# Particular Solution of ODE using Annihilator

1. Dec 10, 2011

### trust

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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

2. Dec 10, 2011

### vela

Staff Emeritus
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

3. Dec 10, 2011

### trust

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.

4. Dec 10, 2011

### vela

Staff Emeritus
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?

5. Dec 11, 2011

### trust

((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

6. Dec 11, 2011

### vela

Staff Emeritus
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.