AMath350.W13.A4.Sol

# With this we have 0 yp and a cos x 00 yp a sin x

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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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