Math 308 Differential Equations
Solutions to 4.3/2 and 4.3/10
4.3/2. The equation is
d2y
dt2+ 9y= 5 sin 2t.
First find yh. The homogeneous equation is d2y
dt2+ 9y= 0, so when we guess yh=est, we find
s2+ 9 = 0. Thus s=±3i, and we have
yh=k1cos 3t+k2sin 3t.
Now we find the particular solution yp. The first guess is yp=Acos 2t+Bsin 2t, and neither
term in ypsolves the homogeneous equation, so this will work. We find
y0
p=−2Asin 2t+ 2Bcos 2t
y00
p=−4Acos 2t−4Bsin 2t.
Now put these into the original equation:
−4Acos 2t−4Bsin 2t+ 9Acos 2t+ 9Bsin 2t= 5 sin 2t
(5A) cos 2t+ (5B) sin 2t= 5 sin 2t,
and this implies A= 0 and B= 1. Thus yp= sin 2t. The general solution is
y(t) = yh+yp=k1cos 3t+k2sin 3t+ sin 2t.
4.3/10. The initial value problem is
d2y
dt2+ 4y= 3 cos 2t, y(0) = 0, y0(0) = 0.
First find yh; following the usual procedure, we find yh=k1cos 2t+k2sin 2t.
Now we find the particular solution yp. Our first guess is yp=Acos 2t+Bsin 2t. This will
not work, because each term in this guess also solves the homogeneous problem. So we multiply
by tto get the next guess: yp=At cos 2t+Bt sin 2t. Neither term solves the homogeneous
problem, so this will work. We find
y0
p= (B−2At) sin 2t+ (A+ 2Bt) cos 2t
y00
p= 4Bcos 2t−4At cos 2t−4Asin 2t−4Bt sin 2t.
Put these into the original equation:
4Bcos 2t−4At cos 2t−4Asin 2t−4Bt sin 2t+ 4At cos 2t+ 4B t sin 2t= 3 cos 2t,
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