$$y''+2y'+y=xe^{-x}$$
$$r^2+2r+1=0$$
$$y_c = c_1 e^{-x} + c_2 x e^{-x}$$
$$y_p=Ae^{-x}+Bxe^{-x}$$
$$y_p'=-Ae^{-x}-Be^{-x}(x-1)$$
$$y_p''=Ae^{-x}+Be^{-x}(x-2)$$
something is wrong here because I get x=0

Question

$$y''+2y'+y=xe^{-x}$$
$$r^2+2r+1=0$$
$$y_c = c_1 e^{-x} + c_2 x e^{-x}$$
$$y_p=Ae^{-x}+Bxe^{-x}$$
$$y_p'=-Ae^{-x}-Be^{-x}(x-1)$$
$$y_p''=Ae^{-x}+Be^{-x}(x-2)$$
something is wrong here because I get x=0

anonymous · Answer

someone said I should have a Cx there somewhere....something something...didn't quite get it

anonymous · Answer

"There's a quantitative way to justify it, but I can't recall how it goes. The point is xe^-x is the forcing term and one of the solutions to the homogeneous. So you need another term in your yp if you're ever going to get an actual solution that satisfies the original non-homogeneous equation."

turingtest · Answer

yes you are missing an extra x it seems

turingtest · Answer

normally the particular for the non-homogeneous problem you have here would be$$Y_p=(Ax+B)e^{-x}$$but you must remember that all your solutions must remain linearly independent from each other. since your guess for the particular already appears in the complimentary, you must multiply by another x, making your guess for the particular$$Y_p=x(Ax+B)e^{-x}$$that should fix your problem

anonymous · Answer

so there is no C term?

turingtest · Answer

the problem is the $c_2xe^{-x}$ term in the complimentary is the same as the $xe^{-x}$ in the particular
and no, there is no C term

anonymous · Answer

ok I'll try it again, let's see...

anonymous · Answer

$$y_p=Axe^{-x}+Bxe^{-x}$$
$$y_p'=-Ae^{-x}(x-1)-Be^{-x}(x-1)$$
$$y_p''=Ae^{-x}(x-2)+Be^{-x}(x-2)$$
$$(Ae^{-x}(x-2)+Be^{-x}(x-2))+2(-Ae^{-x}(x-1)-Be^{-x}(x-1))+$$$$(Axe^{-x}+Bxe^{-x})=xe^{-x}$$

anonymous · Answer

I somehow cancel out all of the x-terms...I guess I'm doing something wrong

anonymous · Answer

oh I see what I did, let's try again

anonymous · Answer

I did the distributive property wrong

turingtest · Answer

...and you missed an x in the guess...

turingtest · Answer

$$y_p=(Ax^2+Bx)e^{-x}$$or is that what you meant?

anonymous · Answer

doesn't it disappear in the first and second derivative?

turingtest · Answer

but then where is the 2A ?

anonymous · Answer

oh there is an x^2, yeah that's what I did wrong

turingtest · Answer

yeah, one more time ;) good luck, I don't wanna do it on paper if I can help it :P

anonymous · Answer

yeah I should be fine =P
thanks!