Help on differential equations please!! Answer in the back of the book is different from the one I got :(
The original equation is y''+3y'+2y=1/(1+e^x)
So I have yh= (c1)e^(-x)+(c2)e^(-2x)
where c1 and c2 are constants
I used wronskian to solve for yp
and yp=ln(1+e^x)(e^(-x)+e^(-2x))-e^(-x)
with u1=ln(1+e^x)
and u2=ln(1+e^x)-e^x
so the answer is yg=yh+yp
but in the answer in the back of the book the last part of yp (the -e^-x portion) is left out and I'm not sure why.

Question

anonymous · Answer

I have y(x) using Mathematica on the screen in front of me. I could not solve this with paper and pencil if my life depended on it. Do you want the answer now?

mikaylas · Answer

The answer in the book says y=(c1)e^(-x)+(c2)e^(-2x)+ln(1+e^x)(e^(-x)+e^(-2x))

I was thinking that maybe I'm supposed to set yp equal to the right side of the original equation and maybe get rid of the -e^(-x) that way, but it's not working D:

kmeis002 · Answer

Ah here is why:

Your solution is correct, but remember that c1 is arbitrary constant, they simply combined like terms to get:

 $$ y=((C_1e^{-x}-e^{-x})+(C_2)e^{-2x}+ln(1+e^x)(e^{-x}+e^{-2x})$$
factor out and e^-x

$$ y= ((C_1-1)e^{-x}+(C_2)e^{-2x}+ln(1+e^x)(e^{-x}+e^{-2x})$$

Since C1 is arbitrary, we can just rename 
$$ C_1 = C_1-1$$

Since we dont know it yet and when the IC are known, C1 will work itself out. So the solution becomes

$$ ( y=(C_1e^{-x}+(C_2)e^{-2x}+ln(1+e^x)(e^{-x}+e^{-2x})$$

This is done in a lot of problems to simplify stuff, so dont get discouraged. Good work on finding the correct solution.

mikaylas · Answer

Thank you so much!!! :D