Suppose y is a solution of the d.e:
y"+p(x)y'+q(x)y= q(x) on the interval (-1,1) with y(0)=1 and y'(0)= 1.
What is y?
I used the auxiliary equation: m^2+p(x)m+q(x)= q(x)
The Attempt at a Solution
My question is can I do this? I can cancel out the q functions and the right and left and I'm left with m=0 and m=-p(x). I can solve from there and I get a solution of y=1
I am a little uneasy about my solution since I have never seen an auxiliary equation set equal to a function that's not identically zero. Is this legal? Or does my auxiliary equation have to be equal to zero? If that's the case, I don't know how to solve this with the given info since I don't know the particular solution....