Power series solution, differential equation question

[Honey Bee]

I can not find a solid explanation on this anywhere, so forgive me if this has been addressed already.

Given something like y''+y'-(x^2)y=1 or y''+2xy'-y=x, how do I approach solving a differential with a power series solution when the differential does not equal zero?

Would I solve the left hand side as normally with series substitution as if it were equal to zero, then expand it out and equate the coefficients with the right hand side? Or would I just expand it out after substituting the series for y and then equate the coefficients?

Also, if I set it equal to zero and solve it, would this be a complimentary/homogeneous solution, meaning I would have to solve for a particular solution? If so, how should I solve for the particular solution?

I want to thank everyone in advance for their considerations. I know this is a basic question, but I have no clarification on it and it is leaving me frustrated.

 

[pasmith]

I can not find a solid explanation on this anywhere, so forgive me if this has been addressed already.

Given something like y''+y'-(x^2)y=1 or y''+2xy'-y=x, how do I approach solving a differential with a power series solution when the differential does not equal zero?

Would I solve the left hand side as normally with series substitution as if it were equal to zero, then expand it out and equate the coefficients with the right hand side?

Or would I just expand it out after substituting the series for y and then equate the coefficient

Also, if I set it equal to zero and solve it, would this be a complimentary/homogeneous solution, meaning I would have to solve for a particular solution? If so, how should I solve for the particular solution?

I would use the particular solution plus complementary function approach.

You can find particular solutions using variation of parameters or by inspection: for example $y = x$ is a solution of $y'' + 2xy' - y = x$.

 

[maka89]

In theory you can just plug in the series representation of y, and equate the coefficients. Your right hand side will however make sure that you have to split up your sum so that you don't have one nice sum going from zero to infinity. This will make it a bit harder to find a closed form expression for your sum(if it exists).