Name
ID
MATH 308
Exam I
Fall 2000
Section 200
Solutions
P. Yasskin
HAND COMPUTATIONS
Problems 1-5:
Solve the initial value problem using the method appropriate to one of the
following types of first order differential equations:
a. Separable Equation
b. Equation with Homogeneous Coefficients
c. Linear Equation
d. Bernoulli Equation
Hint: k 1
1"s
Exact
Equation
e.
Be sure to identify the equation type. There is one problem for each equation type.
1. (10 points)
dy
4y x
dx
x
Linear Equation.
with
yŸ1 0
Standard form is:
dy
" 4 y x.
dx
x
The integrating factor is
I exp ; " 4 dx expŸ"4 lnx expŸlnx "4 x "4
x
Multiply by the integrating factor and recognize the left as the derivative of a product:
d Ÿx "4 y x "4 dy " 4x "5 y x "3
dx
dx
Integrate and solve:
"2
2
y " x Cx 4
x "4 y ; x "3 dx x C
2
"2
Solve for the constant of integration:
x0 1
y0 0
®
0 "1 C
2
So the solution is:
2
y " x 1 x 4 1 Ÿx 4 " x 2 2
2
2
®
C 1
2
2. (10 points)
2y
dy
" t 6t y
dt
yŸ1 9
with
1
1 2, So let
1"s
1" 1
2
2
2y
dy
dz
2z
" t 6t y " 2zt 6tz
dt
dt
Solve for dz and put the equation into standard linear form:
dt
dz z 3t
dz " z 3t
®
t
t
dt
dt
The integrating factor is
Bernoulli Equation.
s 1,
2
k
;
y z2.
Then
dt
I e t e ln t t
Multiply the equation by the integrating factor and integrate:
®
t dz z 3t 2
tz ; 3t 2 dt t 3 C
dt
Solve for z and then y:
2
®
y t2 C
z t2 C
t
t
Use the initial conditions to find C and substitute back
9 Ÿ1 C 2
®
®
y0 9
C2
x0 1
So the solution is:
2
y t2 2
t
3. (10 points) dx x 12
dt
t
Separable Equation.
dx x 3 1
dt
tx 2
Exponentiate:
e
1
ln x 3 1
3
tx
with
xŸ1 2
Separate the variables:
2
®
dx ; 1 dt
; 3x
t
x 1
e ln|t|C
®
Use the initial conditions to find A:
t0 1
x0 2
®
Substitute back and solve for x:
1/3
3
9 1/3 t
Ÿx 1 ®
1/3
|x 3 1| e C |t|
Ÿ2
3
1 ln|x 3 1| ln|t| C
3
®
®
1 1/3 A
x 3 1 9t 3
®
®
Ÿx
3
1 1/3 At
A 9 1/3
1/3
x Ÿ9t 3 " 1 4. (10 points)
dy
sin x " sin y
xycos
y cos x
dx
with
yŸ0 2
Exact Equation.
Rewrite as a differential 1-form:
y sin x " sin y
y sin x " sin y
®
" x cos y cos x dx dy 0
dy x cos y cos x dx
Multiply by the integrating factor (which is the denominator):
Ÿ"y sin x
sin y dx Ÿx cos y cos x dy 0
Check it is exact:
d Ÿ"y sin x sin y " sin x cos y
d Ÿx cos y cos x cos y " sin x
and
dy
dx
They are equal. So it is exact. Find the scalar potential (first integral) G which satisfies
dG Ÿ"y sin x sin y dx Ÿx cos y cos x dy
or
G "y sin x sin y
x
and
G x cos y cos x
y
So
G y cos x x sin y
and the implicit solution of dG 0 is G C or
y cos x x sin y C
Use the initial conditions to find C:
x0 0
y0 2
So the implicit solution is:
®
2 cos 0 0 sin 2 C
®
C2
y cos x x sin y 2
This cannot be solved for y.
5. (10 points)
2
dy
x2 y
dx
y
x
with
yŸ1 3
Substitute yŸx xvŸx :
Equation with Homogeneous Coefficients.
2
dy
x dv v x 2 y 12 v
®
x dv 12
dx
dx
dx
y
v
v
x
Separate variables and integrate:
v 3 lnx C
®
; v 2 dv ; 1x dx
3
The initial condition yŸ1 3 says vŸ1 3. Use this to find C:
v0 3
®
9 ln 1 C
x0 1
Substitute back and solve for v and then y:
v 3 ln x 9
®
v Ÿ3 lnx 27 1/3
®
3
®
C9
y xŸ3 ln x 27 1/3
6. (15 points) Set up the differential equation and initial condition for PŸt in the following
problem. Do not solve the equations.
A certain pond can contain 12, 000 ft 3 of water before the dam overflows. Initially, there
are 8, 000 ft 3 of water and 50 gallons of pollution in the pond. Acme Polluters is putting
2 gallons of pollution in the pond a day. Every day, 2, 000 ft 3 of fresh water is pumped
into the pond and 1, 000 ft 3 of polluted water is pumped out. Let PŸt be the gallons of
pollution in the pond after t days. When does the dam overflow and what is the
concentration of the polution in the water when the dam first overflows?
The ft 3 of water in the pond is
Differential Equation:
WŸt 8000 1000t
(The pure water does not add polution to the pond.)
dP 2 gal " PŸt gal 1000 ft 3
dt
day
day
WŸt ft 3
in
out
or
1000
dP 2 "
PŸt 8000 1000t
dt
Initial condition:
PŸ0 50
MAPLE COMPUTATIONS
dy
e t " y.
7. (20 points) Consider the differential equation
dt
a. (4 pts) Find the general solution.
b. (4 pts) Find the specific solution satisfying the initial condition yŸ0 4.
c. (4 pts) Plot the direction field of the differential equation for times "4 t t t 6.
d. (4 pts) Add to the direction field the solutions satisfying each of the initial
conditions
yŸ0 "4, yŸ0 "2, yŸ0 0, yŸ0 2, yŸ0 4, yŸ0 6
Adjust the vertical range to a reasonable value.
e. (4 pts) Looking at the exact solution and the plot, describe the behavior of the
solutions for large times. At about what time does this asymptotic behavior begin?
dy
cos x sin y with yŸ0 1.
dx
a. (10 pts) What happens if you try to solve the equations using dsolve ? Find a
Taylor polynomial approximation about x 0 for the solution to this initial value
problem keeping terms up to and including x 4 .
b. (5 pts) Plot the direction field for this differential equation together with the solution
satisfying the initial condition. Plot the Taylor polynomial approximation. Combine
the two plots into a single plot. On approximately what interval is the Taylor
polynomial a good approximation?
8. (15 points) Consider the initial value problem