This is the problem:
y'' - 2y' -3y = 2 sinx
Yh = C1e^3x + C2e^-x
Ok
Now we have to assume an yp
yp = A sinx + B cosx
yp' = A cosx - b sinx
yp'' = -A sinx - b cosx
ok so far, so good.
Inserting it into the de
(-A sinx - b cosx ) - 2 ( A cosx - b sinx) -3( A sinx + B cosx) = 2 sinx
-Asinx -Bcosx -2Acosx -2Bsinx -3Asinx -3 Bcosx = 2 sinx
sinx ( -A -2b - 3a) + cosx (-B -2A -3B)
I get some problems in solving the rest of the de from here,
how can i go about that? detailed explanation would be appreciated.
y'' - 2y' -3y = 2 sinx
Yh = C1e^3x + C2e^-x
Ok
Now we have to assume an yp
yp = A sinx + B cosx
yp' = A cosx - b sinx
yp'' = -A sinx - b cosx
ok so far, so good.
Inserting it into the de
(-A sinx - b cosx ) - 2 ( A cosx - b sinx) -3( A sinx + B cosx) = 2 sinx
-Asinx -Bcosx -2Acosx -2Bsinx -3Asinx -3 Bcosx = 2 sinx
sinx ( -A -2b - 3a) + cosx (-B -2A -3B)
I get some problems in solving the rest of the de from here,
how can i go about that? detailed explanation would be appreciated.