Monday Quiz 8
Name:
SOLUTIONS
Maths 104 - Calculus I
April 25, 2011
Note: In order to receive full credit, you must show work that justifies your answer.
1. Find the solution of the differential equation
dy
y cos x
=
dx
1 + y2
that satisfies the initial condition y(0) = 1. [You may leave the solution in
implicit form, i.e. you do not need to write y explicitly as a function of x.]
This is separable differential equation, so we can separate variables and
integrate:
Z
Z
1 + y2
dy = cos x dx.
y
Therefore ln |y| + 12 y 2 = sin x + C, and it remains to determine C. For this we
use the initial condition: when x = 0, y = 1, so ln |1| + 21 12 = sin 0 + C, which
means C = 12 . The solution, in implicit form, is thus
1
1
ln |y| + y 2 = sin x + .
2
2
Wednesday Quiz 8
Name:
SOLUTIONS
Maths 104 - Calculus I
April 20, 2011
Note: In order to receive full credit, you must show work that justifies your answer.
1. Evaluate the indefinite integral as an infinite series:
Z
cos x − 1
dx.
x2
P
n x2n
The power series expansion for cos x is cos x = ∞
n=0 (−1) (2n)! . Note that the
P
0
n x2n
first term is (−1)0 x0! = 1, so cos x − 1 = ∞
n=1 (−1) (2n)! (we just leave off the
first term). Therefore
cos x − 1
=
x2
P∞
n x2n
n=1 (−1) (2n)!
x2
=
∞
X
n=1
(−1)n
x2n−2
.
(2n)!
Since the power series for cos x converges everywhere, so does the power series
for cosxx−1
, hence we can integrate term-by-term:
2
Z
cos x − 1
dx =
x2
Z X
∞
∞
2n−2
X
x2n−1
nx
(−1)
(−1)n
dx =
.
(2n)!
(2n − 1)(2n)!
n=1
n=1
Note that it is also possible to express this as a series where we start summing
from 0 instead of 1. Just put k = n − 1, so that n = k + 1, 2n − 1 = 2k + 1, etc,
and as a result,
∞
X
∞
X
x2n−1
x2k+1
k+1
(−1)
=
(−1)
.
(2n − 1)(2n)! k=0
(2k + 1)(2k + 2)!
n=1
n
Keep this in mind when you’re doing a multiple choice exam. You might have
the correct answer, just in a different form to the one on the answer sheet.