FINALEXAMPRACTICEPROBLEMS

Practice Problems FALL 2013 Instructors of M34

FINAL EXAM PRACTICE PROBLEMS

1. Solve the first order linear differential equation.

ty + y = e2t

Solution.

Rewrite

the equation

as

y

+

1 t

y

=

e2t t

,

and

so

the

integrating

factor

is e

1 t

dt

=

t.

Now,

t

y

+

1 t

y

=

t?

e2t t

=

(ty)

=

e2t

=

ty

=

e2t dt

=

1 2

e2t

+ C

y

=

e2t + 2t

C

(C = 2C).

2. Use the method of variation of parameters to find the general solution of the differential equation: y - 5y + 6y = et

Solution. The two solutions of the homogeneous equation are y1 = e2t and y2 = e3t and the

Wronskian is

e2t

det (e2t)

e2t = e5t. (e3t )

Thus, the specific solution is Y (t) = -e2t

e3t ? et e5t

dt

+

e3t

e2t ? et e5t

dt

=

1 et. 2

Therefore, the general solution to the equation is

y

=

C1e2t

+

C2e3t

+

1 2

et.

3. Use the Laplace transform to solve the initial value problem:

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

Solution. Take the Laplace transform to get:

L

y

- L {y} = L {1}

=

s2L {y}

-

s

y(0)

-

y(0)

-

L

{y}

=

1 s

=

(s2

-

1)Y

(s)

=

1 s

+

1

=

s

+ s

1

=

Y

(s)

=

s+1 s(s2 - 1)

=

1 s(s -

1)

=

s

1 -

1

-

1 s

=

L-1 {Y (s)} = L-1

1 s-1

- L-1

1 s

= y = et - 1.

1

4. Use a convolution integral to find the inverse Laplace transform:

L-1

2 (s - 1)(s + 2)

Solution. Consider

(s

-

1 1)(s

+

2)

=

s

1 -

1

?

s

1 +

2

=

L

et

L

e-2t

= L-1

2 (s - 1)(s2 + 4)

t

= et e-2t = et- e-2 d

0

= et

-

1 3

e-3

t 0

=

1 3

-e2t + et

.

5. Find the inverse Laplace transform of L-1 Solution. Compute

2e-8s s2 + 4

.

2e-8s s2 + 4

=

e-8s

?

2 s2 + 4

=

e-8sL {sin 2t}

=

L-1

2e-8s s2 + 4

= u8(t) sin(2(t - 8)).

6. Solve the given initial value problem:

x

=

-3

2 x,

4 -1

2

x(0) = 1

-3 2

Solution. Let A =

. To find the eigenvalues,

4 -1

-3 - det

2 = 2 + 4 - 5 = ( + 5)( - 1) = 0

4 -1 -

1 = -5, 2 = 1.

To find the eigenvector of 1 = -5, consider

[A

-

(-5)I

]

1

=

0

=

2

2 1

=

0

=

1

=

1

.

2

0

4 4 2

0

2

-1

To find the eigenvector of 2 = 1, consider

[A

-

(1)I

]

1

=

0

=

-4

2

1

=

0

=

1

=

1

.

2

0

4 -2 2

0

2

2

Thus, the general solution is

x(t)

=

c1

1

e-5t

+

c2

1

et.

-1

2

2

In order to satisfy the initial condition,

1

1

2

x(0) = c1 + c2 =

=

c1 + c2

=2

-1

2

1

-c1 + 2c2 = 1

= c1 = c2 = 1.

7. Use the Laplace transform to solve the given initial value problem:

y + y = 1, 2,

0t

0

Solution. These two are nonhomogeneous equations. The structural theorem tells the struc-

ture of solutions are y = yc + Y (t) where yc are solutions to the homogeneous ODE and Y (t)

solutions to the nonhomogeneous ODE respectively. The parameters in the general solutions

are determined by initial conditions. Let's here write down the solution to (b) explicitly: Find

the general solution of the differential equation The general solution of the corresponding

homogeneous equation y + 2y + y = 0 is

yc(t) = C1e-t + C2te-t

where

C1, C2

are

arbitrary

constants.

Setting

y1(t) = e-t,

y2(t) = te-t,

g(t) =

e-t 9t

and

using

the

method

of

variation

of

parameters,

we

find

that

a

particular

solution

of

y

+ 2y

+y

=

e-t 9t

is given by

Y (t) = u1(t)y1(t) + u2(t)y2(t)

or

Y = u1y1 + u2y2

where

u1 = -

W

y2 ? g (y1, y2)

dt,

u2 =

y1 ? g W (y1, y2)

dt,

and W (y1, y2)(t)

=

e-t det

-e-t

te-t

e-t - te-t

=

e-2t. Thus, we find that

u1

=

-

1 9

te-t

?

e-t t

e-2t

dt

=

-

1 9

dt

=

-

1 9

t,

u2

=

1 9

e-t

?

e-t t

e-2t

dt

=

1 9

1 t

dt

=

1 9

ln(t),

(t > 0)

and the particular solution is

Y (t)

=

1 9

(-te-t

+

ln(t)te-t).

4

In fact, as a particular solution of

y

+

2y

+y

=

e-t 9t

we

can

take

1 9

ln(t)te-9t

because the

term

-

1 9

te-t

is

included

in

the

solution

space

of

the

corresponding

homogeneous

equation.

It

follows that the general solution of the differential equation

y

+

2y

+

y

=

e-t 9t

(t > 0) is

y(t)

=

C1e-t

+

C2te-t

+

1 9

ln(t)te-t

where C1, C2 are arbitrary constants.

11. Find the general solution to the equation

y - 3y + 3y - y = 0.

See solution of Problem 13. 12. Use the Laplace transform to solve the initial value problem

y - 2y + 2y = 0, y(0) = 0, y(0) = 1

13. Solve the following differential equation y - 4y + y - 4y = 0; y(0) = 1, y(0) = 2, y(0) = 0.

The characteristic equation of the homogeneous r3 - 4r2 + r - 4 = (r2 + 1)(r - 1) = 0. So the roots are 4, i, -i and the general solution is given by the form y(t) = c1e4t + c2cost + c3sint. The constants c1, c2, c3 are determined by the initial conditions.

14. Write the vibration into the oscillation form u(t) = R(t) cos(t - ), i.e., determine the

amplitude R(t), the frequency and the phase shift angle , for the motion u(t)+2u(t)+5u =

0 with initial conditions u(0) = 3 and u(0) = 5. Note: we require the phase shift angle

(-

2

,

2

)

here.

Solution. This problem is a variant of Example 3 in Page 201 of the textbook. We changed the

data to make the computations easy. The general solution to the ODE is u(t) = e-t(c1 cos 2t+

c2 sin 2t). One can determine c1 = 3, c2 = 4 by initial conditions. Hence u(t) = 5e-t cos(2t-)

where

tan

=

4/3

hence

=

arctan

4 3

(

4

,

2

).

15. Find the general solution of the given system of equations

324

x

=

2

0

2 x

423

5

324

Denote

by

A

=

2

0

2

and

consider

its

characteristic

polynomial

423

3- 2 4 PA() = det(A - I3) = 2 - 2

4 2 3-

= (3 - )(-)(3 - ) + 2 ? 2 ? 4 + 4 ? 2 ? 2 - 4 ? 4 ? - 2 ? 2 ? (3 - ) - 2 ? 2 ? (3 - ) = -3 + 62 - + 8 = -( + 1)2( - 8).

2

Hence

the

eigenvalues

are

1

=8

2

= -1,

3

=

-1.

Next

we

observe

that 1

=

1

an

2

-1

-1

eigenvector

for

1

= 8.

Also

one

can

see

that

the

vectors

2

=

2

and

3

=

0

0

1

are two linearly independent eigenvectors for the eigenvalue 2 = 3 = -1. Therefore the

general solution for the system is

2

-1

-1

x(t)

=

c1

1

e8t

+

c2

2

e-t

+

c3

0

e-t.

2

0

1

16. Solve the following initial value problem. Describe the behavior of the solution as t .

x

=

-2

1 x;

-5 4

1

x(0) = . 3

-2 1

Denote by A =

and consider its characteristic polynomial

-5 4

-2 - 1

PA() = det(A - I2) =

= (-2 - )(4 - ) + 5

-5 4 -

= 2 - 2 - 3 = ( - 3)( + 1).

Hence the eigenvalues are 1 = 3 and 2 = -1. Next let's find an eigenvector for 1 = 3; So

assume

that

1

=

11 21

is

such

an

eigenvector,

i.e.,

(A - 1I2)1

=

0.

Plugging

in

the

date

6

above this means that

-5

1 11 = 0 .

-5 1

21

0

This

implies that

-511

+

21

=

0,

and

solving

we

obtain

that

every

eigenvector

of

1

is

of

the

x

1

form 1 = for every real number x. So we ca pick 1 = . Similarly we can

5x

5

x show that every eigenvector 2 of 2 = -1 is of the form 1 = for every real number

x

1 x, so we can pick 2 = . In conclusion the general solution is of the form

1

x(t) = c1 1 e3t + c2 1 e-t

(1)

5

1

Using the initial data we obtain the system

c1 + c2 = 1

5c1 + c2 = 3

and solving we get c1 = c2 = 1/2. Hence the solution is

x(t)

=

1 2

1 5

e3t +

1 2

1 1

e-t.

17. Express the solution of the given initial value problem in terms of a convolution integral. 4y + 4y + 17y = g(t); y(0) = 0, y(0) = 0.

18. Solve the equation

dy dx

=

y(5

-

y)

Also find the equilibrium solutions, classify each one as asymptotically stable or unstable, and

sketch the equilibrium solutions and several other solutions in the xy-plane.

7

To

solve

the

equation,

we

integrate

both

sides

of

dy y(5-y)

=

dx

and

we

obtain:

dy y(5 -

y)

dy

=

dx

1

1

(

5

y

+

5

5

-

y )dy

=

x + C1

1 5

dy y

+

1 5

dy 5-y

=

x + C1

1 5

ln

|y|

-

1 5

ln

|5

-

y|

=

x + C1

ln |y| - ln |5 - y| = 5x + 5C1

ln

|

5

y -

y

|

=

5x + C2

(C2 = 5C1)

eln

|

y 5-y

|

=

e5x+C2

|

5

y -

y

|

=

e5xeC2

(eC2 > 0)

y 5-y

=

C e5x

(C )

y = Ce5x(5 - y)

y = 5Ce5x - Cye5x

y + Cye5x = 5Ce5x

y(1 + Ce5x) = 5Ce5x

y

=

1

5C e5x + Ce5x

.

Thus, the solutions are given by

y(x)

=

c

5e5x + e5x

.

There are two equilibrium solutions to this equation

(c

=

1 C

)

y=0 y=5

unstable, stable.

We omit sketching the equilibrium solutions and several other solutions on the next page.

19. a) Solve the following differential equation

(y cos(x) + 2xey)dx + (sin(x) + x2ey - 1)dy = 0

b) Find the general solution of the following differential equation y - 3y - 4y = 2e-t + 1.

8

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