time t, which is influenced by the forces acting on it according to Newton’s law of motion, expressed as mu’’(t) = f(t) Here, u’’ denotes the mass's acceleration, and f represents the net force acting on the mass, with both u and f being time-dependent functions To ascertain f, we will examine several specific cases.

4.1 Undamped Systems In this case there are two separate forces that must be considered :

1 The weight w = mg of the mass always acts downward

2 The spring force FS is assumed to be proportional to the total elongation L + u of the spring and always acts to restore the spring to its natural position If L + u > 0, then the spring is extended, and the spring force is directed upward In this case

On the other hand, if L + u < 0, then the spring is compressed a distance |L + u|, and the spring force, which is now directed downward, is given by Fs = k|L + u| However, when

L + u < 0, it follows that |L + u| = −(L + u), so Fsis again given by Eq (3) Thus, regardless of where

In calculating δ care must be taken to choose the correct quadrant; this can be done by checking the signs of cos δ and sin δ in Eqs (5)

Figure 4.3 illustrates the graph of Eq (7) and its equivalent Eq (5) under typical initial conditions, depicting a displaced cosine wave This graph represents the periodic or simple harmonic motion of the mass, characterized by a specific period of motion.

The natural frequency of vibration, denoted as ω0 and measured in radians per unit time, is influenced by the mass (m) and spring constant (k) The maximum displacement from equilibrium, known as the amplitude (R), remains constant over time in the absence of damping, indicating that energy is not dissipated The phase angle (δ) measures the wave's displacement from its normal position, with δ = 0 representing equilibrium Importantly, while the system vibrates at a consistent frequency ω0 for a given mass and spring constant, the initial conditions affect the amplitude of the motion Additionally, the period (T) of vibration increases with larger masses, causing slower vibrations, while a stiffer spring (higher k) results in a decreased period, leading to faster vibrations.

FIG 4.3 Simple harmonic motion; = cosu R (ω 0t − δ )

4.2 Damped Systems In this case, beside the two forces (1 The weight and 2 The spring force) as above, we have to consider one more force, that is:

3 The damping or resistive forceFdalways acts in the direction opposite to the direction of motion of the mass This force may arise from several sources: resistance from the air or other medium in which the mass moves, internal energy dissipation due to the extension or compression of the spring, friction between the mass and the guides (if any) that constrain its motion to one dimension, or a mechanical device (dashpot) that imparts a resistive force to the mass In any case, we assume that the resistive force is proportional to the speed |du/dt| of the mass; this is usually referred to as viscous damping If du/dt > 0, then u is increasing, so the mass is moving downward Then F d is directed upward and is given by

The damping force, represented by the equation Fd(t) = −γ u’(t), where γ is a positive damping constant, consistently applies regardless of the mass's motion direction When the rate of change of displacement (du/dt) is negative, indicating that the mass is moving upward, the damping force acts downward and is expressed as Fd = γ |u’(t)| Since the absolute value of the velocity |u’(t)| equals −u’(t) in this context, the damping force remains defined by the same equation, Fd(t) = −γ u’(t), demonstrating the uniformity of the damping force in all scenarios.

The complexity of the damping force raises questions about the adequacy of its modeling by Eq (8) While some dashpots conform to this equation, it may be feasible to disregard other minor sources of dissipation or adjust the damping constant γ to account for them A significant advantage of this assumption is that it results in a linear differential equation, facilitating a straightforward analysis of the system, as demonstrated in the following sections.

Considering the forces at play, we can reformulate Newton's law to express the motion of a mass as mu’’(t) = mg + FS(t) + Fd(t), which simplifies to mu’’(t) + γ u’(t) + ku(t) = 0 under the condition that mg − kL = 0 Here, m, γ, and k are positive constants, indicating that the equation of motion maintains a consistent structure with previous formulations.

We focus on analyzing how changes in the damping coefficient γ impact the system, while keeping the mass m and spring constant k constant The roots of the characteristic equation associated with this system will provide valuable insights into its behavior.

Depending on the sign of γ 2 − 4km, the solution u has one of the following forms:

Given that m, γ, and k are positive constants, the expression γ² - 4km is always less than γ² Therefore, if γ² - 4km is greater than or equal to zero, the roots r₁ and r₂ derived from the formulas will be negative Conversely, if γ² - 4km is less than zero, the roots r₁ and r₂ will be complex with negative real parts In all scenarios, the solution u approaches zero.

FIG 4.4 Damped oscillation; = u Re − 2 γt/ m cos( t à − δ)

In the context of small damping, where γ 2 /4km is minimal, the frequency of oscillation experiences a slight reduction This phenomenon is characterized by the quasi period, denoted as Td = 2π à/, which represents the time interval between successive maxima or minima of the mass's position, or the time taken for the mass to pass through its equilibrium position in the same direction The relationship between the quasi period Td and the overall period T is also established in this framework.

(12) where again the last approximation is valid when γ 2 /4km is small Thus, small damping increases the quasi period

Equations (11) and (12) reinforce the signiﬁcance of the dimensionless ratio γ 2 /4km

The significance of damping in motion analysis is not solely dependent on the magnitude of γ, but rather on the ratio of γ² to 4km When this ratio is small, the impact of damping can be disregarded in the calculations of quasi frequency and quasi period However, for a comprehensive understanding of the mass's motion over time, the damping force must always be considered, regardless of its size.

Nonhomogeneous Equations: Method of Undetermined Coefficients

The nonhomogeneous equation $ y'' + p(t)y' + q(t)y = g(t) $ involves continuous functions $ p, q, $ and $ g $ defined on an open interval $ I $ Its corresponding homogeneous equation, $ y'' + p(t)y' + q(t)y = 0 $, is established by setting $ g(t) = 0 $ while retaining the same functions $ p $ and $ q $ Understanding the structure of solutions for the nonhomogeneous equation is essential, as it lays the groundwork for formulating its general solution.

5.1 Theorem If Y 1 and Y 2 are two solutions of the nonhomogeneous equation (1), then their diﬀerence Y 1 − Y 2 is a solution of the corresponding homogeneous equation (2) If, in addition, y 1 and y 2 are a fundamental set of solutions of Eq (2), then

Proof To prove this result, note that Y 1 and Y 2 satisfy the equations (1), this means that

Y + p(t) Y 1 ' + q(t)Y 1 = g(t) and Y 1 '' + p(t) Y 1 ' + q(t)Y 1 = g(t) (4) Subtracting the second of these equations from the ﬁrst, we have

Equation (5) indicates that the difference Y 1 − Y 2 is a solution to Eq (2) According to Theorem 2.9, all solutions of Eq (2) can be represented as linear combinations of a fundamental set of solutions Therefore, the solution Y 1 − Y 2 can also be expressed in this manner, confirming that Eq (3) is valid, thus completing the proof.

5.2 Theorem The general solution of the nonhomogeneous equation (1) can be written in the form y = φ(t) = c 1 y 1 (t) + c 2 y 2 (t) + Y (t), (6) where y 1 and y 2 are a fundamental set of solutions of the corresponding homogeneous equation

(2), c 1 and c 2 are arbitrary constants, and Y is some speciﬁc solution of the nonhomogeneous equation (1)

The proof of Theorem 5.2 follows quickly from the preceding theorem Note that Eq (3) holds if we identify Y 1 with an arbitrary solution φof Eq (1) and Y 2 with the speciﬁc solution Y From

The equation φ(t) − Y(t) = c₁y₁(t) + c₂y₂(t) represents a general solution to Eq (1), as derived from Eq (3) and equivalent to Eq (6) Since φ is an arbitrary solution, the right side of this equation encompasses all possible solutions of Eq (1).

In somewhat diﬀerent words, Theorem 5.2 states that to solve the nonhomogeneous equation (1), we must do three things:

1 Find the general solution c 1 y 1 (t) + c 2 y 2 (t) of the corresponding homogeneous equation This solution is frequently called the complementary solution and may be denoted by yc(t)

2 Find some single solution Y(t) of the nonhomogeneous equation Often this solution is referred to as a particular solution

3 Add together the functions found in the two preceding steps

In this section, we will concentrate on determining a particular solution Y(t) for the nonhomogeneous equation (1), following our earlier discussion on finding yc(t) when the homogeneous equation (2) has constant coefficients We will explore two methods for this purpose: the method of undetermined coefficients and the method of variation of parameters Each method offers distinct advantages and potential drawbacks.

5.3 Method of Undetermined Coefficients The method of undetermined coefficients requires that we make an initial assumption about the form of the particular solution Y (t), but with the coefficients left unspecified We then substitute the assumed expression into Eq (1) and attempt case, we choose Y= te α t Qn(t) with Qn(t) being a polynomial of degree n whose coefficients are found by substituting Y to Eq (8)

Case III: The constant α is the double root of the characteristic equation k 2 +ak+b = 0 In this case, we choose Y= t 2 e α t Qn(t) with Qn(t) being a polynomial of degree n whose coeﬀicients are found by substituting Y to Eq (8)

1 Consider y’’ +3y’ − 4y = t; (α=0; n=1) (9) The corresponding homogeneous equation is y’’+3y’ −4y =0 with the characteristic equation k 2 +3k-4=0 ⇔ k=1 or -4 Therefore, the general solution of the corresponding homogeneous equation is c1e t +c2e -4t Since α=0 is not a root of characteristic equation, we ﬁnd a particular solution of Eq (9) of the form Y=At+B; Substituting this form into (9) we obtain that -4At+3A-4B=t Identifying the corresponding coeﬀicients of t we have that A=-1/4 and B=-3/16 This yields a particular solution of Eq (9) as Y= 163

− t and hence, the general solution of (9): y= c1e t +c2e -4 t

2 Consider y’’ - y’ = e t (t+1); (α=1; n=1) (10) The corresponding homogeneous equation is y’’ - y’=0 having the general solution as c1e t +c2e -t Since α=1 is a single root of the characteristic equation, we ﬁnd a particular solution of Eq (10) of the form Y=te t (At+B); Substituting this form into (10) we obtain that e t (2At+B+2A)=e t (t+1)

⇒ A=1/2 and B=0 Therefore, the general solution of (10) is y= c1e t +c2e -t 2

3 Consider y’’ - 2y’+y = e t ; (α=1; n=0) (11) The corresponding homogeneous equation is y’’ - 2y’+y=0 having the general solution as

The general solution of the differential equation is given by y = (c1 + c2t)e^t, where α = 1 represents a double root of the characteristic equation To find a particular solution, we assume the form Y = At^2 e^t, and upon substituting this into the equation, we determine that A = 1/2 Thus, the complete solution can be expressed as y = (c1 + c2t)e^t + (1/2)t^2 e^t.

FORM 2: g(t)=Pm(t)cosβt + Qn(t)sinβt, where Pm(t) and Qn(t) are known polynomials of order m and n, respectively For g(t) of this form we consider following cases:

Case I: The constant iβ is not a root of the characteristic equation k 2 +ak+b = 0 In this case, we choose Y= Rl(t)cosβt+Sl(t)sinβt with Rl(t) and Sl(t) being polynomials of degree l = max{m, n} whose coeﬀicients are found by substituting Y to Eq (8) Case II: The constant iβ is a root of the characteristic equation k 2 +ak+b = 0 In this case, we choose Y= t(Rl(t)cosβt+Sl(t)sinβt) with Rl(t) and Sl(t) being polynomials of degree l whose coeﬀicients are found by substituting Y to Eq (8)

Consider y’’ + 4y = 3 cos 2t; (β=2, l = 0 ) (12) Firstly, solve the homogeneous equation y’’+ 4y = 0 (13) that corresponds to Eq (12) The characteristic equation k 2 +4=0 has two complex conjugate root k = ±2i Therefore, the general solution of Eq (13) is c1cos 2t + c2sin 2t

To solve the nonhomogeneous term 3 cos 2t, we assume a particular solution of the form Y(t) = At cos 2t + Bt sin 2t, given that 2i is a root of the characteristic equation By calculating the first and second derivatives, Y’(t) and Y’’(t), and substituting them into the differential equation, we can collect the terms to find the particular solution.

Therefore A = 0 and B 43 , so a particular solution of Eq (20) is Y (t) 4

The general solution of (12) is y(t)= c1cos 2t + c2sin 2t+

In the analysis of the equation g(t) = e^(αt) [Pm(t)cos(βt) + Qn(t)sin(βt)], we apply the substitution y = e^(αt)z This transforms the equation into z’’ + (2α + a)z’ + (α² + aα + b)z = Pm(t)cos(βt) + Qn(t)sin(βt) This resulting equation can be classified under Form 2, allowing us to solve for z By substituting back to y using the initial transformation, we ultimately derive the solution for the original equation.

In the case of Form 3, an alternative method to solve equation (8) involves finding a particular solution of the form Y(t) = e^(αt)[Rl(t)cos(βt) + Sl(t)sin(βt)] when α + iβ is not a root of the characteristic equation k² + ak + b = 0 Conversely, if α + iβ is a root of the characteristic equation, the solution takes the form Y(t) = te^(αt)[Rl(t)cos(βt) + Sl(t)sin(βt)].

Solve the equation y’’− 2y’ +5y = 3e t cos 2t (15) Substituting y= e t z to (15) we obtain: z’’+4z

This equation is precisely the Eq (12) above and has the general solution as

Find a particular solution of y’’ − 3y’ − 4y = 3e 2 t + 2 sint − 8e t cos 2t (19)

By splitting up the right side of Eq (19), we obtain the three equations y’’− 3y’− 4y = 3e 2 t , y’’ − 3y’ − 4y = 2 sint, and y’’ − 3y’ − 4y = −8e t cos 2t.

Solutions of these three equations are Y1=-e 2t /2, Y1=(3 cos t − 5 sin t)/17, and

Therefore a particular solution of Eq (19) is their sum, namely,

Variation of Parameters

This section introduces the variation of parameters method, developed by Lagrange, as an effective approach to finding particular solutions for nonhomogeneous equations Unlike the method of undetermined coefficients, variation of parameters is a versatile technique applicable to any equation without needing specific assumptions about the solution's form In this section, we will derive a formula for a particular solution of an arbitrary second-order linear nonhomogeneous differential equation using this method However, it is important to note that variation of parameters involves evaluating integrals related to the nonhomogeneous term, which can pose challenges.

Again we consider the nonhomogeneous equation y’’+p(t)y’+q(t)y=g(t) (20) and the corresponding homogeneous equation y’’+p(t)y’+q(t)y=0 (21)

As a starting point, we assume that we know the general solution y(t) = c 1 y 1 (t) + c 2 y 2 (t) (22) of the corresponding homogeneous equation (21)

It is important to note that our previous discussions have focused on solving Eq (21) exclusively under the assumption of constant coefficients However, if Eq (21) features coefficients that vary with time (t), it is typically necessary to apply the methods outlined in earlier sections to determine y(t).

To solve the nonhomogeneous equation (20), we replace the constants $ c_1 $ and $ c_2 $ in Eq (22) with functions $ u_1(t) $ and $ u_2(t) $, leading to the expression $ y = u_1(t)y_1(t) + u_2(t)y_2(t) $ as shown in Eq (23) Our goal is to determine the functions $ u_1(t) $ and $ u_2(t) $ to ensure that this expression satisfies the nonhomogeneous equation.

Thus we diﬀerentiate Eq (23), obtaining y’ = u 1 ’(t)y 1 (t) + u 1 (t)y 1 ’(t) + u 2 ’(t)y 2 (t) + u 2 (t)y 2 ’(t) (24)

To find u1 and u2, we can substitute y from Equation (23) into Equation (20) Even without this substitution, we can expect the outcome to be a singular equation that incorporates a combination of u1, u2, and their first two derivatives.

With only one equation and two unknown functions, there are numerous potential combinations of u1 and u2 that can satisfy our requirements By introducing a second condition of our preference, we can create two equations to solve for the unknown functions u1 and u2 We will demonstrate, following Lagrange's approach, that selecting this second condition strategically can significantly enhance computational efficiency.

To establish the conditions for the functions $ u_1'(t) $ and $ u_2'(t) $ in Equation (24), we set the expression $ u_1'(t)y_1(t) + u_2'(t)y_2(t) = 0 $ Consequently, we derive $ y' = u_1(t)y_1'(t) + u_2(t)y_2'(t) $ Further differentiation yields $ y'' = u_1'(t)y_1'(t) + u_1(t)y_1''(t) + u_2'(t)y_2'(t) + u_2(t)y_2''(t) $ By substituting $ y $, $ y' $, and $ y'' $ into Equation (20) using Equations (23), (26), and (27), and rearranging the terms, we arrive at the expression: $ u_1(t)[y_1''(t) + p(t)y_1(t) + q(t)y_1(t)] + u_2(t)[y_2''(t) + p(t)y_2'(t) + q(t)y_2(t)] $.

The equation $ u_1'(t)y_1'(t) + u_2'(t)y_2'(t) = g(t) $ arises from the homogeneous solutions of the differential equation Since both $ y_1 $ and $ y_2 $ satisfy the homogeneous equation, the expression simplifies to $ u_1'(t)y_1'(t) + u_2'(t)y_2'(t) = g(t) $ This leads to a system of two linear algebraic equations for the derivatives $ u_1'(t) $ and $ u_2'(t) $, with the coefficient matrix being a crucial component of the solution process.

Wronskian W(y1,y2)≠0 since y 1 , y 2 are linearly independent By solving the system (25), (29) we obtain u 1 ’(t)) , (

By integrating Eqs (30) we ﬁnd the desired functions u 1 (t) and u 2 (t), namely, u 1 (t) = −∫ W y ( ( y t ) g , y ( t ) ) dt

We derive a specific solution to equation (20) using formula (23), where u1(t) and u2(t) are defined by equation (31) This particular solution Y(t) is expressed in relation to an arbitrary forcing function g(t), making it an excellent foundation for exploring the impact of changes in the forcing function or for analyzing the system's response to various forcing functions.

To find a particular solution for the differential equation y’’ + 4y = 3 csc t, we note that it does not fit the method of undetermined coefficients due to the nonhomogeneous term g(t) = 3 csc t, which involves a quotient of sine and cosine functions The corresponding homogeneous equation, y’’ + 4y = 0, has a general solution expressed as y(t) = c₁ cos 2t + c₂ sin 2t.

To solve the nonhomogeneous equation, we substitute the constants c1 and c2 with functions u1(t) and u2(t) Our goal is to determine these functions so that the expression y = u1(t) cos(2t) + u2(t) sin(2t) becomes a valid solution.

With the additional requirement u 1 ’(t) cos 2t + u 2 ’(t) sin 2t = 0, (35) and substituting y(t) from (34) into (33) we obtain that u 1 and u 2 must satisfy

−2u 1 ’(t) sin 2t + 2u 2 ’(t) cos 2t = 3 csc t (36) Solving the system of linear equations for the two unknown quantities u 1 ’(t) and u 2 ’(t) we ﬁnd u 1 ’(t) = −3 csc t sin 2t = −3 cos t. u 2 ’(t) # csc t − 3 sin t.

After obtaining the derivatives u 1 ’(t) and u 2 ’(t), the next step involves integration to find u 1 (t) and u 2 (t) The results yield u 1 (t) = −3 sin(t) + c1 and u 2 (t) = 23 ln |csc(t) − cot(t)| + 3 cos(t) + c2 Substituting these expressions into Equation (34) leads to the general solution of Equation (33), which is y = -3 sin(t) +

23 ln | csc t − cot t| sin 2t + c1 cos 2t + c2 sin 2t.

Modelling: Forced Oscillation

The mechanical system of free oscillation, commonly known as the mass-spring problem, is governed by the equation mu’’ + γ u’ + ku = 0 In this equation, mu’’ signifies the force of inertia, γ u’ represents the damping force, and ku denotes the spring force.

Forced motions occur when an external force g(t) is applied to a body To derive the model, we incorporate this new force g(t) into the existing forces, resulting in the equation mu’’ + γ u’ + ku = g(t).

Then, g(t) is called the input or driving force, and the corresponding solutions are called an output or a response of the system to the driving force

Of particular interest are periodic input, say g(t)=F 0 cos ωt with ω > 0 Then the equation of motion is mu’’+ γ u’ + ku = F 0 cos ωt (1)

Forced Vibrations without Damping: First suppose that there is no damping (γ=0); then Eq (1) reduces to mu’’+ ku = F 0 cos ωt (2)

Ifω 0 = k/ m ≠ω then the general solution of Eq (2) is

The constants c1 and c2 are established based on the initial conditions, leading to a motion that is typically a combination of two periodic motions with distinct frequencies (ω0 and ω) and amplitudes Notably, there are two specific cases that are particularly intriguing.

Beats.Suppose that the mass is initially at rest, so that u(0) = 0 and u’’(0) = 0 Then it turns out that the constants c 1 and c 2 in Eq (3) are given by

(4) and the solution of Eq (2) is

(5) This is the sum of two periodic functions of diﬀerent periods but the same amplitude

Making use of the trigonometric identities for cos(A ± B) with A = (ω 0 + ω)t/2 and

B = (ω 0 − ω)t/2, we can write Eq (5) in the form

The phenomenon known as a beat arises from motion that exhibits a periodic variation of amplitude, as illustrated by the solution to the equation u’’ + = u 0.5 cos(0.8t), with initial conditions u(0) = 0 and u’(0) = 0, resulting in u = 2.77778 sin(0.1t) sin(0.9t) This effect is commonly observed in acoustics when two tuning forks with nearly identical frequencies are struck simultaneously, making the amplitude variation noticeable to the ear In the field of electronics, this time-dependent amplitude variation is referred to as amplitude modulation Figure 6.1 provides a graphical representation of this behavior as described by the equation.

Resonance occurs when the frequency of the external forcing function matches the system's natural frequency (ω = ω0) In this scenario, the nonhomogeneous term F0 cos(ωt) becomes a solution to the homogeneous equation, leading to a specific solution for the system's behavior as described by Eq (2).

The solution predicts unbounded motion as time approaches infinity due to the term t sin ω0t, regardless of the constants c1 and c2 However, in reality, unbounded oscillations do not occur because the mathematical model becomes invalid when the displacement, u, becomes large, as it relies on the assumption of a linear spring force dependent on small displacements When damping is included in the model, the motion remains bounded, but if the damping is minimal and the frequency ω is close to the natural frequency ω0, the response to the input function F0 cos ωt can be significantly large, a phenomenon known as resonance.

Resonance plays a crucial role in structural design, as it can lead to significant instabilities and potential failures in buildings and bridges For instance, soldiers break step while crossing bridges to avoid creating a periodic force that matches the bridge's natural frequency A notable case of resonance issues occurred with the high-pressure fuel turbopump in the space shuttle main engine, which faced instability and could not operate beyond 20,000 rpm, far below its intended design speed of 39,000 rpm, ultimately contributing to the shutdown of the space shuttle program.

Over a period of six months, with an estimated expenditure of $500,000 per day, resonance can be effectively utilized in the design of instruments like seismographs, which are specifically intended to detect faint periodic incoming signals.

Forced vibrations in a damped spring-mass system can be analyzed effectively, even though the calculations may be complex When considering a forcing function of F0 cos ωt, the motion of the system, characterized by damping (γ≠0), can be determined through a systematic approach The solution to the governing equation provides insights into the behavior of the system under these conditions.

In the context of the characteristic equation linked to Eq (1), the roots r1 and r2 can be either real and negative or complex conjugates with a negative real part Regardless of their nature, both exp(r1 t) and exp(r2 t) tend toward zero as time t approaches infinity Consequently, as t approaches infinity, the overall behavior of the system stabilizes.

The equation u c (t) = c 1 exp(r 1 t) + c 2 exp(r 2 t) represents the transient solution, while U(t), which exhibits steady oscillation at the same frequency as the external force, is known as the steady-state solution or forced response The transient solution allows for the fulfillment of any initial conditions, and over time, the energy introduced by the initial displacement and velocity dissipates through the damping force, leading to a motion that reflects the system's response to the external force.

Without damping, the eﬀect of the initial conditions would persist for all time

Investigating the relationship between the amplitude R of steady-state oscillation and the parameter γ reveals that a smaller γ results in a larger ratio of R max to F 0 Figure 6.3 illustrates representative graphs of Rk/F 0 plotted against ω/ω0 for various values of γ, highlighting this significant correlation.

The phase angle δ is influenced by the frequency ω in distinct ways When ω is close to zero, δ approaches 0, indicating that the response is nearly in phase with the excitation, resulting in their maxima and minima occurring simultaneously As ω approaches the resonant frequency ω0, δ reaches π/2, causing the response to lag behind the excitation by π/2, with peaks and valleys occurring later In contrast, at very high frequencies, δ approaches π, leading to the response being nearly out of phase with the excitation, where the response is at its minimum when the excitation is at its maximum, and vice versa.

FIGURE 6.3: Forced vibration with damping: amplitude of steady-state response versus frequency of driving force;

FIGURE 6.4 A forced vibration with damping; solution of u’’+ 0.125u’ + u = 3 cos2t, u(0) = 2, u'(0) = 0

The graph in Figure 6.4 illustrates the solution to the initial value problem defined by the equation u’’ + 0.125u’ + u = 3 cos 2t, with initial conditions u(0) = 2 and u'(0) = 0 It also includes the graph of the forcing function for comparison Notably, the initial transient motion diminishes over time, while the steady forced response maintains an amplitude of approximately 1 Additionally, there is a phase difference of roughly π between the excitation and the response.

More precisely, we ﬁnd that ∆= 145/4 ≈ 3.0104, so R = F 0 /∆≈ 0.9965 Furthermore, cos δ = −3/∆≈ −0.9965 and sin δ = 1/4∆≈ 0.08305, so that δ ≈3.0585 Thus the calculated values of R and δ are close to the values estimated from the graph.

Power Series Solutions

8.1 Deﬁnition: The function f(x) is called real analytic at a point x 0 if it coincides with its

Taylor’s series of some neighbourhood (x0-R, x0 + R) of x0, i.e., f(x) = 0 n

The positive number R normally coincides with the radius of convergence of the Taylor’s series

Examples: e x , sin x, cos x are real analytic functions at any point x0∈ R

Step 3: Substitute y, y’ and y’’ obtained from Step 2 into (8.1) Then, collect the like powers of t and equate the sum of the coeﬀicients of each occurring power of t to zero, starting from the constant terms, then the terms containing t, the terms containing t 2 , …etc This gives the relations from which we can determine the unknown coeﬀicients in (8.2) successively

In this example, h(t)= (1-t 2 ), p(t)=-2t, q(t)=2, r(t)=0 are already polynomials We now write y=∑ ∞

Next, we substitute y, y’, y’’ into Equation (8.4) to obtain

Collecting the like powers of t we have that:

Equating the coeﬀicients of each occurring power of t to zero, we obtain that a0 +a2=0; a3=0; a n+2 ) 1 )(

− − n n a n n n for n≥2 Therefore, by induction, we derive a 2k+1 =0 and a 2k ) 1 2 (

Hence, we obtain the general solution of (8.4) as y = a1t+ ∑ ∞

Consider the homogeneous equation y’’ + p(t)y’ + q(t)y = 0 (1) with the coeﬀicients being continuous on the interval I

1 Show that two solutions of (1) on I that are zero at the same point in I cannot be linearly independent on I

2 Show that two solutions of (1) on I that have maxima or minima at the same point in I cannot be linearly independent on I

3 Suppose that y1 and y2 are two linearly independent solutions of (1) on I Show that z1=a11y1+a12y2 and z2=a21y1+a22y2 (for some constants ajk) form a basis of the solutions of (1) if and only if the determinant of the coeﬀicients ajk is not zero

4 Show that the equation t 2 y’’ -4ty’ + 6y=0 has y1=t 2 and y2=t 3 as a basis of the solutions for all t Show that W(t 2 , t 3 )=0 at t=0 Does this contradicts with Theorem 2.7 ?

In Problems 5 through 9, demonstrate that the function y1 is a solution to the specified equation Employ the reduction of order method to determine y2, ensuring that y1 and y2 together establish a basis Important: Begin by expressing the equation in standard form before applying the formula from Section 2.11.

In each of Problems 10 through 16 ﬁnd the general solution of the given diﬀerential equation

In each of Problems 17 through 24 ﬁnd the solution of the given initial value problem Sketch the graph of the solution and describe its behavior as t increases

In each of Problems 31 through 34 ﬁnd the Wronskian of two solutions of the given diﬀerential equation without solving the equatio n

In each of Problems 45 through 47 ﬁnd the solution of the given initial value problem Sketch the graph of the solution and describe its behavior for increasing t

48 Consider the initial value problem 3u’’ − u’ + 2u = 0, u(0) = 2, u’(0) = 0.

(a) Find the solution u(t) of this problem

(b) Find the ﬁrst time at which |u(t)| = 10

49 Consider the initial value problem 5u’’ + 2u’ + 7u = 0, u(0) = 2, u’(0) = 1.

(a) Find the solution u(t) of this problem

(b) Find the smallest T such that |u(t)| ≤ 0.1 for all t > T

50 Euler Equations An equation of the form t 2 y’’ + αty’ + βy = 0, t > 0, where α and β are real constants, is called an Euler equation Show that the substitution x = ln t transforms an Euler equation into an equation with constant coeﬀicients Then, using this substitution to solve the following equations a) t 2 y’’ − 3ty’ + 4y = 0, t > 0 b) t 2 y’’ + 2ty’+ 0.25y = 0, t > 0

In each of Problems 61 through 64 solve the given initial value problem Sketch the graph of the solution and describe its behavior for increasing t

65 If a, b, and c are positive constants, show that all solutions of ay’’ + by’ + cy = 0 approach zero as t →∞

66 (a) If a > 0 and c > 0, but b = 0, show that the result of Problem 65 is no longer true, but that all solutions are bounded as t →∞

When a > 0 and b > 0, but c = 0, the conclusions drawn in Problem 65 are invalid However, as time approaches infinity (t → ∞), all solutions converge to a constant that is influenced by the initial conditions Specifically, for the initial conditions y(0) = y0 and y’(0) = y1, this constant can be determined based on these values.

In each of Problems 74 through 80:

(a) Determine a suitable form for Y(t) if the method of undetermined coeﬀicients is to be used

(b) Find a particular solution of the given equation

89 A mass weighing 2 lb stretches a spring 6 in If the mass is pulled down an additional 3 in and then released, and if there is no damping, determine the position u of the mass at any time t Plot u versus t Find the frequency, period, and amplitude of the motion

90 A mass of 100 g stretches a spring 5 cm If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/sec, and if there is no damping, determine the position u of the mass at any time t When does the mass ﬁrst return to its equilibrium position?

91 A mass weighing 3 lb stretches a spring 3 in If the mass is pushed upward, contracting the spring a distance of 1 in., and then set in motion with a downward velocity of 2 ft/sec, and if there is no damping, ﬁnd the position u of the mass at any time t Determine the frequency, period, amplitude, and phase of the motion

92 A series circuit has a capacitor of 0.25 × 10 −6 farad and an inductor of 1 henry If the initial charge on the capacitor is 10 −6 coulomb and there is no initial current, ﬁnd the charge Q on the capacitor at any time t

93 A mass of 20 g stretches a spring 5 cm Suppose that the mass is also attached to a viscous damper with a damping constant of 400 dyne-sec/cm If the mass is pulled down an additional 2 cm and then released, ﬁnd its position u at any time t Plot u versus t Determine the quasi frequency and the quasi period Determine the ratio of the quasi period to the period of the corresponding undamped motion Also ﬁnd the time τ such that |u(t)| < 0.05 cm for all t > τ

94 A mass weighing 4 lb stretches a spring 1.5 in The mass is displaced 2 in in the positive direction from its equilibrium position and released with no initial velocity Assuming that there is no damping and that the mass is acted on by an external force of 2 cos 3t lb, formulate the initial value problem describing the motion of the mass

(b) Plot the graph of the solution

(c) If the given external force is replaced by a force 4 sin ωt of frequency ω, ﬁnd the value of ω for which resonance occurs

95 A mass of 5 kg stretches a spring 10 cm The mass is acted on by an external force of

The system experiences a force of 10 sin(t/2) N and encounters a viscous force of 2 N at a speed of 4 cm/sec When the mass is displaced from its equilibrium position and given an initial velocity of 3 cm/sec, we can formulate the initial value problem that governs its motion.

(a) Find the solution of the initial value problem

(b) Identify the transient and steady-state parts of the solution

(c) Plot the graph of the steady-state solution

(d) If the given external force is replaced by a force 2 cos ωt of frequency ω, ﬁnd the value of ω for which the amplitude of the forced response is maximum

96 If an undamped spring–mass system with a mass that weighs 6 lb and a spring constant

At time t = 0, a mass with a spring constant of 1 lb/in is subjected to an external force of 4 cos(7t) lb, initiating its motion To determine the position of the mass at any given time, we can analyze the system's dynamics and derive the displacement function The resulting equation will allow us to graph the displacement versus time, illustrating the mass's oscillatory behavior influenced by the external force.

97 A mass that weighs 8 lb stretches a spring 6 in The system is acted on by an external force of 8 sin 8t lb If the mass is pulled down 3 in and then released, determine the position of the mass at any time Determine the ﬁrst four times at which the velocity of the mass is zero

98 Find the power series solutions (in powers of x) of the following equations:

Laplace Transform

Tiêu đề	Lecture on Infinite Series and Differential Equations
Tác giả	Nguyen Thieu Huy
Người hướng dẫn	Dr. Tran Xuan Tiep
Trường học	Hanoi University of Science and Technology
Chuyên ngành	Applied Mathematics and Informatics
Thể loại	Lecture
Năm xuất bản	2009
Thành phố	Ha Noi

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Số trang	83
Dung lượng	11,36 MB