Solve $y '' + 3y ' + 2y = \frac{1}{e^{x} + 1} $ I am trying to solve this differential equation.
$$y'' + 3y' + 2y = \frac{1}{e^{x} + 1}$$
I know that I have to solve $$x^2+3x+2=0$$
when the solutions of this equation are $(x_1,x_2) = (-2,-1)$
so $$y_0(x) = c_1 e^{-2x}+c_2e^{-x}$$
Then I am looking for a solution of $y(x) = K \frac{1}{e^x +1}$ and I calculate $y'$ and $y''$.
The problem is that I end up to nowhere. Can someone help me? 
 A: $y''+3y'+2y=\frac1{e^x+1}$
$z=y'+2y,z'=y''+2y'$
$y''+3y'+2y=y''+2y'+y'+2y=z'+z=\frac1{e^x+1}$
$e^xz'+e^xz=\frac{e^x}{e^x+1}$
$e^xz=\int \frac{e^x}{e^x+1}dx=\ln (e^x+1)+c$
$z=e^{-x}\ln(e^x+1)+ce^{-x}$
$y'+2y=e^{-x}\ln(e^x+1)+ce^{-x}$
$e^{2x}y'+2e^{2x}y=e^x\ln(e^x+1)+ce^x$
$e^{2x}y=\int e^x\ln(e^x+1)+ce^xdx$
$t=e^x+1,dt=e^xdx$
$\int e^x\ln(e^x+1)dx=\int \ln tdt=t\ln t-t+c_2=(e^x+1)\ln(e^x+1)-e^x+c_3$
$e^{2x}y=(e^x+1)\ln(e^x+1)-e^x+ce^x+c_3=(e^x+1)\ln(e^x+1)+k_1e^x+k_2$
$y=\frac{e^x+1}{e^{2x}}\ln (e^x+1)+k_1e^{-x}+k_2e^{-2x}$
Unless I made any mistakes in there, that should be the complete solution.
A: Here is one of several methods. Let $G(x)$ (a Green function of your equation) be such that


*

*$G$ solves the homogeneous equation away from $x=0$

*$G$ is continuous and there is a jump of $G'$ by $1$ at $x=0$.
For example we can take $G(x)=-e^{-2|x|}/4$. A solution of the inhomogeneous equation is then
$$y(x)=\int_{-\infty}^\infty G(t) \frac{1}{e^{x-t}+1} dt=-\frac{1}{4}\int_0^\infty\frac{e^{-2t}}{e^{x-t}+1}+\frac{e^{-2t}}{e^{x+t}+1} dt.$$
A: let homogeneous solutions of this problem y1 and y2. and y1=exp(x) and y2=exp(2x)
and let the nonhomogeneous solution of this problem yp.
and let yp u1*exp(x) +u2*exp(2x)
then, wronskian of exp(x) and exp(2x) = exp(3x)
then, differentiation of u1 = -exp(-3x)exp(2x)(1/(exp(x)+1))
and differentiation of u2 = exp(-3x)exp(x)(1/(exp(x)+1))
u can integrate differentiation of u1 and u2 and u can get u1, u2
finally, the solution is c1*y1+c2*y2+yp
