Form of particular solution to inhomogeneous differential equation I want to solve the initial value problem
$$x''(t)+3x'(t)+2x(t)=\frac{1}{1+e^t}, \quad x(0)=2\ln2,~x'(0)=1-3\ln 2$$
considering the homogeneous couterpart
$$x''(t)+3x'(t)+2x(t)=0$$
with the characteristic polynomial $$ \chi(\lambda)=\lambda^2+3\lambda+2=(\lambda+1)(\lambda+2)$$
the general solution to the homogeneous equation is of the form
$$x(t)=c_1e^{-1t}+c_2e^{-2t}$$
using my initial conditions
$$x(0)=c_1+c_2=2\ln 2, \quad x'(0)=-c_1-2c_2=1-3\ln 2$$
I have $$c_1=1+\ln2,~c_2=-1+\ln 2,\quad x(t)=(1+\ln 2)e^{-t}+(-1+\ln 2)e^{-2t}$$
Solution using @Karn Watcharasupat's method:
$u := x+x', \quad u'=x'+x''$ the equation becomes a first order linear differential equation, with $\mu$ integration factor such that $\mu 2=\mu'\Rightarrow \mu  = e^{2t}$
$$u'+2u=\frac{1}{1+e^t}$$
$$(\mu u)'=\mu \frac{1}{1+e^t}\iff \frac{d\left(e^{2t}u\right)}{dt}=\frac{e^{2t}}{1+e^t}$$
$$ e^{2t}u=\int \frac{e^{2t}}{1+e^t}dt=\int \frac{v-1}{v} dv=\int 1 dv - \int \frac{1}{v}dv=v-\ln |v|=1+e^t-\ln|1+e^t|+C_1$$ for $v:=1+e^t$.
And then $$u=\frac{e^t-\ln|1+e^t|+C_2}{e^{2t}}=e^{-t}-e^{-2 t}\ln|1+e^t|+e^{-2t}C_2$$
Plugging back in $$x+x'=C_2e^{-2t}+e^{-t}-e^{-2 t}\ln(1+e^t)$$ which is similarly a first order ODE with $\mu=\mu'\Rightarrow \mu=e^t$
$$\left(x\mu\right)'=\mu\cdot \left(C_2e^{-2t}+e^{-t}-e^{-2 t}\ln(1+e^t)\right)=e^t\left(C_2e^{-2t}+e^{-t}-e^{-2 t}\ln(1+e^t)\right)$$ $$\frac{d\left(xe^t\right)}{dt}=\left(C_2e^{-t}+1-e^{- t}\ln(1+e^t)\right)$$
\begin{align}xe^t&=\int \left(C_2e^{-t}+1-e^{- t}\ln(1+e^t)\right) dt\\&=\int C_2e^{-t} dt+\int 1 dt -\int e^{- t}\ln(1+e^t) dt\\&=-C_2e^{-t}+t-\left[-e^{-t}\ln \left(e^t+1\right)+t-\ln \left|e^t+1\right|+C\right]\\&=-C_2e^{-t}+t+e^{-t}\ln \left(e^t+1\right)-t+\ln \left(e^t+1\right)+C\\&=C_2e^{-t}+e^{-t}\ln \left(e^t+1\right)+\ln \left(e^t+1\right)+C\end{align}
And finally
$$x=\frac{C_2e^{-t}+e^{-t}\ln \left(e^t+1\right)+\ln \left(e^t+1\right)+C}{e^t}=C_2e^{-2t}+e^{-2t}\ln \left(e^t+1\right)+e^{-t}\ln \left(e^t+1\right)+Ce^{-t}$$
$$x(t)=C_1e^{-t}+C_2e^{-2t}+e^{-t}\ln \left(e^t+1\right)+e^{-2t}\ln \left(e^t+1\right)$$
$$x'(t)=-C_1e^{-t}-2C_2e^{-2t}-e^{-t}\ln \left(e^t+1\right)+\frac{1}{e^t+1}-2e^{-2t}\ln \left(e^t+1\right)+\frac{e^{-t}}{e^t+1}$$
Finding $C_1,~C_2$
$$x(0)=C_1+C_2+2\ln(2)=2\ln(2)\Rightarrow C_1=-C_2$$
$$x'(0)=-C_1-2C_2-\ln 2+\frac{1}{2}-2\ln 2+\frac{1}{2}=1-3\ln 2 \Rightarrow C_1=-2C_2$$
So $$C_1=C_2=0$$ and the solution to the IVP is
$$x(t)=e^{-t}\ln \left(e^t+1\right)+e^{-2t}\ln \left(e^t+1\right)$$
 A: You can only apply the guessing process if the left side has constant coefficients and the right side only terms of the form polynomial times exponential.
As your right side is not of this form, you will need to apply the variation of constants method. 
$$
y(t)=c_1(t)y_1(t)+c_2(t)y_2(t)
$$
where
$$
c_1'(t)y_1(t)+c_2'(t)y_2(t)=0\\
c_1'(t)y_1'(t)+c_2'(t)y_2'(t)=f(t)
$$
so that
$$
c_1'(t)=-\frac{f(t)y_2(t)}{W(t)},\\
c_2'(t)=\frac{f(t)y_1(t)}{W(t)},\\
W(t)=y_1(t)y_2'(t)-y_1'(t)y_2(t),
$$
As $W(t)=\det\pmatrix{e^{-t}&e^{-2t}\\-e^{-t}&-2e^{-2t}}=-e^{-3t}$. This gives the coefficient functions as
This involves computing the integrals
$$
c_1(t)=-\int\frac{e^t}{1+e^t}dt\text{ and }c_2(t)=\int\frac{e^{2t}}{1+e^t}dt
$$
which both are symbolically solvable.
A: I will first use the substitution $$u=x+x'$$
So we have
$$x''(t)+3x'(t)+2x(t)=\frac{1}{1+e^t}$$
transforming into
$$u'+2u=\frac{1}{1+e^t}$$
So we have the IF
$$e^{\int 2 \ dt}=e^{2t}$$
So
\begin{align}
\frac{d}{dt}(e^{2t}u)&=\frac{e^{2t}}{1+e^t}\\
e^{2t}u&=\int \frac{e^{2t}}{1+e^t}\ dt\\
&=e^t - \ln(1 + e^t)+C_1\\
u&=C_1e^{-2t}+e^{-t}-e^{-2t}\ln(1 + e^t)\\
\end{align}
Substituting $u=x+x'$ back in,
\begin{align}
x+x'&=C_1e^{-2t}+e^{-t}-e^{-2t}\ln(1 + e^t)\\
\end{align}
This is a first order LODE again with IF $=e^t$.
\begin{align}
\frac{d}{dt}(e^{t}x)&=(e^t)(C_1e^{-2t}+e^{-t}-e^{-2t}\ln(1 + e^t))\\
&=C_1e^{-t}+1 - e^{-t} \ln(1 + e^t)\\
e^{t}x&=\int C_1e^{-t}+ 1 - e^{-t} \ln(1 + e^t) \ dt\\
&=-C_1e^{-t}+ \ln(1 + e^{-t}) + e^{-t} \ln(1 + e^t)+C_2\\
x&=-C_1e^{-2t}+C_2e^{-t}+ e^{-t}\ln(1 + e^{-t}) + e^{-2t} \ln(1 + e^t)
\end{align}
