Homework 4 Solution 1 Sol.

Homework 4 Solution

In Problems 1-12, use the method of `Undetermined Coefficients' to find the general solutions.

1. y + 3y + 2y = 6.

Sol. The characteristic equation m2 + 3m + 2 = (m + 1)(m + 2) = 0 has roots m = -1 and m = -2.

The complementary solution is

yc = C1e-x + C2e-2x.

From the constant function g(x) = 6 we assume a constant function yp = A is a particular solution of the equation. Substituting into the given equation yields

A = 3.

Thus a particular solution is yp = 3, and so the general solution is

y = yc + yp = C1e-x + C2e-2x + 3.

2. y - 10y + 25y = 150x + 15

Sol. The characteristic equation m2 - 5m + 25 = (m - 5)2 = 0 has a root m = 5 with multiplicity

2. The complementary solution is

yc = C1e5x + C2xe5x.

From the linear function g(x) = 150x + 15 we assume a linear function yp = Ax + B is a particular solution of the equation. Substituting into the given equation yields

A = 6, B = 3.

Thus a particular solution is yp = 6x + 3, and so the general solution is

y = yc + yp = C1e5x + C2e5x + 6x + 3.

3.

1 4

y

+ y + y = x2 - 2x Sol. The equation is equivalent to y

+ 4y + 4y = 4x2 - 8x.

The

characteristic equation m2 + 4m + 4 = (m + 2)2 = 0 has a root m = -2 with multiplicity 2. The

complementary solution is

yc = C1e-2x + C2xe-2x.

From the quadratic function g(x) = 4x2 - 8x we assume a quadratic function yp = Ax2 + Bx + C is a particular solution of the equation. Substituting into the given equation yields

7 A = 1, B = -4, C = .

2

Thus

a

particular

solution

is

yp

=

x2

-

4x

+

7 2

,

and

so

the

general

solution

is

y

=

yc

+

yp

=

C1e-2x

+

C2e-2x

+

x2

-

4x

+

7 .

2

4. y - 3y + 2y = e3x

Sol. The characteristic equation m2 - 3m + 2 = (m - 1)(m - 2) = 0 has roots m = 1 and m = 2.

The complementary solution is

yc = C1ex + C2e2x.

1

From the exponential function g(x) = e3x we assume an exponential function yp = Ae3x is a particular solution of the equation. Substituting yp = 3Ae3x and yp = 9Ae9x into the given

equation yields 1

A= . 2

Thus

a

particular

solution

is

yp

=

1 2

e3x,

and

so

the

general

solution

is

y

=

yc

+

yp

=

C1ex

+

C2e2x

+

1 e3x. 2

5. y - y - 2y = 10 sin x

Sol. The characteristic equation m2 - m - 2 = (m + 1)(m - 2) = 0 has roots m = -1 and m = 2.

The complementary solution is

yc = C1e-x + C2e2x.

From the trig. function g(x) = 10 sin x we try a trig. function yp = A cos x + B sin x for a particular solution of the equation. Substituting

into the given equation yields

yp = B cos x - A sin x yp = -A cos x - B sin x

-3A - B = 0, A - 3B = 10.

So we have A = 1 and B = -3. Thus a particular solution is yp = cos x - 3 sin x, and so the general

solution is y = yc + yp = C1e-x + C2e2x + cos x - 3 sin x.

6. y - 2y - 3y = 6xe2x Sol. The characteristic equation m2 - 2m - 3 = (m + 1)(m - 3) = 0 has roots m = -1 and m = 3.

The complementary solution is

yc = C1e-x + C2e3x.

From the function g(x) = 6xe2x we try yp = (Ax + B)e2x for a particular solution of the equation. Substituting

yp = e2x(2Ax + 2B + A) yp = e2x(4Ax + 4B + 2A + 2A)

into the given equation yields

A = -2, B = -4/3

Thus a particular solution is yp = (-2x - 4/3)e2x, and so the general solution is

y = yc + yp = C1e-x + C2e3x + (-2x - 4/3)e2x.

7. y + 3y = -48x2e3x

Sol. The characteristic equation m2 + 3 = 0 has complex roots m = ? 3i. The complementary

solution is

yc = C1 cos 3x + C2 sin 3x.

2

From the function g(x) = -48x2e3x we try yp = (Ax2 + Bx + C)e3x for a particular solution of the equation. Substituting

yp = e3x[3(Ax2 + Bx + C) + 2Ax + B] = e3x[3Ax2 + (2A + 3B)x + B + 3C] yp = e3x[3{3Ax2 + (2A + 3B)x + (B + 3C)} + 6Ax + 2A + 3B] = e3x[9Ax2 + (12A + 9B)x + 2A + 6B + 9C]

into the given equation yields

yp + 3yp = e3x[12Ax2 + (12A + 12B)x + 2A + 6B + 12]

= e3x[-48x2 +

0x +

0].

So we have

4 A = -4, B = 4 C = - .

3

Thus

a

particular

solution

is

yp

=

(-4x2

+

4x

-

4 3

)e3x,

and

so

the

general

solution

is

y

=

yc

+

yp

=

C1

cos

3x

+

C2

sin

3x

+

(-4x2

+

4x

-

4 )e3x. 3

8. y - y = -3

Sol. The characteristic equation m2 - m = 0 has two roots m = 0 and m = 1. The complementary

solution is

yc = C1 + C2ex.

Note that yc already contains constant functions y = A. (One can check y = A can not be a particular solution of the equation) Instead we modify our assumption and try yp = Ax for a particular solution of the equation. Substituting yp = A and yp = 0 into the given equation yields

A = 3.

Thus a particular solution is yp = 3x, and so the general solution is

y = yc + yp = C1 + C2ex + 3x.

9. y + 2y - 3y = ex Sol. The characteristic equation m2 + 2m - 3 = (m - 1)(m + 3) = 0 has two roots m = 1 and

m = -3. The complementary solution is

yc = C1ex + C2e-3x.

Note that yc already contains functions y = Aex. Instead we modify our assumption and try yp = Axex for a particular solution of the equation. Substituting

yp = ex(Ax + A) yp = ex(Ax + A + A)

into the given equation yields

yp + 2yp - 3yp = ex{(A + 2A - 3A)x + 2A + 2A}

= ex{

0x +

+1}.

3

So we have

1 A= .

4

Thus

a

particular

solution

is

yp

=

1 4

xex,

and

so

the

general

solution

is

y

=

yc

+

yp

=

C1ex

+

C2e-3x

+

1 xex. 4

10. y - 4y + 4y = e2x Sol. The characteristic equation m2 - 4m + 4 = (m - 2)2 = 0 has a root m = 2 with multiplicity

2. The complementary solution is

yc = C1e2x + C2xe2x.

Note that yc already contains functions y = Ae2x as well as y = Axe2x. Instead we modify our assumption and try yp = Ax2e2x for a particular solution of the equation. Substituting

yp = e2x(2Ax2 + 2Ax) yp = e2x{2(2Ax2 + 2Ax) + 4Ax + 2A} = e2x{4Ax2 + 8Ax + 2A}

into the given equation yields

yp - 4yp + 4yp = e2x{(4A - 8A + 4A)x2 + (8A - 8A)x + 2A}

= e2x{

0x2 +

0x + 1}.

So we have

1 A= .

2

Thus

a

particular

solution

is

yp

=

1 2

x2ex,

and

so

the

general

solution

is

y

=

yc

+

yp

=

C1e2x

+

C2xe2x

+

1 x2ex. 2

11. y

-y

+

1 4

y

=

3

+

e

1 2

x

Sol.

The

characteristic

equation

m2 - m +

1 4

=

(m -

1 2

)2

=

0

has

a

root

m

=

1 2

with

multiplicity

2. The complementary solution is

yc

=

C1

e

1 2

x

+

C2xe

1 2

x

.

In view of Superposition Principle, we seek a particular solution yp = yp1 + yp2 where yp1 and yp2 are particular solutions of

1 y -y + y=3

and

y

-y

+

1 y

=

ex/2

4

4

respectively. As in Problem #1, one can find yp1 = 12.

Note

that

yc

already

contains

functions

y

=

Ae

1 2

x

as

well

as

y

=

Axe

1 2

x

.

Instead

we

modify

our

assumption

and

try

yp

=

Ax2

e

1 2

x

for

a

particular

solution

of

the

equation.

Substituting

yp

=

e

1 2

x

(

1

Ax2

2

+

2Ax)

yp

=

e

1 2

x

{

1

(

1

Ax2

22

+

2Ax)

+

Ax

+

2A}

=

e

1 2

x

{

1

Ax2

4

+

2Ax

+

2A}

4

into the given equation yields

yp

-

yp

+

1 4 yp

=

e

1 2

x{(

1

A

4

-

1 A

2

+

1 A)x2 4

+

(2A

-

2A)x

+

2A}

=

e

1 2

x{

0x2 +

0x + 1}.

So we have

1 A= .

2

Thus

a

particular

solution

is

yp

=

1 2

x2

e

1 2

x

,

and

so

the

general

solution

is

y

=

yc

+

yp

=

C1e

1 2

x

+

C2

xe

1 2

x

+

12

+

1

x2

e

1 2

x

.

2

12. y - 8y + 20y = 100x2 - 2 - 13xex

You can use Superposition Principle as discussed during the class. (or simply yp = Ax2 + Bx + C + (Dx + E)ex also works.)

In Problems 13-16, use the method of `Variation of Parameters' to find the general solutions

13. .y - y - 2y = 2e-x Sol. The characteristic equation m2 - m - 2 = (m + 1)(m - 2) = 0 has two roots m = 2 and m = -1. Let y1 = e2x and y2 = e-x. Wronskian W (y1, y2) is

W (y1, y2) =

e2x 2e2x

e-x -e-x

= -e2xe-x - 2e-xe2x = -3ex.

We

seek

a

particular

solution

yp

=

u1y1

+ u2y2

where

u1

=

-y2 ? g(x) W

and

u2

=

y1 ? g(x) . W

So

u1 = -

e-x ? 2e-x

2

-3ex

dx = 3

e-3xdx = - 2 e-3x, 9

u2 = Therefore a particular solution is

e2x ? 2e-x

2

-3ex

dx = - 3

2 1dx = - x.

3

yp

=

u1y1

+

u2y2

=

- 2 e-3x 9

?

e2x

-

2 x

3

?

e-x

=

- 2 e-x 9

-

2 x

3

?

e-x,

and the the general solution is

y

=

yc

+

yp

=

C1e2x

+

C2e-x

-

2 e-x 9

-

2 x

3

?

e-x

=

C1e2x

+

C2e-x

-

2 x

3

?

e-x.

14. y + 2y + y = 3e-x Sol. The characteristic equation m2 + 2m + 1 = (m + 1)2 = 0 has a root m = -1 with multiplicity 2. Let y1 = e-x and y2 = xe-x. Wronskian W (y1, y2) is

W (y1, y2) =

e-x -e-x

xe-x (1 - x)e-x

= e-2x(1 - x + x) = e-2x.

5

We

seek

a

particular

solution

yp

=

u1y1

+ u2y2

where

u1

=

-y2 ? g(x) W

and

u2

=

y1 ? g(x) . W

So

u1 = -

xe-x ? 3e-x e-2x dx = -3

xdx = - 3 x2, 2

u2 = Therefore a particular solution is

e-x ? 3e-x e-2x dx = 3 1dx = 3x.

yp

=

u1y1

+

u2y2

=

- 3 x2 2

?

e-x

+

3x

?

xe-x

=

3 x2e-x, 2

and the the general solution is

y

=

yc

+

yp

=

C1e-x

+

C2xe-x

+

3 x2e-x 2

15. 4y

- 4y

+

y

=

16e

1 2

x

Sol. The standard form of the equation is y

-y

+

1 4

y

=

4e

1 2

x

and

g(x)

=

4e

1 2

x.

The characteristic

equation

m2

-

m

+

1 4

=

(m

-

1 2

)2

=

0

has

a

root

m

=

1 2

with multiplicity 2.

Let

y1

=

e

1 2

x

and

y2

=

xe

1 2

x.

Wronskian

W (y1, y2)

is

e

1 2

x

W (y1, y2) =

1 2

e

1 2

x

xe

1 2

x

(1

+

1 2

x)e

1 2

x

=

ex(1

+

1 x

-

1 x)

=

ex.

22

We

seek

a

particular

solution

yp

=

u1y1

+ u2y2

where

u1

=

-y2 ? g(x) W

and

u2

=

y1 ? g(x) . W

So

u1 = -

xe

1 2

x

?

4e

1 2

x

ex dx = -4

xdx = -2x2,

u2 = Therefore a particular solution is

e

1 2

x

?

4e

1 2

x

ex dx =

4dx = 4x.

yp

=

u1y1

+

u2y2

=

-2x2

?

e

1 2

x

+

4x

?

xe

1 2

x

=

2x2

e

1 2

x

,

and the the general solution is

y

=

yc

+

yp

=

C1

e

1 2

x

+

C2xe

1 2

x

+

2x2

e

1 2

x

16. y + y = sec x Sol. The characteristic equation m2 + 1 = 0 has complex roots m = ?i. Let y1 = cos x and y2 = sin x. Wronskian W (y1, y2) is

cos x W (y1, y2) = - sin x

sin x = cos2 x + sin2 x = 1. cos x

6

We

seek

a

particular

solution

yp

=

u1y1

+ u2y2

where

u1

=

-y2 ? g(x) W

and

u2

=

y1 ? g(x) . W

So

sin x

u1 = -

sin x ? sec xdx = -

dx = ln | cos x|, cos x

u2 = cos x ? sec xdx = Therefore a particular solution is

1 cos x ? dx =

cos x

1dx = x.

yp = u1y1 + u2y2 = ln | cos x| ? cos x + x ? sin x, and the the general solution is

y = yc + yp = C1 cos x + C2 sin x + ln | cos x| cos x + x sin x

In Problems 17-18, solve the given differential equations subject to the initial condition.

17. y + y = 4x + 1,

y(0)

=

1,

y

(0)

=

4 5

Sol. The characteristic equation m2 + m = m(m + 1) = 0 has roots m = 0 and m = -1. The

complementary solution is

yc = C1e0x + C2e-x = C1 + C2e-x.

Note that yc and y = Ax + B share constant terms. (One can check the equation does not

have a particular solution of the form y = Ax + B). Instead we modify our assumption and try yp = x(Ax + B) = Ax2 + Bx for a particular solution of the equation. Substituting yp = 2Ax + B and yp = 2A into the given equation yields

A = 2, B = -3

Thus a particular solution is yp = x(2x - 3) = 2x2 - 3x, and so the general solution is

y = yc + yp = C1 + C2e-x + x(2x - 3).

Since y

= -C2e-x + 4x - 3, the initial conditions y(0) = 1,

y (0) =

4 5

imply

C1 + C2 = 1 and - C2 - 3 = 4/5.

Now

C1

=

24 5

and

C2

=

-19 5

,

and

the

unique

solution

is

y = 24 - 19 e-x + x(2x - 3). 55

18. y + 4y + 5y = 35e-4x, y(0) = -3, y (0) = 1

Sol. The characteristic equation m2+4m+5 = 0 has complex roots m = -2?i. The complementary

solution is

yc = e-2x(C1 cos x + C2 sin x).

7

Our assumption is that yp = Ae-4x is a particular solution for some A. Substituting yp = -4Ae-4x and yp = 16Ae-4x into the given equation yields

A=7 Thus a particular solution is yp = 7e-4x, and so the general solution is

y = yc + yp = e-2x(C1 cos x + C2 sin x) + 7e-4x. Since y = e-2x{-2(C1 cos x + C2 sin x) - C1 cos x + C2 cos x} - 28e-4x, the initial conditions y(0) = -3, y (0) = 1 imply

C1 + 7 = -3 and - 2C1 + C2 - 28 = 1. Now C1 = -10 and C2 = 9, and the unique solution is

y = e-2x(-10 cos x + 9 sin x) + 7e-4x.

8

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