Ma 221 Homework Solutions Due 2/20/14

Ma 221 Homework Solutions Due 2/20/14
4.4 Page 182 18 , 19 , 24 , 34
18.
y ′′  4y  8 sin 2t
The characteristic polynomial is pr  r 2  4 so the roots are r  2i and therefore
y h  c 1 sin 2t  c 2 cos 2t
Consider a companion equation
v ′′  4v  8 cos 2t
Multiplying the first equation by i and adding it to the second equation yields
v ′′  iy ′′  v  iy  8cos 2t  i sin 2t  8e 2it
Let w  v  iy. Then we have
w ′′  4w  8e 2it
Since p2i  0, but p ′ 2i  4i ≠ 0, then
2it
2it
 2 te
w p  8te
i
4i
y p  Im w p
2it
w p  2 te
 i  −2ite 2it
i
i
 −2itcos 2t  i sin 2t
Thus
y p  −2t cos 2t
19.
4y ′′  11y ′ − 3y  −2te −3t
Consider the homogeneous equation 4y ′′  11y ′ − 3y  0 first.
pr  4r 2  11r − 3  0
−11  121 − 44−3
−11  169

 −11  13  −3, 1
16
8
28
28
−3t
Thus e is a homogeneous solution.
y p  tA 1 t  A 0 e −3t  A 1 t 2  A 0 t e −3t
r
y ′p  −3A 1 t 2  2A 1 − 3A 0 t  A 0 e −3t
y ′′p  9A 1 t 2  9A 0 − 12A 1 t  2A 1 − 6A 0 e −3t
Substituting into the DE yields after some algebra
1
−26A 1 t  8A 1 − 13A 0 e −3t  −2te −3t
Thus
− 26A 1  −2
8A 1 − 13A 0  0
Hence A 0 
8
169
, A1 
1
13
and
yp 
t  8 te −3t
169
13
24.
y ′′  y  4x cos x
Since cos x and sin x are homogeneous solutions, the we let
y p  A 1 x 2  A 0 x cos x  B 1 x 2  B 0 x sin x
Thus
y ′p  B 1 x 2  B 0  2A 1 x  A 0 cos x  −A 1 x 2  2B 1 − A 0 x  B 0 sin x
and
y ′′p  −A 1 x 2  4B 1 − A 0 x  2B 0  A 1  cos x  −B 1 x 2  −4A 1 − B 0 x  2B 1 − A 0  sin x
Substituting into the DE and combining yields
4B 1 x  2B 0  A 1  cos x  −4A 1 x  2B 1 − A 0  sin x  4x cos x
Hence
4B 1  4 or B 1  1
− 4A 1  0 so A 1  0
2B 0  A 1   0 so B 0  −A 1  0
2B 1 − A 0   0 so A 0  B 1  1
Thus
y p  x cos x  x 2 sin x
34.
2y ′′′  3y ′′  y ′ − 4y  e −t
The characteristic equation is
pr  2r 3  3r 2  r − 4  0
p−1  −2  3 − 1 − 4  −4 ≠ 0. Thus e −t is not a homogeneous solution and
y p  Ae −t
−t Substituting into the DE we have
y ′p  −Ae −t , y ′′p  Ae −t , y ′′′
p  −Ae
−2A  3A − A − 4Ae −t  e −t
or −4A  1 so
y p  − 1 e −t
4
2

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