1. Concordia University March 20, 2009
Applied Differential Equations
Section J
Exam II (B)
ANSWER KEY
(1) (8 points) We evaluate the Wronskian of the three functions (by developing along the second
column)
 
1 + x2 x x2
 2x 2x 2x 1 + x2 x2
det 1 2x = −x · det + 1 · det
2 2 2 2
2 0 2
= x · (4x − 4x) + (2 + 2x2 − 2x2 ) = 2 = 0,
for any x, hence the functions are linearly independent on any interval of the real line, in
particular on (1, ∞).
(2) (15 points) Consider first the homogeneous equation y + 2y + 2y = 0. Its auxiliary equation
√
is r 2 + 2r + 2 = 0 and has the roots −2 ± 4 − 8 = −2 ± 2i = −1 ± i. This makes the
2 2
complementary part of the general solution of the nonhomogenous equation equal to
yc (x) = C1 e−x cos x + C2 e−x sin x, C1 , C2 arbitrary constants.
We will now continue with the method of undetermined coefficients to find a particular
solution to the nonhomogenous equation. Set yp (x) = Ae−2x , thus yp (x) = −2Ae−2x and
yp (x) = 4Ae−2x . By substitution in the nonhomogenous ODE, we obtain 2Ae−2x = e−2x , or
1
A = 1/2. So, we obtain that yp (x) = e−2x .
2
Therefore the general solution of the nonhomogenous ODE is
1 −2x
ygen (x) = yc (x) + yp (x) = C1 e−x cos x + C2 e−x sin x + e , C1 , C2 arbitrary constants.
2
We’ll find the constants from the initial conditions. First, y(0) = 0 ⇒ 0 = C1 + 1 ⇒ 2
C1 = − 1 . Then we find the first derivative of ygen : ygen (x) = C1 (−e−x cos x − e−x sin x) +
2
1
C2 (−e−x sin x+e−x cos x)−e−2x . Since y (0) = 0, we get 0 = −C1 +C2 −1 ⇒ C2 = C1 +1 = 2 .
Thus the solution to the given initial value problem is the function
1 1 1
y(x) = − e−x cos x + e−x sin x + e−2x .
2 2 2
1

2. 2
(3) (12 points) Like in the previous problem, consider the homogeneous ODE associated: y + y =
0. Its auxiliary equation is r2 + 1 = 0 with the roots ±i, hence the general solution of
the homogeneous ODE (and the complementary of the nonhomogenous ODE) is yc (x) =
C1 cos x + C2 sin x, C1 , C2 arbitrary constants.
By denoting y1 (x) = cos x, y2 (x) = sin x, we proceed now with the method of variation of
parameters to find yp . Its form is yp = u1 y1 + u2 y2 .
cos x sin x 0 sin x sin x
W = det = 1, W1 = det = − sec2 x sin x = −
− sin x cos x sec2 x cos x cos2 x
and
cos x 0
W2 = det = sec2 x cos x = sec x,
− sin x sec2 x
1
due to the definition of sec x = cos x .
Thus u2 = sec x ⇒ u2 = ln | sec x + tan x| (following the formula provided). On the
other hand, u1 = − cos2x ⇒ u1 = − cos2x dx. We may find this antiderivative by using
sin
x
sin
x
the substitution v = cos x, dv = − sin x dx. Since, v −2 dv = −v −1 + C (we actually chose
C = 0), we have that u1 (x) = −1/ cos x = − sec x.
So, yp (x) = cos x · (− sec x) + sin x · (ln | sec x + tan x|) = −1 + sin x ln | sec x + tan x|.
Finally we can state the general solution of the given ODE:
ygen (x) = C1 cos x + C2 sin x + sin x · (ln | sec x + tan x|) − 1, C1 , C2 arbitrary constants.
(4) (5 points) After multiplication by x, this ODE becomes an Euler-Cauchy equation. Hence we
seek solutions of the form xm . This leads to the auxiliary equation m(m − 1) − 3m = 0 ⇒
m2 − 4m = 0 ⇒ m = 0 and m = 4.
Therefore the general solution of the ODE is ygen (x) = C1 +C2 x4 , C1 , C2 arbitrary constants.
This is not the only method possible. One can also denote w = y and reduce the second
order ODE to a first order linear ODE. After finding w, we integrate once more to find y,
which is the antiderivative of w, and arrive at the same answer.