Particular Solution of ODE using Annihilator

In summary: Then add this particular solution to the general solution of the homogeneous equation to get the full general solution. Does that make sense?
  • #1
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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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  • #2
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
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
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
((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
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.
 

1. What is an annihilator in the context of ODEs?

An annihilator in the context of ODEs is a mathematical operator that can be applied to a given function to produce a new function that satisfies a specific differential equation. It essentially "annihilates" the original function, leaving behind a particular solution to the differential equation.

2. How is an annihilator used to find the particular solution of an ODE?

An annihilator is used by first finding the general solution to the ODE, then applying the appropriate annihilator to it. This results in a new differential equation that can be solved to find the particular solution.

3. Can an annihilator be used to find the general solution of an ODE?

No, an annihilator is specifically used to find the particular solution of an ODE. The general solution is typically found first using other methods, such as separation of variables or integrating factors.

4. Are there different types of annihilators for different types of ODEs?

Yes, there are different types of annihilators for different types of ODEs. For example, there are specific annihilators for first-order linear ODEs, second-order linear ODEs, and more complex ODEs.

5. Are there any limitations to using an annihilator to find the particular solution of an ODE?

Yes, there are certain limitations to using an annihilator. It may not work for all types of ODEs, and it may not always produce a unique solution. It is important to check the validity of the solution obtained using an annihilator method.

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