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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The idea is that because (D2+2D+2)(e-xsin x) = 0, it follows that

(D2+2D+2)[(D2+2D+2)y] = (D2+2D+2)[2e-xsin x] = 0

In other words, you want to solve the homogeneous equation

(D2+2D+2)(D2+2D+2)y = 0
 
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.
 
It's a good first guess, but the problem is you have repeated roots because the annihilator is the identical to the differential operator for the original equation. What do you do when you have repeated roots?
 
((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 y_p = A xe^x\cos x + B xe^x\sin x. So now you plug this back into the original differential equation and solve for constants A and B.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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