Name
NetID
MATH 308
Section 511
Problems 1-3:
Exam 1
Spring 2008
Hand Computations
P. Yasskin
Do 2 of the 3 problems, only.
Identify the differential equation as one of the
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following types:
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a. Separable Equation
b. Equation with Homogeneous Coefficients
c. Linear Equation
d. Bernoulli Equation
e. Exact Equation
Then solve the initial value problem.
dy
y
+ x = 2x 2 y 2
dx
y1 = 1
2
dy
Bernoulli Equation.
Standard form is:
+ Pxy = Qxy n
dx
dy
Here, n = 2 and we set
v = y 1−n = y −1 .
So
y = v −1
and
= −v −2 dv .
dx
dx
dv − 1 v = −2x 2
The equation becomes:
−v −2 dv + 1x v −1 = 2x 2 v −2
or
x
dx
dx
∫ Pdx = e −ln x = 1
This is linear.
P = −1
I=e
x
x
1
dv
1
d  v  = −2x
We multiply by the integrating factor:
or
x dx − x 2 v = −2x
dx x
v = −x 2 + C
1
We integrate:
v = −x 3 + Cx
y = 1v =
x
3
−x + Cx
Apply the initial condition:
x = 1,
y = 1,
v = 2 = −x 3 + Cx = −1 + C
2
1
C=3
y=
−x 3 + 3x
1. (15 points)
with
2. (15 points) −e −x + e y  dx + xe y + e y  dy = 0
Exact Equation.
Check it is exact:
with
y1 = 0
d −e −x + e y  = e y
dy
and
d xe y + e y  = e y
dx
Since they are equal, it is exact. Find the scalar potential:
∂F = −e −x + e y

F = e −x + xe y + fy
∂x
∂F = xe y + e y

F = xe y + e y + gx
So
F = xe y + e y + e −x = C
∂y
Apply the initial conditions:
x = 1,
y=0

y
y
−x
−1
−1
F = xe + e + e = 1 + 1 + e = C = 2 + e
So the implicit solution is:
xe y + e y + e −x = 2 + e −1
Solve for y:
x + 1e y = 2 + e −1 − e −x
−1
−x
y = ln 2 + e − e
x+1
1
3. (15 points)
2
dy
y
= x2 + x
dx
y
y1 = 3
with
Equation with Homogeneous Coefficients.
dy
y
We set
v= x
or
y = xv
and
= x dv + v
dx
dx
The equation becomes:
x dv + v = 12 + v
or
x dv = 12
dx
dx
v
v
3
dx
v
2
This is separable. We separate:
= ln|x| + C
∫ v dv = ∫ x
3
y
y = 3,
v= x =3
Apply the initial conditions:
x = 1,
v 3 = 9 = ln|x| + C = ln 1 + C = C
v 3 = ln|x| + 9
3
3
1/3
y = xv = x3 ln|x| + 27 1/3
v = 3 ln|x| + 27
4. (5 points) Which of the following is the direction field of the differential equation:
dy
= xy.
dx
a.
2
Circle the correct answer:
2
2
y(x) 1
y(x) 1
0
1
1
x
b.
2
2
0
1
1
1
2
2
2
2
2
Correct because it is
y(x)
1
x
y(x) 1
1
the only plot with all
c.
2
0
1
1
x
2
d.
positive slopes in the
2
0
1
1
x
2
1
1
first quadrant.
2
2
5. (5 points) On the following direction field, draw the solution curve with the initial condition
y1 = 2.
5
5
4
4
3
3
y(x)
5
4
3
2
y(x)
2
2
1
1
0
1
1
1
2
3
x
4
5
5
4
3
2
0
1
1
2
2
3
3
4
4
5
5
1
2
3
4
5
x
2
6. (10 points) Find the general solution of the differential equation
Try
y = xr :
x 2 rr − 1x r−2 + 6xrx r−1 + 6x r = 0
r 2 + 5r + 6 = 0
r = −2, −3
r + 3r + 2 = 0
y = c 1 x −2 + c 2 x −3
x2
dy
d2y
+ 6x
+ 6y = 0
2
dx
dx
rr − 1 + 6r + 6 = 0
7. (15 points) Consider the initial value problem
2
d2y
dy
+8
+ 26y = 0 with y0 = 2 and
2
dt
dt
dy
0 = −1
dt
a. (8 pts) Find the general solution of the differential equation.
Try
y = e rt :
General solution:
−8 ± 64 − 208
= −8 ± 12i = −2 ± 3i
4
4
y = c 1 e −2t cos3t + c 2 e −2t sin3t
2r 2 + 8r + 26 = 0
r=
b. (7 pts) Find the solution satisfying the initial conditions.
y = c 1 e −2t cos3t + c 2 e −2t sin3t
dy
= c 1 −2e −2t cos3t − 3e −2t sin3t + c 2 −2e −2t sin3t + 3e −2t cos3t
dt
y0 = c 1 = 2
dy
3c 2 = −1 + 2c 1 == 3
c2 = 1
0 = c 1 −2 + c 2 3 = −1
dt
−2t
−2t
y = 2e cos3t + e sin3t
8. (5 points) The solution of an initial value problem of the form
dy
d2y
+b
+ ky = 0 with y0 = 2 and
2
dt
dt
is graphed below.
dy
0 = 1
dt
2
1
0
x
1
What can you say about the signs and relative sizes of the coefficients, b and k?
Since it oscillates, there are sines and cosines. Since it is damped, there are exponentials with
negative powers.
y = c 1 e αt sinβt + c 2 e αt cosβt
The complex roots are r = α ± iβ =
with α < 0
−b ± b 2 − 4k
2
To have α < 0, we need b > 0.
To have complex roots, we need 4k > b 2 .
3