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# MATH 204 PRACTICE MIDTERM II

Please work out each of the given problems on your own paper.  Credit will be based on the steps that you show towards the final answer.  Show your work.

Problem 1

Solve the following differential equations.

A.  y'' - 2y' + 2y  =  0,  y(0)  =  0,  y'(0)  =  5

Solution

The characteristic equation is

r2 - 2r + 2  =  0

r  =  1 + i       or     r  =  1 - i

The general solution is

y  =  et (c1 cos t + c2 sin t)

The first initial condition gives

0  =  (1)(c1)        c1  =  0

So that

y  =  et (c2 sin t)

and

y'  =  c2 (et sin t + et cos t)

Now use the second initial condition

5  =  c2 (0 + 1)     c2  =  5

The final solution is

y  =  5 et sin t

B.  y''' - 3y'' + 3y' - y  =  0

Solution

The characteristic equation is

r3 - 3r2 + 3r - 1

We can factor this either by a calculator or by noting that r  =  1 is a root and dividing by r - 1.  Either way we get

(r - 1)3 =  0

which has a root of order 3 at x  =  1.

The solution is therefore

y  =  c1et + c2tet + c3t2et

C.  y(iv) - y  =  0

The characteristic equation is

r4 - 1  =  0

This factors as

(r - 1)(r + 1)(r2 + 1)  =  0

So the roots are

r  =  1,      r  =  -1,      r  =  i,      r  =  -i

The general solution is thus

y  =  c1et + c2e-t + (c1cos t  +  c2 sin t)

Problem 2  (15 Points each)
Consider the differential equation

y''' - y'  =  e2t + e3t

A.     Solve this differential equation using the method of UC functions

Solution

First find the homogeneous solution

y''' - y'  =  0

The characteristic equation is given by

r3 - r  =  0

This factors as

r(r + 1)(r - 1)  =  0

r  =  0,     r  =  1,      r  =  -1

The homogeneous solution is

yh  =  c1 + c2et + c3e-t

Next find the particular solution using UC functions.  We have

yp  =  Ae2t + Be3t Notice that these do not coincide with yh

yp'  =  2Ae2t + 3Be3t

yp''  =  4Ae2t + 9Be3t

yp'''  =  8Ae2t + 27Be3t

Plugging into the original differential equation gives

(8Ae2t + 27Be3t ) - (2Ae2t + 3Be3t)  =  e2t + e3t

(8A - 2A)e2t + (27B - 3B)e3t ) =  e2t + e3t

6A  =  1    and     24A  =  1

A  =  1/6    and    B  =  1/24

The final solution is

y  =  yh + yp  =  c1 + c2et + c3e-t + 1/6 e2t + 1/24 e3t

B.     Solve this differential equation using the method of variation of parameters.

We have already solved the homogeneous system

yh  =  c1 + c2et + c3e-t

Thus the final solution is

y  =  v1 + v2et + v3e-t

Now compute the Wronskian

The third column of the adjoint is

Divide by the Wronskian to arrive at the third column of the inverse:

We get the three differential equations

v1'  =  (-1)(e2t + e3t)    v2'  =  1/2 e-t (e2t + e3t)    v3'  =  1/2 et (e2t + e3t)

v1'  =  -e2t - e3t    v2'  =  1/2 (et + e2t)    v3'  =  1/2 (e3t + e4t)

Now integrate to get

v1  =  -1/2 e2t - 1/3 e3t + C1

v2  =  1/2 (et + 1/4 e2t)  + C2

v3  =  1/2 (1/3e3t + 1/4e4t) + C3

We have

y  =  (-1/2 e2t - 1/3 e3t + c1) + (1/2 (et + 1/4 e2t)  + c2)et +(1/2 (1/3e3t + 1/4e4t) + c3)e-t

=  c1 + c2et + c3e-t + (-1/2 + 1/2 + 1/6)e2t + (-1/3 + 1/4 + 1/8)e4t

=  c1 + c2et + c3e-t + 1/6 e2t + 1/24 e4t

Problem 3  (15 Points)

A 2 kg mass stretches a spring 1.96 meters.  The mass is attached to a viscous damper that exerts a force of 4N when the velocity is 0.1 m/sec.  The mass is then pulled down 0.5m and released.

A.     Determine the equation of motion for this system.

Solution

We use the formula

mu'' + gu' + ku  =  F

We have

m  =  2    and    F  =  0

Use hooks law to get

(2)(9.8)  =  k(1.96)

Which produces

k  =  10

To find g, we use

4  =  g(.1)

g  =  40

This gives us the differential equation

2u'' + 40u' + 10u  =  0    u(0)  =  0.5

or

u'' + 20u' + 5u  =  0    u(0)  =  0.5

The characteristic equation for this differential equation is

r2 + 20r + 5  =  0

and has solution approximately equal to -19.7 and -0.25.  The general solution is

y  =  c1e-19.7t + c2e-0.25t

Now use the initial conditions to find the constants.  We have

0.5  =  c1 + c2

Taking the derivative and plugging in 0, we get

0  =  -19.7c1 - 0.25c2

These two equations give

c1  =  -0.01            c2  =  0.51

The final solution is

y  =  -0.01e-18.66t + 0.51e-1.34t

B.  Describe (qualitatively) the difference between replacing the viscous damper with an external force of F  =  3sin( t) and replacing the damper with an external force of F  =  2cos(3t).

If there is no viscous damper and the external force is

F  =  3sin( t)

Then the differential equation becomes

2u''  + 10u  =  3sin( t)        u(0)  =  0.5

The homogenous equation is

2u'' + 10u  =  0

or

u'' + 5u  =  0

Which has solution

yh  =  c1cos( t) + c2 sin( t)

Since the external force is part of the homogeneous solution, hence the motion exhibits resonance and the motion is unbounded.  In reality, the spring will eventually break.

For the second force, the equation is

2u''  + 10u  =  2cos(3t)        u(0)  =  0.5

Since the force is not part of the homogeneous solution, there is no resonance.  Instead the motion will have a beat.

Problem 4  (15 Points each)

Please answer the following true or false.  If false, explain why or provide a counter-example.  If true, explain why.

A.     The differential equation

(t - 1)y'' + cos t y' + (t - 1)y  =  et,    y(0)  =  3,     y'(0)  =  4

has a unique solution defined for all real numbers.

Solution

False,  after dividing by t - 1, p(t)  is not continuous at t  =  1

B.     Let y1  =  t + 1y2  =  4t,  be solutions of the differential equation

y''' + p(t)y'' + q(t)y' + r(t)y  =  s(t)

with p, q, r, and s all continuous.  Then

y3  =  sin t

cannot also be a solution of this differential equation.

Solution

We compute the Wronskian

since -sin t  =  0 at t  =  0, these three function cannot be a solution set for the differential equation.  So True.

Two Important Formulas:

1.    mu'' + gu' + ku  =  F

2.   LQ'' + RQ' + 1/C Q  =  E'(t)