What's the mistake in finding a particular solution for a differential equation?

[kasse]

Homework Statement

y''-2ay'+a^2y=e^ax

Find a general solution

2. The attempt at a solution

I've found the general solution of the homogeneous eq: Ce^ax+Dxe^ax

Next, I must find a particular solution on the form Be^ax (*), right?

The derivative of (*) is Bae^ax and the 2nd derivative is B(a^2)e^ax

so that

y''-2ay'+a^2y=0

e^ax can never be 0, so I must have made a mistake...

 

[EnumaElish]

How do you go from the general sol. Ce^ax+Dxe^ax to the particular sol. Be^ax? E.g., why not Bxe^ax?

 

[kasse]

How do you go from the general sol. Ce^ax+Dxe^ax to the particular sol. Be^ax? E.g., why not Bxe^ax?

Bacause r(x)=e^ax, and my textbook tells me that the particular solution is then on the form Be^ax

 

[proton]

how can the particular soln be of the form Be^ax, when Ce^ax satisfied the homogeneous soln?

also, the particular soln can't be of the form Bxe^ax since Dxe^ax also satisfied the homogeneous soln

therefore, the particular soln must be of the form...?

 

[kasse]

how can the particular soln be of the form Be^ax, when Ce^ax satisfied the homogeneous soln?

also, the particular soln can't be of the form Bxe^ax since Dxe^ax also satisfied the homogeneous soln

therefore, the particular soln must be of the form...?

Ah, use of the modification rule twice?

B(x^2)e^ax?

 

[EnumaElish]

I was thinking that Be^ax = Ce^ax + Dxe^ax when B = C and D = 0.

 

[kasse]

My book operates with the so-called modification rule. I got the corect solution this time.