Unformatted text preview: e 0
yp = and A cos x 00
yp = A sin x + B cos x
B sin x. Substituting this into the DE gives
A cos x B sin x
i.e. A sin x + B cos x (B 6A cos x 7A) cos x + ( A 6B sin x = 5 cos x, 7B ) sin x = 5 cos x. Therefore B 7A = 5, and A+7B = 0, from which we ﬁnd that A = 7/10
and B = 1/10. The general solution to the given DE is thus found to be
y = C1 e 3x 7
1
cos x +
sin x
10
10 + C 2 e 2x c) y 00 + 4y 0 + 3y = x2 + e x
As usual, we start by ﬁnding the complementary function. The characteristic equation of the homogeneous version of the DE is m2 + 4m + 3 = 0,
or (m + 3) (m + 1) = 0, so m = 3, 1, and
yp = C1 e 3x + C2 e x . Now, for the inhomogeneous term x2 we ususally seek a particular solution
of the form Ax2 + Bx + C . For the term e x we would guess instead the
form Dxe x (since e x is a solution to the homogeneous equation). By
the Principle of Superposition, we can simply add these together to obtain
our trial function:
yp = Ax2 + Bx + C + Dxe
0
yp = 2Ax + B + De
00
yp = 2A x 2De x x Dxe x + Dxe x . Substitute these into the DE:
2A 2De x...
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 Winter '14
 Heterogeneity, Homogeneity, yp

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