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4 Nonhomogeneous Equations with Constant Coefficients: The Method of Undetermined Coefficients

# 4 Nonhomogeneous Equations with Constant Coefficients: The Method of Undetermined Coefficients

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4.4 Nonhomogeneous Equations with Constant Coefﬁcients

3. If fi x

xm eΑx cos Βx or fi x

S

223

xm eΑx sin Βx, the associated set of functions is

eΑx cos Βx, xeΑx cos Βx, x2 eΑx cos Βx, . . . , xm eΑx cos Βx,

eΑx sin Βx, xeΑx sin Βx, x2 eΑx sin Βx, . . . , xm eΑx sin Βx .

For each function fi x in f x , determine the associated set of functions S. If any

of the functions in S appears in the general solution to the corresponding homogeneous equation, yh x , multiply each function in S by xr to obtain a new set xr S,

where r is the smallest positive integer so that each function in xr S is not a solution

of the corresponding homogeneous equation. A particular solution is obtained by

taking the linear combination of all functions in the associated sets where repeated

functions should appear only once in the particular solution.

4.4.1 Second-Order Equations

EXAMPLE 4.4.1: Solve the nonhomogeneous equations (a) y

+ 6y 2ex and (b) y

5y 6y 3e 2x .

SOLUTION: (a) The corresponding homogeneous equation y

6y 0 has general solution yh c1 e 2x c2 e 3x .

In[566]:= homsol

Out[566]=

DSolve y x

5y x

0, y x , x

3x

2x

C 1

C 2

y x

5y

5y

6y x

Next, we determine the form of y p x . We choose S

ex because f x

x

x

2e . Notice that e is not a solution to the homogeneous equation, so we

take y p x to be the linear combination of the functions in S. Therefore,

Aex .

yp x

In[567]:= yp x

Out[567]= a

a Exp x

x

Substituting this solution into y

Aex

5Aex

In[568]:= eqn

yp

Out[568]= 12 a

x

6Aex

x

2

5y

2ex , we have

6y

12Aex

5yp x

2ex .

6yp x

x

Equating the coefﬁcients of ex then gives us A

1/ 6.

2 Exp x

224

Chapter 4 Higher-Order Differential Equations

In[569]:= aval

Out[569]=

SolveAlways eqn, x

1

, x

6

a

1 x

6e ,

Hence, a particular solution is y p x

In[570]:= yp x

1

6

a Exp x /. a

x

Out[570]=

6

and a general solution of the nonhomogeneous equation is

y

yp

yh

In[571]:= gensol

e

c1 e

3x

2x

C 1

c2 e

e

2x

1 x

e.

6

3x

C 2

yp x

x

Out[571]=

3x

6

C 1

2x

C 2

In this case, we ﬁnd the same general solution with DSolve.

In[572]:= gensol

5y x

6y x

2 Exp x ,

DSolve y x

y x , x, GeneratedParameters c

x

Out[572]=

y x

3x

6

c 1

2x

c 2

We then graph the general solution for various values of the arbitrary

constants in the same way as in other examples. See Figure 4-19.

In[573]:= toplot

grays

Table gensol 1, 1, 2 ,

c 1 , 1, 1 , c 2 , 1, 1

Table GrayLevel i ,

i, 0, 0.7, 0.7/ 8

Plot Evaluate toplot , x, 3, 5 ,

PlotStyle grays, PlotRange > 3, 5 ,

AspectRatio > 1

(b) In this case, we see that f x

3e 2x so the associated set is S

e 2x .

2x

However, because y e is a solution to the corresponding homogeneous equation, we must multiply each element of this set by xr so that

no element is a solution of the corresponding homogeneous equation.

We multiply the element of S by x to obtain xS

xe 2x because xe 2x is

5y 6y 0. Hence, y p x

Axe 2x . Differentiating

not a solution of y

y p x twice

4.4 Nonhomogeneous Equations with Constant Coefﬁcients

225

5

4

3

2

1

-2

4

2

-1

-2

-3

Figure 4-19

Various solutions of y

In[574]:= yp x

ax Exp

5y

2ex

6y

2x

yp x

x

yp

2x

Out[574]= a

Out[574]=

2x

2a

2x

4a

4a

x

2x

x

and substituting into the nonhomogeneous equation yields

y

5y

6y

Ae

2x

In[575]:= eqn

Out[575]=

so A

3 and y p x

4a

3

3xe

In[576]:= aval

Out[576]=

2x

4Ae

a

4Axe

3e

yp

5 Ae

2x

10 a

5yp x

2x

x

6yp x

5 a

2x

2x

.

SolveAlways eqn, x

3

2Axe

2x

6Axe

2x

x

2x

2x

2x

3 Exp

2a

2x

2x

x

2x

226

Chapter 4 Higher-Order Differential Equations

In[577]:= yp x

Out[577]= 3

2x

A general solution of y

y

yh

yp x /.aval 1

x

5y

6y

yp

c1 e

2x

3e

2x

is

c2 e

3x

3xe

2x

.

As in (a), we can use DSolve to obtain equivalent results. For example,

entering

In[578]:= gensol

5y x

6y x

3 Exp

DSolve y x

y 0

a, y 0

b ,y x ,x

Out[578]=

2x ,

y x

3x

3

2a

b

3

x

3a

x

b

x

3

x

x

solves the equation subject to the initial conditions y 0

a and y 0

and names the resulting output gensol. Thus entering

3

2

1

-2

-1

1

2

3

-1

-2

Figure 4-20 Various solutions of y

5y

6y

3e

2x

b

4.4 Nonhomogeneous Equations with Constant Coefﬁcients

In[579]:= toplot

grays

227

Table gensol 1, 1, 2 , a, 1, 1 ,

b, 1, 1

Table GrayLevel i ,

i , 0 , 0. 7 , 0 . 7 / 8

Plot Evaluate toplot , x, 2, 3 ,

PlotStyle grays, PlotRange > 2, 3 ,

AspectRatio > 1

deﬁnes toplot to be the set consisting of nine functions correspond5y 6y 3e 2x that satisfy the initial coning to the solutions of y

b for a

1, 0, and 1 and b

1, 0, and 1;

ditions y 0

a and y 0

and then graphs the set of functions toplot on the interval 2, 3 in

Figure 4-20.

EXAMPLE 4.4.2: Solve

4

d2y

dt 2

y

t

2

5 cos t

e

t/ 2

.

SOLUTION: The corresponding homogeneous equation is 4y

with general solution yh c1 e t/ 2 c2 et/ 2 .

In[580]:= DSolve 4y

Out[580]=

y t

e

t

t/ 2

0

0, y t , t

y t

C 1

y

e

t/ 2

C 2

A fundamental set of solutions for the corresponding homogeneous

equation is S

e t/ 2 , et/ 2 . The associated set of functions for t 2 is

1, t , the associated set of functions for 5 cos t is F2

cos t, sin t ,

F1

t/ 2

t/ 2

is F3

e

. Note that

and the associated set of functions for e

e t/ 2 is an element of S so we multiply F3 by t resulting in tF3

te t/ 2 .

Then, we search for a particular solution of the form

yp

A

Bt

C cos t

D sin t

Ete

t/ 2

,

where A, B, C, D, and E are constants to be determined.

In[581]:= yp t

Out[581]= a

b t

a b t c Cos t

d Sin t

e t Exp t/2

e e t/2 t c Cos t

d Sin t

No element of F1 is

contained in S and no

element of F2 is contained

in S.

228

Chapter 4 Higher-Order Differential Equations

Computing y p and y p

In[582]:= dyp

yp t

d2yp

Out[582]= b

Out[582]=

yp

e e

1

e e

4

t/ 2

e e

t

1

e e

2

t/ 2

t/ 2

t/ 2

t

t

d Cos t

c Cos t

c Sin t

d Sin t

and substituting into the nonhomogeneous equation results in

A

Bt

5C cos t

5D sin t

4Ee

t/ 2

t

2

5 cos t

e

t/ 2

.

In[583]:= eqn

Out[583]=

4 yp t

yp t

t 2 5 Cos t

Exp t/2

a b t e e t/2 t c Cos t

d Sin t

1

e e t/2 t c Cos t

4

e e t/ 2

4

d Sin t

2 e t/2 t 5 Cos t

Equating coefﬁcients results in

A

so A

2

2, B

B

1, C

1

5C

1, D

0, and E

In[584]:= cvals

Solve

4e

Out[584]=

a

y p is then given by y p

5D

4E

1

1/ 4.

2, b

a

0

1, 5c

5, 5d

0,

1

2, b

2

5

t

1, c

cos t

1, d

0, e

1

4

1

t/ 2

4 te

In[585]:= yp t /.cvals 1

1

e t/2 t Cos t

Out[585]= 2 t

4

and a general solution is given by

y

yh

yp

c1 e

t/ 2

c2 et/ 2

2

t

cos t

1

te

4

t/ 2

.

Note that A Bt 5C cos t 5D sin t 4Ee t/ 2 t 2 5 cos t e t/ 2 is

true for all values of t. Evaluating for ﬁve different values of t gives us

ﬁve equations that we then solve for A, B, C, D, and E, resulting in the

4.4 Nonhomogeneous Equations with Constant Coefﬁcients

In[586]:= e1

eqn/.t > 0

Out[586]=

c

a

In[587]:= e2

Out[587]=

Out[587]=

Out[587]=

4

c

e

eqn/.t > Π/2

eqn/.t > Π

e4

eqn/.t > 1

e5

eqn/.t > 2

b Π 1

e e

d

2

2

1

8

a

1

4

a

Π/ 4

e e

Π

b Π

c

Π/ 2

e e

e

e

b

d Sin 1

a

2 b

d Sin 2

Π/ 4

2

e e

Π

e

Π/ 2

Π

3

e

Π

4

Π/ 4

4

Π/ 2

d

Π

2

c e e

e e

Π/ 4

Π/ 2

Π

c Cos 1

3 e

4 e

4

d Sin 1

Out[587]=

8

e3

a

229

1

e

1

2 e

e

5 Cos 1

c Cos 2

e

2 e

4

1

e

d Sin 2

c Cos 1

c Cos 2

5 Cos 2

In[588]:= Solve e1, e2, e3, e4, e5 ,

a, b, c, d, e //Simplify

Out[588]=

0, b

d

1, a

2, c

1

4

1, e

Last, we check our calculation with DSolve and simplify.

In[589]:= sol2

y t

DSolve 4y t

Exp t/2 , y t , t

Out[589]=

t

2

5 Cos t

y t

e

t/ 2

C 1

2 Cos t

et/ 2 C 2

4 Sin t

t 1 t/ 2

e

t

4 2

et/2 Sin t

1

4

e

e

t/ 2

t/ 2

1 t/ 2

e

Cos t

2

2 t

2 et / 2

230

Chapter 4 Higher-Order Differential Equations

In[590]:= Simplify sol2

Out[590]=

y t

1

e t/ 2 1

4

4 et C 2

8 et / 2

t

4 et / 2 t

4 C 1

Cos t

In order to solve an initial-value problem, ﬁrst determine a general solution and

then use the initial conditions to solve for the unknown constants in the general

solution.

EXAMPLE 4.4.3: Solve y

4y

cos 2t, y 0

0, y 0

0.

SOLUTION: A general solution of the corresponding homogeneous

c1 cos 2t c2 sin 2t. For this equation, F

cos 2t,

equation is yh

sin 2t . Because elements of F are solutions to the corresponding homogeneous equation, we multiply each element of F by t resulting in tF

t cos 2t, t sin 2t . Therefore, we assume that a particular solution has the

form

y p At cos 2t Bt sin 2t,

where A and B are constants to be determined. Proceeding in the same

manner as before, we compute y p and y p

In[591]:= yp t

yp t

a t Cos 2t

b t Sin 2 t

yp t

Out[591]= a Cos 2 t

2 b t Cos 2 t

b Sin 2 t

2 a t Sin 2 t

Out[591]= 4 b Cos 2 t

4 a t Cos 2 t

4 a Sin 2 t

4 b t Sin 2 t

and then substitute into the nonhomogeneous equation.

In[592]:= eqn

yp

t

4yp t

Cos 2t

Out[592]= 4 b Cos 2 t

4 a t Cos 2 t

4 a Sin 2 t

4 b t Sin 2 t

4 a t Cos 2 t

b t Sin 2 t

Cos 2 t

0 and B

1/ 4. Alternatively,

remember that 4A sin 2t 4B cos 2t

cos 2t is true for all values of t.

4.4 Nonhomogeneous Equations with Constant Coefﬁcients

Evaluating for two values of t and then solving for A and B gives the

same result.

In[593]:= e1

eqn/.t > 0

eqn/.t > Π/4

e2

e1, e2

cvals Solve

Out[593]= 4 b

1

Out[593]=

4 a

Out[593]=

a

It follows that y p

1

4t

0

0, b

1

4

sin 2t and y

c1 cos 2t

In[594]:= yp t /.cvals

1

t Sin 2 t

Out[594]=

4

In[595]:= y t

c2 sin 2t

1

4t

sin 2t.

1

c2 Sin 2t 1/4 t Sin 2t

1

t Sin 2 t

c2 Sin 2 t

4

c1 Cos 2t

Out[595]= c1 Cos 2 t

Applying the initial conditions

In[596]:= y t

In[597]:= cvals

Out[597]=

results in y

1

4t

c1

Solve

0, c2

1

Sin 2 t

4

1

t Cos 2 t

2

Out[596]= 2 c2 Cos 2 t

2 c1 Sin 2 t

y 0

0, y 0

0

0

sin 2t, which we graph with Plot in Figure 4-21.

In[598]:= y t /.cvals 1

1

t Sin 2 t

Out[598]=

4

In[599]:= Plot Evaluate y t /.cvals

1

, t, 0, 16Π

We verify the calculation with DSolve.

In[600]:= Clear y

DSolve

4y t

Cos 2t , y 0

y t

y 0 ,

0 y t ,t

1

y t

t Sin 2 t

Out[600]=

4

0,

231

232

Chapter 4 Higher-Order Differential Equations

10

5

10

20

30

40

50

-5

-10

Figure 4-21

The forcing function causes the solution to become unbounded as t

Initial-value problems and boundary-value problems can exhibit dramatically different behavior.

EXAMPLE 4.4.4: Show that the boundary-value problem

4y

4y

y0

37y

cos 3x

has inﬁnitely many solutions.

SOLUTION: First, we ﬁnd a general solution of the corresponding

homogeneous equation.

In[601]:= Clear x, y

homsol

4y x

37y x

0,

DSolve 4y x

y x ,x

x/ 2

x/ 2

Out[601]=

y x

C 2 Cos 3 x

C 1 Sin 3 x

Using the Method of Undetermined Coefﬁcients, we ﬁnd a particular

A cos 3x

solution to the nonhomogeneous equation of the form y p

B sin 3x. Substitution into the nonhomogeneous equation yields.

In[602]:= yp x

capa Cos 3x

step1

4yp x

4yp x

Simplify

capb Sin 3x

37yp x

Cos 3x //

4.4 Nonhomogeneous Equations with Constant Coefﬁcients

Out[602]=

1

capa 12 capb Cos 3 x

12 capa capb Sin 3 x

233

0

This equation is true for all values of x. In particular, substituting x

and x Π/ 6 yields two equations

In[603]:= eq1

Out[603]=

1

step1 /. x > 0

capa

In[604]:= eq2

Out[604]=

0

12 capb

0

step1 /. x > Π/6

12 capa

capb

0

that we then solve for A and B

In[605]:= vals

Out[605]=

to see that A

Solve eq1, eq2

12

1

, capb

capa

145

145

1/ 145 and B

In[606]:= yp x

12/ 145.

yp x /. vals

In[607]:= y x

x/ 2

e

yp

Out[607]=

1

C 2 Cos 3x

x

1

Cos 3 x

145

12

Sin 3 x

145

x/ 2

e

x/ 2

C 1 Sin 3 x

2

145 1 e

Π/ 2

1

145

C 2

In[609]:= y Π

1

Out[609]=

145

so c2

C 1 Sin 3x

C 2 Cos 3 x

Applying the boundary conditions indicates that

In[608]:= y 0

1

Out[608]=

145

x/ 2

Π/ 2

C 2

; c1 is arbitrary.

Solve y 0

y Π

2 Π/ 2

C 2

Π/ 2

145 1

In[610]:= cval

Out[610]=

In[611]:= N cval

Out[611]=

C 2

0.0114193

In[612]:= y x

y x /. cval 1

Π x

2 2 2 Cos 3 x

1

Cos 3 x

Out[612]=

Π/ 2

145

145 1

12

x/ 2

Sin 3 x

C 1 Sin 3 x

145

c2

1

145

e

Π/ 2

c2

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