Solution Guide, differential equation

Lecture Notes in Mathematics

Arkansas Tech UniversityDepartment of Mathematics

A Second Course in ElementaryDifferential Equations

Solution Guide

Marcel B. FinancAll Rights Reserved

2015 Edition

Contents

1 nth Order Ordinary Differential Equations 51.1 Calculus of Matrix-Valued Functions of a Real Variable . . . . 51.2 Uniform Convergence and Weierstrass M-Test . . . . . . . . . 131.3 nth Order Linear Differential Equations: Exsitence and Unique-

ness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 The General Solution of nth Order Linear Homogeneous Equa-

tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.5 Fundamental Sets and Linear Independence . . . . . . . . . . 321.6 nth Order Homogeneous Linear Equations with Constant Co-

efficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361.7 Review of Power Series . . . . . . . . . . . . . . . . . . . . . . 411.8 Series Solutions of Linear Homogeneous Differential Equations 47

3

4 CONTENTS

Chapter 1

nth Order Ordinary DifferentialEquations

1.1 Calculus of Matrix-Valued Functions of a

Real Variable

Problem 1.1.1Consider the following matrices

A(t) =

[t 1 t2

2 2t+ 1

], B(t) =

[t 10 t+ 2

], c(t) =

[t+ 11

](a) Find 2A(t) - 3tB(t).(b) Find A(t)B(t) - B(t)A(t).(c) Find A(t)c(t).(d) Find det(B(t)A(t)).

Solution.(a) We have

2A(t) 3tB(t) = 2[t 1 t2

2 2t+ 1

] 3t

[t 10 t+ 2

]=

[2t 2 2t2

4 4t+ 2

][

3t2 3t0 3t2 + 6t

]=

[2t 2 3t2 2t2 + 3t

4 2 2t 3t2]

5

6CHAPTER 1. NTH ORDER ORDINARY DIFFERENTIAL EQUATIONS

(b) We have

A(t)B(t)B(t)A(t) =[t 1 t2

2 2t+ 1

] [t 10 t+ 2

][t 10 t+ 2

] [t 1 t2

2 2t+ 1

]=

[t2 t t3 + 2t2 t+ 1

2t 2t2 + 5t

][t2 t 2 t3 2t 1

2t+ 4 2t2 + 5t+ 2

]=

[2 2t2 + t+ 24 2

](c) We have

A(t)c(t) =

[t 1 t2

2 2t+ 1

] [t+ 11

]=

[(t 1)(t+ 1) + t2(1)2(t+ 1) + (2t+ 1)(1)

]=

[11

](d) We have

det(B(t)A(t)) = det

([t 10 t+ 2

] [t 1 t2

2 2t+ 1

])= det

([t2 t 2 t3 2t 1

2t+ 4 2t2 + 5t+ 2

]= (t3 + 3t2 + 2t)

Problem 1.1.2Determine all values t such that A(t) is invertible and, for those t-values,find A1(t).

A(t) =

[t+ 1 tt t+ 1

].

Solution.We have det(A(t)) = 2t + 1 so that A is invertible for all t 6= 1

2. In this

case,

A1(t) =1

2t+ 1

[t+ 1 tt t+ 1

]Problem 1.1.3Determine all values t such that A(t) is invertible and, for those t-values,find A1(t).

A(t) =

[sin t cos tsin t cos t

].

1.1. CALCULUS OFMATRIX-VALUED FUNCTIONS OF A REAL VARIABLE7

Solution.We have det(A(t)) = 2 sin t cos 2 = sin 2t so that A is invertible for all t 6= n

2

where n is an integer. In this case,

A1(t) =1

sin 2t

[cos t cos t sin t sin t

]Problem 1.1.4Find

limt0

[sin tt

t cos t 3t+1

e3t sec t 2tt21

].

Solution.

limt0

[sin tt

t cos t 3t+1

e3t sec t 2tt21

]=

[1 0 31 1 0

]Problem 1.1.5Find

limt0

[tet tan tt2 2 esin t

].

Solution.

limt0

[tet tan tt2 2 esin t

]=

[0 02 1

]Problem 1.1.6Find A(t) and A(t) if

A(t) =

[sin t 3tt2 + 2 5

].

Solution.

A(t) =

[cos t 32t 0

]A(t) =

[ sin t 0

2 0

]

8CHAPTER 1. NTH ORDER ORDINARY DIFFERENTIAL EQUATIONS

Problem 1.1.7Express the system

y1 = t2y1 + 3y2 + sec t

y2 = (sin t)y1 + ty2 5

in the matrix form

y(t) = A(t)y(t) + g(t).

Solution.

y(t) =

[y1(t)y2(t)

], A(t) =

[t2 3

sin t t

], g(t) =

[sec t5

]Problem 1.1.8Determine A(t) where

A(t) =

[2t 1

cos t 3t2

], A(0) =

[2 51 2

].

Solution.Integrating componentwise we find

A(t) =

[t2 + c11 t+ c12

sin t+ c21 t3 + c22

]Since

A(0) =

[2 51 2

]=

[c11 c12c21 c22

]by equating componentwise we find c11 = 2, c12 = 5, c21 = 1, and c22 = 2.Hence,

A(t) =

[t2 + 2 t+ 5

sin t+ 1 t3 +2

]Problem 1.1.9Determine A(t) where

A(t) =

[1 t0 0

], A(0) =

[1 12 1

], A(0) =

[1 22 3

].



A(t) =

[t+ c11

t2

2+ c12

c21 c22

]Integrating again we find

A(t) =

[t2

2+ c11t+ d11

t3

6+ c12t+ d12

c21t+ d21 c22t+ d22

]But

A(0) =

[1 12 1

]=

[d11 d12d21 d22

]so by equating componentwise we find d11 = 1, d12 = 1, d21 = 2, andd22 = 1. Thus,

A(t) =

[t2

2+ c11t+ 1

t3

6+ c12t+ 1

c21t+2 c22t+ 1

]Since

A(0) =

[1 22 3

]=

[c11 c12c21 c22

]we find c11 = 1, c12 = 2, c21 = 2, c22 = 3. Hence,

A(t) =

[t2

2 t+ 1 t3

6+ 2t+ 1

4 3t+ 1

]Problem 1.1.10Calculate A(t) =

t0

B(s)ds where

B(s) =

[es 6s

cos 2s sin 2s

].


A(t) =

[ t0esds

t0

6sds t0

cos 2sds t0

sin 2sds

]=

[et 1 3t2sin 2t2

1cos 2t2

]

10CHAPTER 1. NTH ORDERORDINARYDIFFERENTIAL EQUATIONS

Problem 1.1.11Construct a 2 2 nonconstant matrix function A(t) such that A2(t) is aconstant matrix.

Solution.Let

A(t) =

[0 t0 0

]Then

A2(t) =

[0 t0 0

] [0 t0 0

]=

[0 00 0

]Problem 1.1.12(a) Construct a 2 2 differentiable matrix function A(t) such that

d

dtA2(t) 6= 2A d

dtA(t).

That is, the power rule is not true for matrix functions.(b) What is the correct formula relating A2(t) to A(t) and A(t)?

Solution.(a) Let

A(t) =

[1 tt2 0

]. Then

A2(t) =

[1 + t3 tt2 t3

]so that

d

dtA2(t) =

[3t2 12t 3t2

].

On the other hand,

2A(t)d

dtA(t) = 2

[1 tt2 0

] [1 02t 0

]=

[2 + 4t2 0

2t2 0

](b) The correct formula is d

dtA2(t) = A(t)A(t) + A(t)A(t)


Problem 1.1.13Transform the following third-order equation

y 3ty + (sin 2t)y = 7et

into a first order system of the form

x(t) = Ax(t) + b(t).

Solution.Let x1 = y, x2 = y

, x3 = y. Then by letting

x =

x1x2x3

, A = 0 1 00 0 1 sin 2t 3t 0

, b = 00

7et

the differential equation can be presented by the first order system

x(t) = Ax(t) + b(t)

Problem 1.1.14By introducing new variables x1 and x2, write y

2y+ 1 = t as a system oftwo first order linear equations of the form x + Ax = b.

Solution.By letting x1 = y and x2 = y

we have

x =

[x1x2

], A =

[0 12 0

], b =

[0

t 1

]Problem 1.1.15Write the differential equation y + 4y + 4y = 0 as a first order system ofthe form x(t) + A(t)x(t) = b(t).


we have

x =

[x1x2

], A =

[0 14 4

], b =

[00

]


Problem 1.1.16Write the differential equation y+ ky+ (t 1)y = 0 as a first order systemof the form x(t) + A(t)x(t) = b(t)..


we have

x =

[x1x2

], A =

[0 1

t 1 k

], b =

[00

]Problem 1.1.17Change the following second-order equation to a first-order system.

y 5y + ty = 3t2, y(0) = 0, y(0) = 1Solution.If we write the problem in the matrix form

x + Ax = b, x(0) = y0

then

A =

[0 1t 5

]x =

[x1x2

]=

[yy

], b =

[0

3t2

], y0 =

[01

]Problem 1.1.18Consider the following system of first-order linear equations.

x =

[3 21 1

]x.

Find the second-order linear differential equation that x satisfies.

Solution.The system is

x1 = 3x1 + 2x2x2 = x1 x2

It follows that

x1 + 2x2 = 5x1 or x1 =

x1+2x2

5

so we let w = x1+2x25

so that w = x1. Thus, x1 = 3x1+2x2 = 3x1+(5wx1) =

2x1 + 5w. Hence, x1 = 2x

1 + 5w

= 2x1 + 5x1 or x1 2x1 5x1 = 0.

Likewise, x22x25x2 = 0. Hence, x1 and x2 satisfy the differential equationy 2y 5y = 0

1.2. UNIFORM CONVERGENCE AND WEIERSTRASS M-TEST 13

1.2 Uniform Convergence and Weierstrass M-

Test

Problem 1.2.1Define fn : [0, 1] R by fn(x) = xn. Define f : [0, 1] R by

f(x) =

{0 if 0 x < 11 if x = 1

(a) Show that the sequence {fn}n=1 converges pointwise to f.(b) Show that the sequence {fn}n=1 does not converge uniformly to f. Hint:Suppose otherwise. Let = 0.5 and get a contradiction by using a point(0.5)

1N < x < 1.

Solution.(a) For all 0 x < 1 we have limn fn(x) = limn xn = 0. Also,limn fn(1) = 1. Hence, the sequence {fn}n=1 converges pointwise to f.(b) Suppose the contrary. Let = 1

2. Then there exists a positive integer N

such that for all n N we have

|fn(x) f(x)| 0 be given. Let N be a positive integer such that N > 1

. Then for

n N x xnn x = |x|nn < 1n 1N < .

Thus, the given sequence converges uniformly (and pointwise) to the functionf(x) = x


Problem 1.2.7Let fn(x) =

xn

1+xnfor x [0, 2].

(a) Find the pointwise limit f(x) = limn fn(x) on [0, 2].(b) Does fn f uniformly on [0, 2]?

Solution.(a) The pointwise limit is

f(x) =

0 if 0 x < 112

if x = 11 if 1 < x 2.

(b) The convergence cannot be uniform because if it were, f would have tobe continuous since each fn is continuous on [0, 2]

Problem 1.2.8For each n N define fn : R R by fn(x) = n+cosx2n+sin2 x .(a) Show that fn 12 uniformly.(b) Find limn

72fn(x)dx.

Solution.(a) Let > 0 be given. Note that

|fn(x)1

2| =

2 cosx sin2 x2(2n+ sin2 x) 34n.

Since limn34n

= 0 we can find a positive integer N such that if n Nthen 3

4n< . Thus, for n N and all x R we have

|fn(x)1

2| 3

4n< .

This shows that fn 12 uniformly on R and also on [2, 7].(b) We have

limn

72

fn(x)dx =

72

limn

fn(x)dx =

72

1

2dx =

5

2

Problem 1.2.9Show that the sequence defined by fn(x) = (cosx)

n does not converge uni-formly on [

2, 2].

1.2. UNIFORM CONVERGENCE AND WEIERSTRASS M-TEST 17

Solution.We have proved in Problem 1.2.5 that this sequence converges pointwise tothe discontinuous function

f(x) =

{0 if

2 x < 0 and 0 < x

2

1 if x = 0.

Therefore, uniform convergence cannot occur for this given sequence

Problem 1.2.10Let {fn}n=1 be a sequence of functions such that

sup{|fn(x)| : 2 x 5} 2n

1 + 4n.

(a) Show that this sequence converges uniformly to a function f to be found.

(b) What is the value of the limit limn 52fn(x)dx?

Solution.(a) Using the squeeze rule we find

limn

sup{|fn(x)| : 2 x 5} = 0.

Let > 0 be given. Then there is a positive integer N such that for n N,we have

sup{|fn(x)| : 2 x 5} < .Thus,

|fn(x)| sup{|fn(x)| : 2 x 5} < for all x [2, 5]. Thus, {fn}n=1 converges uniformly to the zero function.(b) We have

limn

52

fn(x)dx =

52

limn

fn(x)dx =

52

0dx = 0

Problem 1.2.11Show that the series

n=1

12n

cos (3nx) converges uniformly on R.

Solution.We have 12n cos (3nx)

12n = Mn.Since

n=1

12n

= 12

11 1

2

= 1, by the Weierstrass M-test the given series con-

verges uniformly on R



n=1

nn4+x4

converges uniformly on R.

Solution.Since n4 + x4 n4 for all x R, we have nn4 + x4

nn4 = 1n3 = Mn.Since

n=1

1n3

is a pseries with p = 3 > 1 so it is convergent. By theWeierstrass M-test the given series converges uniformly on R


n=1

1n2+x2

converges uniformly on R.

Solution.Since n2 + x2 n2 for all x R, we have 1n2 + x2

1n2 = Mn.Since

n=1

1n2

is a pseries with p = 2 > 1 so it is convergent. By theWeierstrass M-test the given series converges uniformly on R

Problem 1.2.14Suppose that

n=1 |an|

1.3. NTH ORDER LINEAR DIFFERENTIAL EQUATIONS: EXSITENCE ANDUNIQUENESS19

1.3 nth Order Linear Differential Equations:

Exsitence and Uniqueness

Problem 1.3.1Convert the initial value problem

y 1t2 9

y + ln (t+ 1)y + (cos t)y = 0, y(0) = 1, y(0) = 3, y(0) = 0

into the matrix form

x(t) = A(t)x(t) + b(t), x(t0) = x0.

Solution.Letting x1 = y, x2 = y

, and x3 = y we find

A(t) =

0 1 00 0 1 cos t ln (t+ 1) 1

t29

x(t) =

x1x2x3

, b(t) =00

0

, x(0) =13

0


y +1

t+ 1y + (tan t)y = 0, y(0) = 0, y(0) = 1, y(0) = 2


x(t) = A(t)x(t) + b(t), x(t0) = x0.



A(t) =

0 1 00 0 1 tan t 1

t+10


x(t) =

x1x2x3

, b(t) =00

0

, x(0) =01

2


y 1t2 + 9

y+ln (t2 + 1)y+(cos t)y = t3+t5, y(0) = 1, y(0) = 3, y(0) = 0


x(t) = A(t)x(t) + b(t), x(t0) = x0.



A(t) =

0 1 00 0 1 cos t ln (t2 + 1) 1

t2+9

x(t) =

x1x2x3

, b(t) = 00t3 + t 5

, x(0) =13

0

Problem 1.3.4Transform the following third-order equation

y 3ty + (sin 2t)y = 7et

into a first order system of the form

x(t) = Ax(t) + b(t)

Solution.Let x1 = y, x2 = y

, x3 = y. Then by letting then the differential equation

can be presented by the first order system

x(t) = Ax(t) + b(t)


Problem 1.3.5If y(t) = 14

33e4t + 13

15e2t 16

55e7t is the solution to the initial value problem

y 5y 22y + 56y = 0, y(0) = 1, y(0) = 2, y(0) = 4

then what is the solution to the corresponding matrix system

x(t) = A(t)x(t) + b(t), x(t0) = x0?

Solution.The solution to the matrix system is

x(t) =

yyy

= 1433e4t + 1315e2t 1655e7t56

33e4t + 26

15e2t 112

55e7t

22433e4t + 52

15e2t 784

55e7t

Problem 1.3.6Suppose that the vector

x(t) =

et + et + cos t+ sin tet et sin t+ cos tet + et cos t sin tet et + sin t cos t

is a solution to the matrix system

x(t) =

0 1 0 00 0 1 00 0 0 11 0 0 0

x.Find the ordinary differential equation with solution y(t) = et + et + cos t+sin t.

Solution.We have y(4) = x

(4)1 = x

4 = (e

tet+sin tcos t) = et+et+cos t+sin t = y.Thus, y satisfies the equation y(4) y = 0

Problem 1.3.7Suppose that the vector

x(t) =

1 + cos (5t) + sin (5t)5 sin (5t) + 5 cos (5t)25 cos (5t) 25 sin (5t)


is a solution to the matrix system

x(t) =

0 1 00 0 10 25 0

x.Find the ordinary differential equation with solution y(t) = 1 + cos (5t) +sin (5t).

Solution.We have y = x

(3)1 = x

3 = (25 cos (5t) 25 sin (5t)) = 125 sin (5t)

125 cos (5t) = 25y. Thus, y satisfies the equation y + 25y = 0

Problem 1.3.8Find the first two Picards iterations in Problem 1.3.1.

Solution.We have

x(t) = x(0) +

t0

A(s)x(s)ds.

Thus,

x0(t) =

130

x1(t) =

130

+ t0

0 1 00 0 1 cos s ln (s+ 1) 1

s29

130

ds=

130

+ t0

30 cos s 3 ln (s+ 1)

ds=

130

+ 3t0 sin t 3(t+ 1) ln (t+ 1) + 3t

=

1 + 3t3 sin t 3(t+ 1) ln (t+ 1) + 3t


Problem 1.3.9Find the first two Picards iterations in Problem 1.3.3.

Solution.We have

x(t) = x(0) +

t0

A(s)x(s)ds.

Thus,

x0(t) =

130

x1(t) =

130

+ t0

0 1 00 0 1 cos s ln (s2 + 1) 1

s2+9

130

ds=

130

+ t0

30 cos s 3 ln (s2 + 1)

ds=

130

+ 3t0 sin t 3t ln (t2 + 1) + 6t 6 tan1 t

=

1 + 3t3 sin t 3t ln (t2 + 1) + 6t 6 tan1 t

Problem 1.3.10Prove the inquality (1.3.6) by induction on N 1.

Solution.Basis of Induction: For N = 1, we have

||x1 x0|| M(t t0) = MK11(t t0)1

1!.

Induction Hypothesis: Suppose

||xk xk1|| MKk1(t t0)k

k!.


For k = 1, 2, , N.Induction Step: We want to show that

||xN+1 xN || MKN(t t0)N+1

(N + 1)!.

Indeed, we have

||xN+1 xN || K tt0

||xN xN1||ds

K tt0

MKN1(s t0)N

N !ds

=MKN(t t0)N+1

(N + 1)!

Problem 1.3.11Let f(t) be a continuous function such that f(t) C +K

taf(s)ds for some

constants C and K 0. Show that f(t) CeK(ta).

Solution.This follows from Gronwalls lemma with h(t) = K

Problem 1.3.12Let f(t) be a non-negative continuous function such that f(t)

taf(s)ds

for all t a. Show that f(t) 0 for all t a.

Solution.This follows from Gronwalls lemma with C = 0 and h(t) = 1

For Problems 1.3.12 - 1.3.15, use Theorem 1.3.1 to find the largest inter-val a < t < b in which a unique solution is guaranteed to exist.

Problem 1.3.13

y 1t2 9

y + ln (t+ 1)y + (cos t)y = 0, y(0) = 1, y(0) = 3, y(0) = 0

Solution.The coefficient functions are all continuous for t 6= 3, 3 and t > 1. Sincet0 = 0, the largest interval of existence is 1 < t < 3


Problem 1.3.14

y +1

t+ 1y + (tan t)y = 0, y(0) = 0, y(0) = 1, y(0) = 2

Solution.The coefficient functions are all continuous for t 6= 1 and t 6= (2n + 1)

2

where n is an integer. Since t0 = 0, the largest interval of existence is1 < t <

2

Problem 1.3.15

y 1t2 + 9

y + ln (t2 + 1)y + (cos t)y = 0, y(0) = 1, y(0) = 3, y(0) = 0

Solution.The coefficient functions are all continuous for t so that the interval of exis-tence is < t 0, we must have 0 = r3 2r2 r + 2 = r2(r 2) (r 2) =(r 2)(r2 1) = (r 2)(r 1)(r + 1). Hence, r = 1, 1, 2


1.4 The General Solution of nth Order Linear

Homogeneous Equations

In Problems 1.4.1 - 1.4.3, show that the given solutions form a fundamentalset for the differential equation by computing the Wronskian.

Problem 1.4.1

y y = 0, y1(t) = 1, y2(t) = et, y3(t) = et.

Solution.We have

W (t) =

1 et et

0 et et0 et et

= (1)

et etet et et 0 et0 et

+ et 0 et0 et = 2 6= 0

Problem 1.4.2

y(4) + y = 0, y1(t) = 1, y2(t) = t, y3(t) = cos t, y4(t) = sin t.

Solution.We have

W (t) =

1 t cos t sin t0 1 sin t cos t0 0 cos t sin t0 0 sin t cos t

= cos2 t+ sin2 t = 1 6= 0

Problem 1.4.3

t2y + ty y = 0, y1(t) = 1, y2(t) = ln t, y3(t) = t2.

1.4. THE GENERAL SOLUTION OFNTH ORDER LINEAR HOMOGENEOUS EQUATIONS27

Solution.We have

W (t) =

1 ln t t2

0 1t

2t0 1

t22

= (1)

t1 2tt2 2 ln t 0 2t0 2

+ t2 0 t10 t2 =

=3t1 6= 0, t > 0

Use the fact that the solutions given in Problems 1.4.1 - 1.4.3 for a funda-mental set of solutions to solve the following initial value problems 1.4.4 -1.4.6.

Problem 1.4.4

y y = 0, y(0) = 3, y(0) = 3, y(0) = 1.

Solution.The general solution is y(t) = c1 + c2e

t + c3et. With the initial conditions

we havey(0) = 3 = c1 + c2 + c3 = 3y(0) = 3 = c2 c3 = 3y(0) = 1 = c2 + c3 = 1

Solving these simultaneous equations gives c1 = 2, c2 = 1 and c3 = 2 andso the unique solution is

y(t) = 2 et + 2et

Problem 1.4.5

y(4) + y = 0, y(

2) = 2 + , y(

2) = 3, y(

2) = 3, y(

2) = 1.

Solution.The general solution is y(t) = c1 + c2t + c3 cos t + c4 sin t. With the initialconditions we have

y(2) = 2 + = c1 + 2 c2 + c4 = 2 +

y(2) = 3 = c2 c3 = 3

y(2) = 3 = c4 = 3

y(2) = 1 = c3 = 1


Solving these simultaneous equations gives c1 = ( + 1), c2 = 4, c3 = 1and c4 = 3. Hence, the unique solution is

y(t) = ( + 1) + 4t+ cos t+ 3 sin t

Problem 1.4.6

t2y + ty y = 0, , y(1) = 1, y(1) = 2, y(1) = 6.

Solution.The general solution is y(t) = c1 + c2 ln t + c3t

2. With the initial conditionswe have

y(1) = 1 = c1 + c3 = 1y(1) = 2 = c2 + 2c3 = 2y(1) = 6 = c2 + 2c3 = 6

Solving these simultaneous equations gives c1 = 2, c2 = 4 and c3 = 1 andso the unique solution is

y(t) = 2 + 4 ln t t2

Problem 1.4.7In each question below, show that the Wronskian determinant W (t) behavesas predicted by Abels Theorem. That is, for the given value of t0, show that

W (t) = W (t0)e

tt0pn1(s)ds.

(a) W (t) found in Problem 1.4.1 and t0 = 1.(b) W (t) found in Problem 1.4.2 and t0 = 1.(c) W (t) found in Problem 1.4.3 and t0 = 2.

Solution.(a) For the given differential equation pn1(t) = p2(t) = 0 so that Abelstheorem predicts W (t) = W (t0). Now, for t0 = 1 we have W (t) = W (1) =constant. From Problem 1.4.1, we found that W (t) = W (1)e

t1 0ds = 2.

(b) For the given differential equation pn1(t) = p3(t) = 0 so that Abelstheorem predict W (t) = W (t0). Now, for t0 = 1 we have W (t) = W (1) =

constant. From Problem 1.4.2, we found that W (t) = W (1)e t1 0ds = 1.

(c) For the given differential equation pn1(t) = p2(t) =1t

so that Abels

theorem predict W (t) = W (2)e t2

dss = W (2)eln (

2t) = 2

tW (2). From Problem

1.4.3, we found that W (t) = 3t

= W (2)e t2

dss where W (2) = 3

2

1.4. THE GENERAL SOLUTION OFNTH ORDER LINEAR HOMOGENEOUS EQUATIONS29

Problem 1.4.8Determine W (t) for the differential equation y+(sin t)y+(cos t)y+2y = 0such that W (1) = 0.

Solution.Here pn1(t) = p2(t) = sin t. By Abels Theorem we have

W (t) = W (1)e t1 sin sds 0

Problem 1.4.9Determine W (t) for the differential equation t3y2y = 0 such that W (1) =3.

Solution.Here pn1(t) = p2(t) = 0. By Abels Theorem we have

W (t) = W (1)e t1 0ds = W (1) = 3

Problem 1.4.10Consider the initial value problem

y y = 0, y(0) = , y(0) = , y(0) = 4.

The general solution of the differential equation is y(t) = c1 + c2et + c3e

t.(a) For what values of and will limt y(t) = 0?(b) For what values and will the solution y(t) be bounded for t 0, i.e.,|y(t)| M for all t 0 and for some M > 0? Will any values of and produce a solution y(t) that is bounded for all real number t?

Solution.(a) Since y(0) = , c1 + c2 + c3 = . Since y

(t) = c2et c3et and y(0) = ,

c2c3 = . Also, since y(t) = c2et+c3et and y(0) = 4 we have c2+c3 = 4.Solving these equations for c1, c2, and c3 we find c1 = 4, c2 = /2 + 2and c3 = /2 + 2. Thus,

y(t) = 4 + (/2 + 2)et + (/2 + 2)et.

If limt y(t) = 0, then we must have 4 = 0 and 2 + 2 = 0. In this case, = 4 and = 4. Thus,

y(t) = 4et.


(b) In the expression of y(t) we know that et is bounded for t 0 (et 1for t 0) whereas et is unbounded for t 0. Thus, for y(t) to be boundedwe must choose

2+ 2 = 0 or = 4. The number can be any number.

Now, for the solution y(t) to be bounded on < t < we must havesimultaneously /2+2 = 0 and /2+2 = 0. But there is no that satisfiesthese two equations at the same time. Hence, y(t) is always unbounded forany choice of and

Problem 1.4.11Consider the differential equation y + p2(t)y

+ p1(t)y = 0 on the interval

1 < t < 1. Suppose it is known that the coefficient functions p2(t) and p1(t)are both continuous on 1 < t < 1. Is it possible that y(t) = c1 + c2t2 + c3t4is the general solution for some functions p1(t) and p2(t) continuous on 1

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