Can someone show me to get the solution to y''+y'=e^(-x) ?? I have that y_c is r=0 and r=-1 but I can't figure out y_p. Any help would be greatly appreciated.

Question

zepdrix · Answer

$$\Large\rm y''+y'=e^{-x}$$So your homogeneous part of your solution is given by the characteristic equation:$$\Large\rm r^2+r=0$$Ok your roots look good.
$$\Large\rm y_c=e^{0x}+e^{-x}$$

zepdrix · Answer

Our particular solution will match the form of the right side,$$\Large\rm y_p=Ae^{-x}$$We just need to know what this A value is.

anonymous · Answer

I got that part as well. I am using the form of Ae^(-x) for the y_p and have made it down to Ae^(-x) - Ae^(-x) = e^(-x), but I'm lost after this.

anonymous · Answer

I can see by observation that A = 1, but I'm just not quite sure how to get the finishing touches on the answer that Wolfram Alpha has.

zepdrix · Answer

Hmm yah I was looking at their solution also :o hmmm

zepdrix · Answer

Well we're getting A-A=1, yes?
I'm not sure how we get A=1 from that D: Hmm

anonymous · Answer

oh wait..you are right! yes A is not 1.

zepdrix · Answer

Mmmm,
so remember that when you have repeated roots, you multiply by an x each time yah?

I guess that's what happening here? I didn't know that could happen with your particular solution :U mmm im rusty...

anonymous · Answer

yes yes you are completely right. Just as you are saying that I am digging back in my notes and see that! Thank you very much for your help.