y'' - 4y = -4e^x
a) find the complementary function
b) find the particular integral
c) write down the solution, given y(0) = 2 and y'(0) = 0
2. The attempt at a solution
Yesterday I learned how to do simpler differential equations of the form y'' + ay' + by = 0.
I Tried some others this morning and solved them quite happily without using any reference material.
Here is what I have so far for this problem, I'm quite sure I'm going about it the wrong way:
r^2*e^rx - 4e^rx = -4e^x
This seemed to be the the most straightforward way to determine r. Next I factorised the left hand side and divided through by e^x as follows:
(r^2 - 4)e^rx = -4e^x
(r^2 - 4)e^r = -4
I am unsure where to go from there. Would I take the natural logarithm? It's been a long time since I've used logs so if that is the correct way to find r I'd be grateful if somebody took me through it. I'm still not quite clear on what exactly is meant by the particular integral, if someone could show me this too it would be very helpful. Thanks in advance.