LectnotesCh3

8/3/2019 LectnotesCh3

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Contents

Chapter 1. First-Order Differential Equations 5

1. Notation and Definitions 5

2. An Example of Modeling by First-Order DEs 9

3. The Geometry of First-Order DEs 9

4. Several Types of Solvable First-Order Differential Equations 10

4.1. Separable DEs 11

4.2. First-order linear equations 12

4.3. Bernoullis equations 164.4. Exact DEs 18

4.5. Non-exact DEs with integrating factors 21

4.6. Homogenous DEs 23

5. The Existence and Uniqueness Theorem 25

6. Reducible Second-Order DEs 27

6.1. Dependent variable y missing 27

6.2. Independent variable x missing 28

7. Solutions to (Partial) Exercises 31

Chapter 2. Second-Order Linear Equations 371. Basic Theoretical Results 37

1.1. Existence and uniqueness of solutions to IVP 38

1.2. Principle of superposition 38

1.3. Linear dependence/independence and Wronskian 39

2. Reduction of Order 45

3. Second-Order Homogeneous Linear DE with Constant Coefficients 49

4. Nonhomogeneous DE: Method of Undetermined Coefficients 52

5. Nonhomogeneous DE: Variation of Parameters 57

6. Solutions to Selected Exercises 60

Chapter 3. Higher Order Linear Differential Equations 63

1. Basic Theoretical Results 63

2. Reduction of Order 67

3. Linear Homogeneous DE with Constant Coefficients 68

4. Nonhomogeneous DE: Method of Undetermined Coefficients 71

3


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4 CONTENTS

5. Nonhomogeneous DE: Method of Variation-of-Parameters 73

6. Solutions to Selected Exercises 76


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CHAPTER 3

Higher Order Linear Differential Equations

The theory and solution techniques developed in the preceding chapter for second

order linear equations can be extended directly to linear equations of higher order.

In this chapter, we just briefly discuss these extensions, and leave most the proofs as

exercises.

Please read the textbook Page 219-224, and refer to Chapter Two for the proofs!

1. Basic Theoretical Results

Consider the nth-order linear DE of the form

L[y] := y(n) + p1(x)y(n1) + + pn(x)y = f(x), (1.1)

where p1, p2, , pn, f are continuous in some interval I. The associated homogeneousDE of (1.1) is

L[y] := y(n) + p1(x)y(n1) + + pn(x)y = 0. (1.2)

The following theorem is of crucial importance.

Theorem 1.1. (Existence and Uniqueness) Consider the IVP:y(n) + p1(x)y

(n1) + + pn(x)y = f(x),y(x0) = y0, y

(x0) = y1, , y(n1)(x0) = yn1.(1.3)

If p1, , pn, f are continuous in an interval I containingx0, then (1.3) has a uniquesolution y(x) in I.

As before, an important property of the homogeneous equation (1.2) is the prin-

ciple of superposition.

Theorem 1.2. If {yk}nk=1 are solutions to (1.2), then any linear combination

z(x) =n

k=1

ckyk(x), x I

is also a solution of (1.2) for any arbitrary constants {ck}nk=1.65


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66 3. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS

As with the second-order linear homogeneous DE, the main result to be estab-

lished is that every solution of (1.2) can be expressed as a linear combination

of its n linear independent solutions {yk}nk=1. That is,V = y C

n(I) : L[y] = 0 = span{y1, , yn}.In the first place, we need to extend the definition of linear dependence/independenceof two functions to n functions.

Definition 1.1. The functions f1, f2, , fn are said to be linearly dependenton I if there exists a set of constants c1, c2, , cn, which are not all zero, such that

c1f1(x) + c2f2(x) + + cnfn(x) = 0, (1.4)for allx I, otherwise, they are said to be linearly independent, in other words,the equation (1.4) holds, only when c1 = c2 = = cn = 0.

Example 1.1. Determine whether the following functions are linearly dependent:

f1(x) = 2x 3, f2(x) = x2 + 1, f3(x) = 2x2 1.

Solution: Set the equation

c1f1 + c2f2 + c3f3 = 0 c1(2x 3) + c2(x2 + 1) + c3(2x2 1) = 0. (1.5)That is

(c2 + 2c3)x2 + 2c1x + (3c1 + c2 c3) = 0,

which should be valid for all x. Therefore, we have

c2 + 2c3 = 0, 2c1 = 0, 3c1 + c2 c3 = 0.

This system has only zero solution: c1 = c2 = c3 = 0. Hence, these three functionsare linearly independent.

As before, we may use the Wronskian to determine the linear dependence/independence.

Definition 1.2. Suppose that y1, y2, , yn are differentiable up to derivatives oforder n 1. Then the Wronskian is defined by

W(y1, y2, , yn)(x) =

y1 y2 yny1 y

2 yn

...... ...

y(n1)1 y

(n1)2 y(n1)n

.

Exercise 1.1. Compute the Wronskian W(1, x , x2, x3) and W(1, x, , xn).

Exercise 1.2. Show that if there exists x0 I such that W(y1, y2, , yn)(x0) =0, then y1, y2, , yn are linearly independent in I. Equivalently, if y1, y2, , yn arelinearly dependent, then the Wronskian W(y1, y2, , yn)(x) 0, for all x I.


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1. BASIC THEORETICAL RESULTS 67

Question 1.1. Suppose that y1, y2, , yn are linearly independent, can we claimthat there always exists a point x0 I, such that W(y1, y2, , yn)(x0) = 0? Explainwhy? (Recall the conclusion for two functions in Chapter Two).

We next generalize the Abels formula to higher order DE (see Problem 20 on

Page 225 of the textbook). We start with the third-order equations.

Example 1.2. Lety1, y2 and y3 be solutions to the third-order equation

y + p1(x)y + p2(x)y

+ p3(x)y = 0. (1.6)

LetW(x) = W(y1, y2, y3)(x) be the Wronskian. Show that W satisfies the equation

W(x) = p1(x)W(x). (1.7)Therefore, we have the Abels formula

W(y1, y2, y3)(x) = c exp p1(x)dx. (1.8)

It follows that W is either always zero or nowhere zero in I.

Solution: To prove (1.7), we differentiate W(x) and recall that the derivative of a 3-

by-3 determinant is the sum of three 3-by-3 determinants obtained by differentiating

the first, second and third rows, respectively. Therefore, we have

W(x) = det

y1 y2 y3y1 y2 y3

y1 y2 y

3

. (1.9)

Since y1, y2 and y3 are solutions of (1.6), we substitute yi by p1(x)yi p2(x)yi

p3(x)yi for i = 1, 2, 3. Using the addition property of the determinants, we find that

W(x) = det

y1 y2 y3y1 y2 y3p1y1 p1y2 p1y3

= p1(x)W(x). (1.10)

Thus, (1.8) follows.

Using the same argument, we can derive the Alels formula for the n-order linear

homogeneous DE.

Exercise 1.3. Generalize the Abels formula to the nth order equation

y(n) + p1(x)y(n1) + + pn(x)y = 0

with solutions y1, y2, , yn. That is to establish the Abels formulaW(y1, y2, , yn)(x) = c exp

p1(x)dx

.

As with the second-order equation, we can prove that

Theorem 1.3.


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(i) If W(y1, y2, , yn)(x0) = 0 for some x0 I, then y1, y2, , yn arelinearly independent in I.

(ii) If they are linearly independent, and are solutions of

y(n) + p1(x)y(n1) +

+ pn(x)y = 0

in I, then the Wronskian is nowhere zero in I.

Proof. The proof is similar to that for two functions in Chapter Two. You may

also refer to Pages 220-221 of the textbook for the proof.

We now state the main result on the solution structure of the nth order linear

homogeneous equation.

Theorem 1.4. If p1, p2, , pn C(I), and the functions y1, y2, , yn are nlinearly independent solutions (i.e., W(y1, y2, , yn)(x) = 0) of

y(n) + p1

(x)y(n1) +

+ pn(x)y = 0, (1.11)

then every solution can be expressed as a linear combination of the solutions y1, y2, , yn,which forms a fundamental set of solutions.

For the nonhomogeneous equation

y(n) + p1(x)y(n1) + + pn(x)y = f(x), (1.12)

the general solution

y(x) = yp + yc = yp + c1y1 + c2y2 + + cnyn,where yp is a particular solution of (1.12), andyc is the complementary function, i.e.,

the general solution of (1.11).


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2. REDUCTION OF ORDER 69

2. Reduction of Order

The method of reduction of order has proven to be a useful approach to find

another linear independent solution based on a given solution. Although such a

method can be extended to higher-order DE, it can only be used to reduce the order

of DE one order each time.

Example 2.1. Show that if y1 is a solution of

y + p1(x)y + p2(x)y

+ p3(x)y = 0,

then the substitution y = y1(x)v(x) leads to the following second-order equation for

v :

y1v + (3y1 + p1y1)v

+ (3y1 + 2p1y1 + p2y1)v

= 0. (2.1)

Note: v satisfies the above second-order equation, so we need to one solution of this

equation to proceed this method.

Use the method to solve

(2 x)y + (2x 3)y xy + y = 0, x < 2; y1(x) = ex. (2.2)

Proof: As the targeted solution is y1v, we have

(y1v) + p1(x)(y1v)

+ p2(x)(y1v) + p3(x)(y1v) = 0.

Using the fact that y1 is a solution, we obtain (2.1).

Applying this method to (2.2), we find if y = exv is a solution, then v satisfies

v +3 x2 xv

= 0.

Let u = v. We solve u and find that

u = c1(x 2)ex.Then we integrate twice to find v :

v = c1xex + c2x.

Thus, the general solution is

y(x) = c1x + c2xex + c3e

x.


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3. Linear Homogeneous DE with Constant Coefficients

In the previous chapter, we saw that the second-order homogeneous equations with

constant coefficients have solutions of the form y(x) = erx. This observation enabled us

to derive two linearly independent solutions needed to determine the general solution

the underlying DE. We next apply this idea to the nth-order DE.

Please read Pages 226-238 of the textbook!

Consider the nth order homogeneous equation

L[y] = a0y(n) + a1y

(n1) + + any = 0, (3.1)where a0, a1, , an are real constants and a0 = 0. We expect that y = erx is asolution, and therefore

L[erx] = erxa0rn + a1r

n1 + + an = erxZ(r), (3.2)

where

Z(r) = a0rn + a1r

n1 + + an. (3.3)Consequently, we conclude that y(x) = erx is a solution to (3.1) if and only if r

is a root of the characteristic equation

Z(r) = a0rn + a1r

n1 + + an = 0, (3.4)where Z(r) is referred to as the characteristic polynomial as before.

The solution structure is characterized by the following theorem.

Theorem 3.1. Consider the DE

L[y] = a0y(n) + a1y(n1) + + any = 0. (3.5)Letr1, r2, , rk be the distinct roots of the characteristic equation, and

Z(r) = (r r1)m1(r r2)m2 (r rk)mk

where mi 1 is the multiplicity of the root ri, andk

i=1 mi = n.

(i). If allri are real and distinct, i.e., all mi = 1, then we have n distinct linearly

independent solutions er1x, er2x, , ernx, which form a fundamental set ofsolutions;

(ii). If ri with mi > 1 is real, then the functions erix, xerix, , xmi1erix are mi

LI solutions corresponding to this repeated root;(iii). If rj is complex, say, rj = a + bi, with multiplicity mj 1, then its conjugate

is a root as well, and the 2mj LI solutions are

eax cos bx, xeax cos bx, , xmj1eax cos bx,eax sin bx, xeax sin bx, , xmj1eax sin bx.


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3. LINEAR HOMOGENEOUS DE WITH CONSTANT COEFFICIENTS 71

Example 3.1. Find the general solution to

y y + y y = 0.

Solution: The characteristic equation is

r3 r2 + r 1 = 0 (r 1)(r2 + 1) = 0 r1 = 1, r2,3 = i.Therefore the general solution is

y(x) = c1ex + c2 sin x + c3 cos x.

Example 3.2. Solve the following equation:

y + 2y + 3y + 2y = 0.


r3 + 2r2 + 3r + 2 = 0 (r3 + 2r2 + r) + (2r + 2) = r(r + 1)2 + 2(r + 1)

= (r + 1)(r(r + 1) + 2) = (r + 1)(r2

+ r + 2) = 0.Therefore the roots are

r1 = 1, r2,3 = 1

7i

2.

Hence, the general solution is

y(x) = c1ex + e

x2

c2 cos(

7

2x) + c3 sin(

7

2x)

.


y(4) + 2y + y = 0.


r4 + 2r2 + 1 = 0.

The roots are

r = i, i,i,i.Therefore, the general solution is

y = c1 cos x + c2 sin x + c3x cos x + c4x sin x.

Exercise 3.1. Find the general solution to the given DE:

(i) y 6y + 11y 6y = 0 (y = c1ex

+ c2e2x

+ c3e3x

)(ii) y(4) 9y + 20y = 0 (y = c1e2x + c2e2x + c3e

5x + c4e

5x))

(iii) y6y + 2y + 36y = 0 (y = c1e2x + e4x(c2 cos

2x + c3 sin

2x)

(iv) y(4) + 8y + 24y + 32y + 16y = 0 (y = c1e2x + c2e2x + c3x2e2x +

c4x3e2x)


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Example 3.4. Find the general solution to a forth-order linear homogenous DE

for y(x) with real numbers as coefficients if one solution is known to be x3e4x.

Solution: If x3e4x is a solution, then x2e4x, xe4x, and e4x are solutions as well. We

now have four linearly independent solutions to a fourth-order linear, homogenous

DE. Hence, the general solution is

y(x) = (c1 + c2x + c3x2 + c4x

3)e4x.

Exercise 3.2. Find the general solution to a third-order linear homogenous equa-

tion for y(x) with real numbers as coefficients if two solutions are known to be e2x

and sin3x. [Key : y(x) = c1e2x + c2 cos3x + c3 sin3x].

Example 3.5. Solve y(4) 4y(3) 5y(2) + 36y 36y = 0, if one solution is xe2x.

Solution: If xe2x is a solution, then so is e2x, which implies that (r 2)2 is a factor

of the characteristic equationr4 4r3 5r2 + 36r 36 = 0.

Thus,r4 4r3 5r2 + 36r 36

(r 2)2 = r2 9.

The other two roots of the characteristic equation are r = 3. Finally, the generalsolution is

y(x) = c1e2x + c2xe

2x + c3e3x + c4e

3x.


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4. NONHOMOGENEOUS DE: METHOD OF UNDETERMINED COEFFICIENTS 73

4. Nonhomogeneous DE: Method of Undetermined Coefficients

We now turn our attentions to nonhomogeneous DE and illustrate how the meth-

ods for solving second-order DE introduced in Chapter 2 can be extended to the

general nth-order DE.

In this section, we consider the method of undetermined coefficients. We therefore

restrict our attentions to determining a particular solution to

a0y(n) + a1y

(n1) + + an1y + any = f(x)where f(x) is limited to one of the following types

(1) f(x) =

Aeax (exponential function)A0 + A1 + + Akxk (polynomial)A cos bx + B sin bx (trigonometric polynomial)

(2) sum or product of functions given in (1)

In Chapter 2, we used the following usual trial solutions corresponding to each

type

yp(x) =

A0eax,

B0 + B1x + B2x2 + + Bkxk,

A0 cos bx + B0 sin bx.

However, if the usual trial solution contains one term that solves the associated

homogeneous DE, then we have to modify the trial solution by multiplying by xm,

until no any term solves the homogeneous DE. The basic rule is summarized below:

Multiply the usual trial solution by xm, where m is the smallest positive inte-ger such that the resulting trial solution has No Term solves the associatedhomogeneous DE.

Example 4.1. Determine a trial solution for y(5) y(4) + 2y(3) 2y + y y =4cos x.

Solution: The characteristic equation of the homogeneous DE is

r5 r4 + 2r3 2r2 + r 1 = 0

r4(r 1) + 2r2(r 1) + (r 1) = (r 1)(r4 + 2r2 + 1) = 0 (r 1)(r2 + 1)2 = 0 roots : 1,i ,i,i,i.

Then the general solution to the homogeneous DE is

yc(x) = c1ex + c2 cos x + c3 sin x + c4x cos x + c5x sin x.


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The usual trial solution corresponding to the nonhomogeneous term f(x) = 4 cos x is

yp(x) = A0 cos x + B0 sin x.

However, we see that this yp solves the homogeneous DE. Hence, we need to modify

it by multiplying x2, and obtain that

yp(x) = x2(A0 cos x + B0 sin x)

The constants A0 and B0 can be determined by substituting the proposed trial solu-

tion into the given DE.

Example 4.2. Determine the form of a trial solution to the DE with the char-

acteristic equation (r2 + 2r + 5)(r 1)3 = 0 and nonhomogeneous term f(x) =5ex + 7ex sin2x.

Solution: The general solution to the associated homogeneous DE is

yc(x) = ex

(c1 cos2x + c2 sin2x) + c3ex

+ c4xex

+ c5x2

ex

The usual trial solution corresponding to the nonhomogeneous term f1(x) = 5ex is

yp1(x) = A0ex. However, this coincides with part of yc(x). According to the mod-

ification rule, we need to modify the trial solution by multiplying it by x3, which

gives

yp1(x) = A0x3ex.

The usual trial solution corresponding to the nonhomogeneous term f2(x) = 7ex sin2x

is yp2(x) = ex(A1 cos2x + B1 sin2x). Once more, we need to modify it as

yp2(x) = xex(A1 cos2x + B1 sin2x).

Consequently, an appropriate trial solution for the given DE is

yp(x) = yp1(x) + yp2(x) = A0x3ex + xex(A1 cos2x + B1 sin2x).

Exercise 4.1. Find a particular solution to the following DE.

(i) y 3y + 3y y = 4ex.(ii) y(4) + 2y + y = 3 sin x 5cos x.

(iii) y 4y = x + 3 cos x + e2x

Key:

(i) yp(x) =23 x

3ex.

(ii) yp(x) = 38 x2 sin x + 58 x2 cos x.(iii) yp(x) = 18 x2 35 sin x + 18 xe2x


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5. NONHOMOGENEOUS DE: METHOD OF VARIATION-OF-PARAMETERS 75

5. Nonhomogeneous DE: Method of Variation-of-Parameters

We now consider the generalization of the variation-of-parameter method to linear

nonhomogeneous DE of arbitrary order n :

L[y] = y

(n)

+ p1(x)y

(n

1)

+ p2y

(n

2)

+ + pn(x)y = f(x), (5.1)where we assume that the functions p1, p2, , pn and f are continuous in I. Asbefore, to use this method, it is necessary to solve the corresponding homogeneous

differential equation. In general, this may be difficult unless the coefficients are con-

stants. However, the method of variation-of-parameter can be applied to nonhomo-

geneous DE with f(x) in a general form as long as we are able to solve the associated

homogeneous DE.

Please read Pages 239-242 of the textbook!

Suppose that we know a fundamental set of solutions y1, y2, , yn of the homo-geneous equation

L[y] = 0. (5.2)

Then the general solution of (5.2) is

yc(x) = c1y1(x) + c2y2(x) + + cnyn(x). (5.3)We now vary the constants and look for a particular solution to (5.1) of the form

yp(x) = u1(x)y1(x) + u2(x)y2(x) + + un(x)yn(x). (5.4)To determine the unknown functions u1, u2,

, un, we substitute yp(x) into equation

(5.1), but this will only give one equation. Therefore, we enforce u1, u2, , un tosatisfy n 1 additional equations so that the derivatives of order 2 of ui will notappear. More precisely, the unknown functions u1, u2, , un satisfy

y1u1 + y2u

2 + + ynun = 0,

y1u1 + y2u

2 + + ynun = 0,

......

......

y(n2)1 u

1 + y

(n2)2 u

2 + + y(n2)n un = 0,

y(n1)1 u

1 + y

(n1)2 u

2 + + y(n1)n un = f(x).

(5.5)

The matrix form is of the form

y1 x2 . . . yny1 y

2 . . . y

n

......

. . ....

y(n1)1 y

(n1)2 . . . y

(n1)n

u1u2...

un

=

00...

f(x)

. (5.6)


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We see that the determinant of the coefficient matrix is precisely the Wronskiam

W(y1, y2, , yn)(x) = 0 for all x, since y1, y2, , yn are LI solutions of (5.2). Usingthe Cramers rule, we can write the solution of (5.6) in the form

um(x) =

f(x)Wm(x)

W(x) , m = 1, , n, (5.7)where W(x) = W(y1, y2, , yn)(x), and Wm is the determinant obtained from W byreplacing the mth column by the column (0, 0, , 0, 1).

Therefore, by (5.7), we have

um(x) =

f(x)Wm(x)

W(x)dx,

and the particular solution is

yp(x) =n

m=1

ym(x)f(x)Wm(x)W(x)

dx. (5.8)

Finally, the general solution of (5.1) becomes

y(x) =n

m=1

f(x)Wm(x)W(x)

dx + cm

ym(x) (5.9)


y 3y + 3y y = 36ex ln x, x > 0.

Solution: The characteristic equation of the associated homogeneous equation is

r3 3r2 + 3r 1 = 0 (r 1)3 = 0.

The LI solutions are

y1(x) = ex, y2(x) = xe

x, y3(x) = x2ex

According to the variation-of-parameters method, the particular solution is of the

form

yp(x) = exu1(x) + xe

xu2(x) + x2exu3(x),

where u1, u2, u3 satisfy (after divided by ex)

u1 + xu

2 + x

2u3 = 0,

u1 + (x + 1)u2 + (x

2 + 2x)u3 = 0,

u1 + (x + 2)u2 + (x

2 + 4x + 2)u3 = 36ln x.


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5. NONHOMOGENEOUS DE: METHOD OF VARIATION-OF-PARAMETERS 77

To solve this system, we reduce its augmented matrix to the row-echelon form 1 x x2 01 x + 1 x2 + 2x 0

1 x + 2 x2 + 4x + 2 36 l n x

1 x x2 00 1 2x 0

0 2 4x + 2 36 l n x

1 x x2 00 1 2x 0

0 0 2 36 ln x

1 x x2

00 1 2x 00 0 1 18 ln x

.(5.10)

Consequently, we find that

u1 = 18x2 ln x, u2 = 36ln x u3 = 18ln x.

Using integration by parts yields

u1(x) = 18

x2 ln xdx = 2x3(3ln x 1),

u2(x) = 36

x ln xdx = 9x2(1 2 ln x),

u3(x) = 18

ln xdx = 18x(ln x 1),where we have set the integrating constants to be zero. So

yp(x) = exu1 + xe

xu2 + x2exu3 = x

3ex(6ln x 11).The general solution is

y(x) = exx3(6ln x 1) + c1 + c2x + c3x2.

Exercise 5.1. Find the general solution to the following DE.

(i) y + y = sec x.

Key : y(x) = c1 + c2 cos x + c3 sin x + ln |sec x + tan x| x cos x + (sin x) ln |cos x| .

(ii) y 3y + 2y = ex1+ex .Key : y(x) = 1

2(ex + 1)2 ln(ex + 1) + c1 + c2e

x + c3e2x

.

(iii) Given that x, x2, and 1x

are the solutions of the homogeneous equation cor-

responding to

x3y + x2y 2xy + 2y = 2x4, x > 0,determine a particular solution.

Key : yp

(x) =x4

15.


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6. Solutions to Selected Exercises

Exercise 4.1

(i) We find that the general solution to the associated homogeneous equation is

yc(x) = (c1 + c2x + c3x2)ex.

For f(x) = 4ex, the trial solution should be

yp(x) = Ax3ex.

We plug it into the original equation and find that A = 2/3.

(ii) We find that the general solution to the associated homogeneous equation is

yc(x) = (c1 + c2x)cos x + (c3 + c4x)sin x.

For f(x) = 3 sin x 5cos x, the trial solution should beyp(x) = x

2(A cos x + B sin x).

We plug it into the original equation to find A = 3/8 and B = 5/8.(iii) We find that the general solution to the associated homogeneous equation is

yc(x) = c1 + c2e2x + c3e

2x.

For f(x) = x + 3 cos x + e2x, the trial solution should be

yp(x) = x(A + Bx) + Ccos x + D sin x + Exe2x.

We plug it into the original equation to find

A = 0, B = 18

, C = 0, D = 35

, E =1

8.

Exercise 5.1

(i) We find that the general solution to the associated homogeneous equation is

yc(x) = c1 + c2 cos x + c3 sin x.

We look for a particular solution of the form

yp(x) = u1(x) + u2(x)cos x + u3(x)sin x.

Then, we have the system:

u1 + cos x u2 + sin x u

3 = 0,

sin x u2 + cos x u3 = 0, cos x u2 sin x u3 = sec x.


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6. SOLUTIONS TO SELECTED EXERCISES 79

This gives

u1 = sec x, u2 = 1, u3 = tan x.

Therefore,

u1 = ln |sec x + tan x| , u2 = x, u3 = ln | cos x|.The GS is

y(x) = c1 + c2 cos x + c3 sin x + ln |sec x + tan x| x cos x + (sin x) ln |cos x| .(ii) We find that the general solution to the associated homogeneous equation is

yc(x) = c1 + c2ex + c3e

2x.


yp(x) = u1

(x) + u2

(x)ex + u3

(x)e2x.


u1 + ex u2 + e

2x u3 = 0,

ex u2 + 2e2x u3 = 0,

ex u2 + 4e2x u3 =

ex

1+ex .

This gives

u1 =1

2

ex

1 + ex, u2 =

1

1 + ex, u3 =

1

2

ex

1 + ex.

Therefore,

u1 =ex

2 1

2ln(1 + ex), u2 = ln(1 + ex), u3 = 1

2 1

2ln(1 + ex).

Correspondingly,

yp(x) = 12

(ex + 1)2 ln(ex + 1) +ex

2+

e2x

2.

The GS is

y(x) = 12

(ex + 1)2 ln(ex + 1) + c1 + c2ex + c3e

2x.

(iii) We write the equation in the standard form

y +1

xy 2

x2y +

2

x3y = 2x.


yp(x) = xu1 + x2u2 +

1

xu3.


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xu1 + x2 u2 +

1x

u3 = 0,

u1 + 2x u2 1x2 u3 = 0,

2u2 +2x3

u3 = 2x.

This gives

u1 = x2, u2 =2

3x, u3 =

x4

3.

Therefore,

u1 = x3

3, u2 =

x2

3, u3 =

x5

15.

Correspondingly,

yp(x) =x4

15.

Note: This equation is a third-order Euler equation, so we may use the

transform t = ln x to convert it to a third-order equation with constant

coefficients. Then the method of undetermined coefficients may be appliedto find the particular solution.

LectnotesCh3

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