y'' - 2y' -3y = 2 sinx

[Riazy]

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.

 

[galactus]

You nearly have it. Just finish.

Start with \(\displaystyle y_{p}=Acos(x)+Bsin(x)\)

\(\displaystyle y'=-Asin(x)+Bcos(x)\)

\(\displaystyle y''=-Acos(x)-Bsin(x)\)

Subbing these into the DE and simplifying gives:

\(\displaystyle (-4A-2B)cos(x)+(2A-4B)sin(x)=2sin(x)\)

Equate coefficients:

There is no cos term on the right, so we set the cos terms equal to 0.

\(\displaystyle -4A-2B=0\)

\(\displaystyle 2A-4B=2\)

Solving gives \(\displaystyle A=\frac{1}{5}, \;\ B=\frac{-2}{5}\)

This gives the general solution as:

\(\displaystyle C_{1}e^{-x}+C_{2}e^{3x}+\frac{1}{5}cos(x)-\frac{2}{5}sin(x)\)