MATH 204 PRACTICE EXAM I

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.     (25 points)                 x + y
y'  =
x + 4y

Solution

This is a homogeneous differentiable equation.  Let

v  =  y/x        y  =  xv        y'  =  v + xv'

x + xv                  1 + v
v + xv'  =                      =
x + 4xv
1 + 4v

1 + v - v(1 + 4v)
xv'  =
subtracting v and placing under a common denominator
1 + 4v

1 - 4v2
=
1 + 4v

1 + 4v
dv    =  dx/x
Since v'  = dv/dx and then separate and factor.
(1 - 2v)(1 + 2v)

Now use partial fractions

1 + 4v                             A                    B
=
(1 - 2v)(1 + 2v)                1 - 2v              1 + 2v

A(1 + 2v)  + B(1 - 2v)  =  1 + 4v

v  =  1/2:  2A  =  3        A  =  3/2

v  =  -1/2:  2B  =  -1     B  =  -1/2

This yeilds

3/2                -1/2
+                     dv    =  dx/x
1 - 2v            1 + 2v

Now integrate both sides to get

-3/4 ln|1 - 2v| - 1/4 ln|1 + 2v|  =  ln|x| + C1

ln|(1 - 2v)3(1 + 2v)|  =  ln|x-4| + C2

|(1 - 2v)3(1 + 2v)|  =  C x -4

|(1 - 2y/x)3(1 + 2y/x)|  =  C x -4

|(x - 2y)3(x + 2y)|  =  C

B.     (25 points)    (ey + e -x)dx + (ey + 2ye-x)dy  =  0

Solution

We find an integrating factor

My  =  ey           Nx  =  -2y e -x

My - Nx             ey + 2y e -x
=                             =  1
N
ey + 2y e -x

We find the integrating factor by solving

dm/dx =  (1)m        m  =  ex

Now multiply the original differential equation by the integrating factor

(ex+y + 1)dx + (ex+y + 2y)dy  =  0

This transformed differential equation is exact.  We can find the potential function.

fx  =  ex+y + 1

Integrate with respect to x to get

f(x,y)  =  ex+y + x + C(y)

Now take the partial derivative with respect to y to get

fyex+y + C'(y)  =  ex+y + 2y

C'(y)  =  2y        C(y)  =  y2

Finally the solution is

ex+y + x + y2 = C

C.     (25 points)                             y
y'  =  A -

B - Ct

.
A(B - Ct)

Solution:                    y  =                         + k(B - Ct)1/C

1 - C

Solution

We can rewrite as

y
y'  +                 =  A

B - Ct

This is a first order linear differential equation.  We find the integrating factor

Multiplying by the integrating factor gives

(B - Ct)-1/Cy' + (B - Ct)-1/C - 1y  =  A(B - Ct)-1/C

((B - Ct)-1/Cy)'  =  A(B - Ct)-1/C

Now integrate both sides

.
-A(B - Ct)-1/C + 1                      AC(B - Ct)-1/C + 1
(B - Ct)-1/Cy  =                                    + k  =                                      + k

C(-1/C + 1)                                 1 - C

Now solve for y

A(B - Ct)

y  =                        + k(B - Ct)1/C

1 - C

Problem 2

Currently, Lower Angora Lake has 300 million gallons of unpolluted water in it.  Due to a gas leak in a motor boat in Upper Angora Lake, there is an inflow of 5000 gallons per hour that contains 2 grams per gallon of MTBE.  Since it is spring, the outflow of water from Lower Angora Lake is only 4000 gallons per hour.  Assume that the MTBE is completely mixed in the water as soon as it enters the lower lake.

A.     Set up a differential equation that describes the rate of change of MTBE in Lower Angora Lake.

Solution

This is

Rate  =  Rate In - Rate Out

question.  First find the Rate In.

Rate In  =  (5000gal/hr)(2grams/gal)  =  10,000 grams/hr

Now find the Rate Out.  Notice that the number of gallons in the lake is given by

Gallons =  300,000,000 + 1000t

We use this to calculate the Rate Out.

x grams                           4000 gal
Rate Out  =                                               .
300,000,000 + 1000t                 1  hr

Subtracting we get

dx                                   4000 x grams
=  10,000 -
dt                                 300,000,000 + 1000t

B.     Use the result of Problem 1 Part C to determine the amount of MTBE in Lower Angora Lake one day after the leak began.

Solution

We have

A  =  10,000    B  =  300,000,000/4000 =  75,000    C  =  1000/4000  =  0.25

.

10,000(75,000 - 0.25t)
y  =                                               + k(
75,000 - 0.25t)4

1 - 0.25

We find k by noticing that y(0)  =  0

0  =  750,000,000/.75 + 75,0004k  =  109 + 3.164 x 1019 k

k  =  3.16 x 10-11

Now plug in 1 for t to get

9999.93

Problem 3

The Tahoe area has enough resources to support up to 500 black bears, however if the bear population declines below 50, the bears will not be able to find mates and will eventually become extinct.

A.  Set up an autonomous differential equation that models this situation.

Solution

y'  =  -y(1 - y/50)(1 - y/500)

B.  Sketch the direction field for this differential equation and sketch the solution that corresponds to the initial values:  y(0) = 45, y(0) = 100, and y(0) = 600. Assume the proportionality constant is 1.

Problem

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.  3y'' - 2y' -8y = 0, y(0) = 1, y'(0) = -2

Solution

First solve the characteristic equation

3r2 - 2r - 8  =  0

This has solution

B.  y'' - 4y' + 13y = 0, y(0) = 1, y'(0) = 6

Solution

First find the characteristic equation

r2 - 4r + 13  =  0

The quadratic formula or calculator gives

r  =  2 + 3i     or     r  =  2 - 3i

Thus the general solution is

y  =  e2t (c1cos(3t) + c2 sin(3t))

Now use the initial conditions.

e0 (c1cos(0) + c2 sin(0)) = c1 = 1

Next find the derivative

y'  =  2e2t (cos(3t) + c2 sin(3t)) + e2t (-3sin(3t) + 3c2 cos(3t))

Now plug in 0 to get

2e0 (cos(0) + c2 sin(0)) + e0(-3sin(0) + 3c2 cos(0))  =  3c2  =  6

So that

c2  =  3

Thus the solution is

y  =  e2t (cos(3t) + 3 sin(3t))

C.  y'' + 12y' + 36y = 0  y(0) = 0, y'(0) = 1

Solution

The characteristic equation is

r2 + 12r + 36  =  0

which has repeated root solution

r  =  -6

Thus the general solution is

y  =  c1e-6t + c2te-6t

The first initial condition gives

0  =  c1

The derivative is

y' = c2 e-6t(1 - 6t)

Plugging in 0 gives

c2  =  1

Thus the solution to the initial value problem is

y  =  te-6t

Problem 5

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

A.      If a differential equation is autonomous then it is separable.

True,  if

dy/dx  =  f(y)

then

dy/f(y)  =  dx

separates the variable.

B.      The initial value problem y'  -  sin(t2) + et y  =  0,   y(1)  =  4  is guaranteed to have a unique solution defined for all values of t.

Solution

True,  this is a first order linear differential equation with continuous p(t) and q(t).  Therefore there is a unique solution for all values of t.

C.      The difference equation yn+1  =  kyn has solution yn = y0ekn

Solution

False, the solution is

yn  =  y0kn

D.            d3y                       dy
x            + ln(1 - x2          =  x cos x
dx3                       dx

is a third order linear differential equation.

True,  the highest derivative is y''', and the coefficients in front of y''' and y' are both functions of x.

E.   The function      y = t sin t

cannot be a solution of the differential equation.

y'' + p(t)y' + q(t)y  =  0

with p,and q continuous

Solution

True

First note that

y(0)  =  0

and that

y'  =  sin t  + t cos t

so that

y'(0) = 0

If

y  =  t sin t

is a solution and y2 is another solution such that the two solutions are linearly independent, then the Wronskian matrix at t = 0 would have its first column 0, hence the determinant would be zero.  But is the Wronskian is zero for one value of t, then it is 0 for all values of t.  But linearly independent solutions do not have a 0 Wronskian.  Thus y = t sin t cannot be a solution.

Extra Credit:  Write down one thing that your instructor can do to make the class better and one thing that you want to remain the same in the class.

(Any constructive remark will be worth full credit.)