MTH 240 Midterm: MTH240 Term Test 2 2011 Winter
Ryerson University
Department of Mathematics
MTH 240 Winter 2011 – Test II
LAST NAME: FIRST NAME:
(Please print) (Please print)
I.D. NUMBER: SIGNATURE:
Date: March 25, 2011 Duration: 90 minutes
Professor (circle one)
Kim Poliakov
Tasi´c Wang
Section:
INSTRUCTIONS:
•Verify that your exam has 6 pages including this
page.
•The use of notes, formula sheets, books or cal-
culators is not allowed.
•For full-answer questions:
Give full justification for your answers; correct
answers alone may be worth nothing. Cross out
or erase all rough work not relevant to your so-
lution. Write your solutions in the space pro-
vided. If you need more space, use the back of
the page. Indicate this fact on the original page,
making sure that your solution cannot be con-
fused with any rough work which may be there.
For markers’ use only:
Page Value Mark
2 10
3 12
4 10
5 10
6 8
Total 50
1
MTH 240 Test II
1. (10 pts.) Solve the initial value problem:
y0−3x2y=x2, y(0) = 0
Solution: This is a linear D.E. with p(x) = −3x2and q(x) = x2.
The integrating factor I(x) is
I(x) = eR−3x2dx =e−x3
Multiplying both sides of the D.E., we get
e−x3y0−3x2e−x3y=x2e−x3
That is,
e−x3y0=x2e−x3
Integrating both sides of the above, we get
e−x3y=Zx2e−x3dx =−1
3e−x3+C
So, y=Cex3−1
3
To determine C, we use the condition y(0) = 0. So,
0 = Ce03−1
3
So, C=1
3,
and the solution to the above D.E. is y=1
3ex3−1
3
2
Document Summary
Section: verify that your exam has 6 pages including this. Instructions: page: the use of notes, formula sheets, books or cal- culators is not allowed, for full-answer questions: Give full justi cation for your answers; correct answers alone may be worth nothing. Cross out or erase all rough work not relevant to your so- lution. Write your solutions in the space pro- vided. If you need more space, use the back of the page. Indicate this fact on the original page, making sure that your solution cannot be con- fused with any rough work which may be there. Integrating both sides of the above, we get e x3y = x2e x3dx = 1. Solve the initial value problem: y(cid:48) 3x2y = x2, y(0) = 0. Solution: this is a linear d. e. with p(x) = 3x2 and q(x) = x2. The integrating factor i(x) is (cid:82) 3x2dx = e x3. Multiplying both sides of the d. e. , we get.
