Solve $y '' + 3y ' + 2y = \frac{1}{e^{x} + 1} $ I am trying to solve this differential equation.
$$y&#39;&#39; + 3y&#39; + 2y = \frac{1}{e^{x} + 1}$$

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 $y&#39;$ and $y&#39;&#39;$.

The problem is that I end up to nowhere. Can someone help me?
$y&#39;&#39;+3y&#39;+2y=\frac1{e^x+1}$

$z=y&#39;+2y,z&#39;=y&#39;&#39;+2y&#39;$

$y&#39;&#39;+3y&#39;+2y=y&#39;&#39;+2y&#39;+y&#39;+2y=z&#39;+z=\frac1{e^x+1}$

$e^xz&#39;+e^xz=\frac{e^x}{e^x+1}$

$e^xz=\int \frac{e^x}{e^x+1}dx=\ln (e^x+1)+c$

$z=e^{-x}\ln(e^x+1)+ce^{-x}$

$y&#39;+2y=e^{-x}\ln(e^x+1)+ce^{-x}$

$e^{2x}y&#39;+2e^{2x}y=e^x\ln(e^x+1)+ce^x$

$e^{2x}y=\int e^x\ln(e^x+1)+ce^xdx$

$t=e^x+1,dt=e^xdx$

$\int e^x\ln(e^x+1)dx=\int \ln tdt=t\ln t-t+c_2=(e^x+1)\ln(e^x+1)-e^x+c_3$

$e^{2x}y=(e^x+1)\ln(e^x+1)-e^x+ce^x+c_3=(e^x+1)\ln(e^x+1)+k_1e^x+k_2$

$y=\frac{e^x+1}{e^{2x}}\ln (e^x+1)+k_1e^{-x}+k_2e^{-2x}$

Unless I made any mistakes in there, that should be the complete solution.
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