Particular Solution of ODE using Annihilator

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Homework Statement



By using the method of differential operators, solve

y''+2y'+2y=2e-xsinx

1. Determine what is the annihilator of the inhomogeneous term.

2. Find a particular solution.

3. Write the general solution for the equation.

Homework Equations



xneaxsin(bx) --> annihilated by --> (D2-2aD+(a2+b2))n+1

The Attempt at a Solution



1. No problem with this. Annihilated by D2+2D+2I

2. and 3. Not sure how to get these.

I multiple the annihilator by both sides of the equation. I then get ((D+1)2+1)2=0 From here I'm not sure what to do.

Thanks
 
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This is how I think I should solve it from my notes:

((D+I)2+I)((D+I)2+I)y=0

Basis for kernel:

((D+I)2+I): {excos(x), exsin(x)}
((D+I)2+I): {excos(x), exsin(x)}

yp= Aexcos(x) + Bexsin(x)
yH= c1excos(x) +c2exsin(x)

y=yp+yH

y= Aexcos(x) + Bexsin(x) + c1excos(x) +c2exsin(x)

However, the answer (was given) for yp is yp=-xe-xcosx and for the general solution. is y=e-x(c1cosx+c2sinx-xcosx). I'm not sure where I am going wrong.
 
((D+I)2+I)2y=0

Basis for kernel:

((D+I)2+I)2: {excos(x), exsin(x), xexcos(x), xexsin(x)}


yp= A(x)excos(x) + B(x)exsin(x)
yH= c1excos(x) +c2exsin(x)

y=yp+yH

y= A(x)excos(x) + B(x)exsin(x) + c1excos(x) +c2exsin(x)

Is this correct? I still think something is wrong because I'm not getting the correct answer
 
I'm not sure if this is what you meant by what you wrote, but the particular solution is [itex]y_p = A xe^x\cos x + B xe^x\sin x[/itex]. So now you plug this back into the original differential equation and solve for constants A and B.