Math 215 S1
Midterm Exam Solutions
Winter 2012
Last Name, First Name (Please, PRINT!)
Your Signature
Your Student ID Number
Instructor’s Name
Aditya Raghavan
• Turn off and put away ALL ELECTRONIC DEVICES.
• This exam is closed book. You may not use any notes.
• In order to receive credit, you must show your work on the exam paper, with some explanation in
English, if appropriate. Do not do computations in your head. Instead, write them out on the exam
paper. IF a problem asks you to use a specific method, you MUST use this method. You will get
zero credit for any other solution, even if it is correct.
• If you need more room, use the back of the previous page and indicate to the grader that you
have done so.
• This exam consists of this title page and 6 pages with 6 problems.
Problem Max Points Score
1
5
2
5
3
5
4
5
5
5
6
5
Total
30
GOOD
LUCK!
Math 215 S1
Midterm Exam Solutions
Page 1 of 6
1. Solve the initial value problem for the first order linear equation given by
1 0
y + 3y = 3,
x2
y(1) = 2.
Solution:
Rewriting the equation in standard form,
y 0 + 3x2 y = 3x2 ,
we observe that P (x) = 3x2 and Q(x) = 3x2 . we can compute the integrating factor, v(x)
v(x) = e
R
P (x)dx
R
=e
3x2 dx
3
= ex .
Multiplying the equation by v(x) and combining derivatives:
3
3
ex y 0 + 3x2 ex y =
i
d h x3
3
e y(x) = 3x2 ex .
dx
Integrating and solving for y(x)
x3
Z
e y(x) =
3
ex 3x2 dx + C
In the integral, we substitute x3 = t and notice that 3x2 dx = dt. Hence, we get
Z
3
x3
e y(x) = et dt + C = ex + C
yielding,
3
y(x) = 1 + Ce−x .
Given the initial condition y(1) = 2, we can solve for C to get the particular solution,
3
y(1) = 1 + Ce(−1) = 2,
or C = e. The solution to the initial value problem is given by,
3
y(x) = 1 + e1−x .
Math 215 S1
Midterm Exam Solutions
Page 2 of 6
2. Use the method of undetermined coefficients to find the general solution of the differential
equation
y 00 + 2y 0 + y = 6 sin 2x.
Solution:
The characteristic equation is r2 + 2r + 1 = (r + 1)2 = 0 with a double root r = −1. Therefore, the
complementary solution is
yc = C1 e−x + C2 xe−x .
To find a particular solution using undetermined coefficients, we start with
yp = A cos 2x + B sin 2x.
Differentiating and substituting into the equation, we get
yp0 = −2A sin 2x + 2B cos 2x,
yp00 = −4A cos 2x − 4B sin 2x,
(−4A cos 2x − 4B sin 2x) + 2(−2A sin 2x + 2B cos 2x) + (A cos 2x + B sin 2x) = 6 sin 2x,
(4B − 3A) cos 2x − (4A + 3B) sin 2x = 6 sin 2x.
To cancel all the terms we need 4B − 3A = 0 and 4A + 3B = −6, so B = 3A/4, 4A + 9A/4 = −6,
25A = −24, A = −24/25 = −0.96, B = −18/25 = −0.72. The general solution is then
y = C1 e−x + C2 xe−x −
24
18
cos 2x −
sin 2x.
25
25
Math 215 S1
Midterm Exam Solutions
Page 3 of 6
3. Find a particular solution for the second order non-homogeneous differential equation by using the
method of variation of parameters:
y 00 + 4y 0 − 5y = x.
Solution:
Finding y1 and y2 that are solutions to the homogeneous part of the differential equation. The
characteristic equation reads,
r2 + 4r − 5 = 0,
yielding roots r1 = 1 and r2 = −5. Hence, y1 = ex and y2 = e−5x . In this method, we seek a
particular solution of the form yp = v1 y1 + v2 y2 by solving these two equations in v10 and v20
(
v10 y1 + v20 y2 = 0,
v10 y10 + v20 y20 = x.
Using Cramer’s formula, we have
−5x 0
e
x −5e−5x 1
0
= xe−x ,
v1 = x
−5x 6
e
ex
e −5e−5x x e 0 x e x 1
0
= − xe5x ,
v2 = x
−5x 6
e
ex
e −5e−5x Integrating to get v1 and v2 ,
Z
Z
1
1
1
−x
−x
−x
v1 =
xe dx =
−xe + e
= − (x + 1)e−x
6
6
6
Z
Z
1
1 1 5x 1
1 5x
1
5x
5x
v2 = −
xe dx = −
xe −
e dx = − e
x−
6
6 5
5
30
5
Hence, the particular solution, yp = v1 y1 + v2 y2 is
1
1
yp = − (x + 1) −
6
30
1
x−
5
x
4
=− − .
5 25
Math 215 S1
Midterm Exam Solutions
Page 4 of 6
4. Sketch the region of integration of the following integral, reverse the order of integration, and
evaluate either of these two (equal) integrals:
Z πZ π
sin y
dy dx.
y
0
x
Solution:
3
3
2.5
2.5
2
2
1.5
1.5
1
1
0.5
0.5
0.5 1 1.5 2 2.5 3
Reversing the order of integration, we get
Z
π
Z
0
0
0.5 1 1.5 2 2.5 3
y
sin y
dx dy.
y
Let’s evaluate this integral:
Z
0
π
Z
0
y
sin y
dx dy =
y
Z
0
π
sin y y
x x=0 dy =
y
Z
0
π
sin y
(y − 0) dy
y
Z π
π
=
sin y dy = − cos y y=0 = 1 + 1 = 2.
0
Math 215 S1
Midterm Exam Solutions
Page 5 of 6
5. Change the following Cartesian integral into an equivalent polar integral:
√1
2
Z
√
Z
√
0
2−x2
(x + 2y ) dy dx.
3x
You do not need to evaluate either of the integrals.
Solution:
First, we sketch the region of integration.
1.4
1.2
1
0.8
0.6
0.4
0.2
0.2 0.4 0.6 0.8 1 1.2 1.4
√
√
The region is bounded by three curves, y = 3 x, y = 2 − x2 and x = 0. The line x = 0 is θ = π2
√
√
in polar coordinates, while the line y = 3 x is θ = π3 in polar coordinates. The curve 2 − x2 is a
√
√
part of the circle of radius r = 2. Hence, that curve is represented by r = 2 in polar coordinates.
On substituting x = r cos θ and y = r sin θ, the integral can be written as,
Z
θ=π/2
Z
√
r= 2
(r cos θ + 2r sin θ) rdrdθ
θ=π/3
r=0
Math 215 S1
Midterm Exam Solutions
Page 6 of 6
6. Sketch the region bounded by the lines y = 1 − x and y = 2 and the curve y = ex , express its area
as an iterated double integral and evaluate the integral.
Solution:
3
2.5
2
1.5
1
0.5
-2 -1.5 -1 -0.5
0.5
1
We can see that the area of this region can be expressed as an iterated integral in any order, but it
is easier to see minimum and maximum values for y: 1 and 2, so we will set up the integral for the
dx dy order. In this case, for each fixed y between 1 and 2, x is changing from the line y = 1 − x to
the curve y = ex . Expressing x in terms of y we get limits 1 − y and ln y. Now we compute:
ZZ
Z
A=
2
Z
ln y
dA =
R
Z
2
dx dy =
1
1−y
Z
=
1
1
ln y
x
x=1−y
Z
2
(ln y + y − 1) dy
dy =
1
2
Z 2
1
y2
ln y dy + 2 − 2 − + 1
=
− y ln y dy +
2
2
1
y=1
Z 2
Z 2
2
1
1
1
1
= +
ln y dy = + y ln y y=1 −
dy = + 2 ln 2 − 1 = 2 ln 2 − .
2
2
2
2
1
1
2