ODE problem using method of undetermined coefficients $$ y'' + 3y' + 2y = \frac{e^x}{1+e^x} $$
My attempt to this problem was finding $y_h$ as usual and let $y_p$ of the form:
$$ y_p = \frac{Ae^x}{B+De^x}$$
I then go ahead and differentiate $y_p$ and substitute into the first equation to get the following:
$$ \frac{AB^3e^x-ABD^2e^{3x}}{(B+De^x)^4} + 3\frac{ABe^x(B+De^x)^2}{(B+De^x)^4}+2\frac{Ae^x(B+De^x)}{(B+De^x)^4} = \frac{e^x}{1+e^x}$$
From there I got the following:
For the 
$$6AB^3=1$$
$$12AB^2D=0$$
$$8ABD^2=0$$
$$2AD^3=0$$
and 
$$B^4=1$$
$$4B^3D=1$$
$$6B^2D^2=0$$
$$4BD^3=0$$
$$D^4=0$$
which yields an invalid result.
Can someone point out my mistakes? Thanks :).
 A: That approach only works for sums of exponentials, but you have here a rational expression. You will need the variation of constants approach, or, as it is simple here, the use of integrating factors. As in
\begin{align}
(e^x(y'+2y))'&=\frac{e^{2x}}{1+e^x}\\
e^x(y'+2y)&=\int\frac{u}{1+u}du=u-\ln(1+u)+C\tag{$u=e^x$}\\
y'+2y&=1-e^{-x}\ln(1+e^x)+Ce^{-x}
\end{align}
and then again with slightly more complicated integrals
\begin{align}
(y^{2x}y)'&=e^{2x}-e^x\ln(1+e^x)+Ce^x\\
e^{2x}y&=\frac12e^{2x}-\int \ln v\,dv +Ce^x+D\tag{$v=1+e^x$}\\
&=\frac12e^{2x}+v-v\ln v+Ce^x+D\\
y&=\frac12-e^{-2x}(1+e^x)\ln(1+e^x)+\tilde Ce^{-x}+\tilde D e^{-2x},...
\end{align}
A: The "method of undetermined coefficients" only works when the right hand side is a linear combination of exponential-type functions ($\exp$, $\cos$, $\sin$) and polynomials.

Let $u = y' + y$. Then, $u' + 2u = \dfrac{e^x}{1+e^x}$.
$(e^{2x} u)' = \dfrac{e^{3x}}{1+e^x}$
$e^{2x} u = \displaystyle \int \dfrac{e^{3x}}{1+e^x} \ \mathrm dx$
$u = \displaystyle e^{-2x} \int \dfrac{e^{3x}}{1+e^x} \ \mathrm dx$
$y' + y = \displaystyle e^{-2x} \int \dfrac{e^{3x}}{1+e^x} \ \mathrm dx$
$(e^x y)' = \displaystyle e^{-x} \int \dfrac{e^{3x}}{1+e^x} \ \mathrm dx$
$e^x y = \displaystyle \int e^{-x} \left( \int \dfrac{e^{3x}}{1+e^x} \ \mathrm dx \right) \ \mathrm dx$
$y = \displaystyle e^{-x} \int e^{-x} \left( \int \dfrac{e^{3x}}{1+e^x} \ \mathrm dx \right) \ \mathrm dx$
$y = \displaystyle -xe^{-x} + \frac12 - e^{-x} \log(e^{-x}+1) - e^{-2x} \log(e^x+1) + Ae^{-x} + Be^{-2x}$
A: You can use variation of parameters here. Note that the homogeneous solutions are $y_1 = e^{-x}$ and $y_2 = e^{-2x}$, so we look for a particular solution of the form
$$ y_p(x) = v_1(x)y_1(x) + v_2(x)y_2(x) $$
The Wronskian is
$$ W(x) = y_1{y_2}' - y_2{y_1}' = -e^{-3x} $$
so the solution is
$$ {v_1}' = -\frac{y_2(x)f(x)}{W(x)} = \frac{e^{2x}}{1+e^x} $$
$$ {v_2}' = \frac{y_1(x)f(x)}{W(x)} = -\frac{e^{3x}}{1+e^x} $$
You can integrate by substituting $u=1+e^x$
\begin{align}
v_1 &= \int \frac{e^{2x}}{1+e^x} dx = \int\frac{u-1}{u}\ du = \frac13 \big(u - \ln u\big) \\ 
v_2 &= -\int \frac{e^{3x}}{1+e^x} dx = -\int \frac{(u-1)^2}{u}\ du = -\frac13 \left(\frac{u^2}{2} - 2u + \ln u\right)
\end{align}
The final solution is
\begin{align} 
y(x) &= \bar{c_1} e^{-x} + \bar{c_2}e^{-2x} \\ 
&\quad - e^{-x}\big[\ln(1+e^x) - (1+e^x)\big] \\ 
&\quad - e^{-2x}\left[\ln(1+e^x) + \frac{(1+e^x)^2}{2} - 2(1+e^x)\right] \\
&\equiv c_1e^{-x} + c_2e^{-2x} + \frac12 - \big(e^{-x}+e^{-2x}\big)\ln(1+e^x) \\
\end{align}
