- #1

ToastIQ

- 11

- 0

I've been stuck on this problem for 2 days. I'm new here and I've just spent another hour on trying to figure out Latex, but it always ends up in a mess so I'll try without. I hope that's okay.

The equation is

y''+6y'+9y=4e

^{-x}

with the value boundaries (I think that's what it's called in english?) y(0)=2 and y'(0)=-2.

I got the complementary solution to y

_{c}=C

_{1}e

^{-3x}+C

_{2}e

^{-3x}.

I'm assuming my mistake is somewhere after this. I set up the particular solution:

y

_{p}=Ae

^{-x}and get

y'

_{p}=-Ae

^{-x}and y''

_{p}=Ae

^{-x}.

I get A=1. Combining complementary and particular solution:

y=C

_{1}e

^{-3x}+C

_{2}e

^{-3x}+e

^{-x}

I can't seem to get the right coefficients. The result should be

y=(1+2x)e

^{-3x}+e

^{-x}.

I've tried with and without the 4 in 4e

^{-x}, I've tried multiplying by x to get a higher degree and several other alternatives. I've been searching the internet for the most of the past 48 hours. I read somewhere that e

^{-x}is a "special case" but no example for this case was given. My textbook doesn't touch upon this at all. It would be very appreciated if someone could help me out.

Thanks in advance!