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Unformatted text preview: ex + C2 xex .
Applying the variation of parameters method, we seek a particular solution in the form
yp = u(x)ex + v (x)xex . a) y 00 2y 0 + y = As shown in the Course Notes, this requires us to solve the system
(1) u0 ex + v 0 xex = 0
u0 ex + v 0 (ex + xex ) = 1x
x (2) 1x
e , so v 0 = x , and hence v = ln x.
v 0 xex = ex , so u0 = 1, and u = x. Subtracting (1) from (2) yields v 0 ex = Meanwhile, (1) gives us u0 ex =
yp = xex + xex ln x. When we write out the general solution, we see that the ﬁrst of these
terms will be absorbed by the C2 term; the solution is
y = C1 ex + C2 xex + xex ln x.
b) y 00 + y = sin2 x
Solution: The complementary function is yh = C1 cos x + C2 sin x, so we
look for a particular solution in the form
yp = u(x) cos x + v (x) sin x. AMATH 350 Page 2 Assignment #5 Solutions - Fall 2013 The unknown functions must satisfy the system of equations
(3) u0 cos x + v 0 sin x = 0 (4) u0 sin x + v 0 cos x = sin2 x....
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- Fall '12