Unformatted text preview: e can see that 1 is a root, so we factor out (m + 1), and
obtain (m + 1) (m2 + 2m + 10) = 0. Equating the quadratic factor to
zero gives (m + 1)2 + 9 = 0 =) m = 1 ± 3i. Therefore the general
solution is
y = c1 e x + e x [c2 cos (3x) + c3 sin (3x)] . 2. Find the general solutions to the following ODEs:
a) y 00 + y 0 6y = 6x
Solution: The characteristic equation of the homogeneous version of this
equation is m2 + m 6 = 0 =) (m + 3) (m 2) = 0 =) m = 3, 2.
Therefore
y h ( x ) = c 1 e 3x + c 2 e 2x .
0
Now, for a particular solution, assume yp = Ax + B . Then yp = A, and
00
yp = 0. Substitution into the DE gives A 6 (Ax + B ) = 6x. Equating
coeﬃcients, we can see that A = 1, and (since A 6B = 0), B = 1/6.
1
Thus we ﬁnd that yp = x
, and the general solution to the full
6
inhomogeneous equation is y = c1 e 3x + c 2 e 2x x 1
.
6 AMATH 350 Page 2 Assignment #4 Solutions Winter 2013 b) y 00 + y 0 6y = 5 cos x
As in #2a, we have yh = C1 e 3x + C2 e2x .
For the particular solution, the appropriate trial function is
yp = A cos x + B sin x.
With this, we hav...
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 Winter '14
 Heterogeneity, Homogeneity, yp

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