(a) Find Wronskian $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x)$, $y_2(x)$ for ODE
$$-y''(e^{x}+e^{-x}+2)+y'(e^{x}-e^{-x})+2y=0.$$
Dividing but sides by $-(e^{x}+e^{-x}+2)$, we get
$$L[y]=y''-y'\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}-y\frac{2}{e^{x}+e^{-x}+2}=0,$$
where $p(x)=-\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}$, and $q(t)=-\frac{2}{e^{x}+e^{-x}+2}$.
By Abel's Theorem,
$$\begin{align}W(y_1,y_2)(x)&=c\exp(\int-{p(x)dx})\\&=c\exp(\int\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2}dx)\\&=c (e^x+e^{-x}+2).\end{align}$$
Let $c=1 \Rightarrow W(y_1,y_2)(x)=e^x+e^{-x}+2$.
b) Check that $y_1(x)=e^{x}+1$ is a solution and find another linearly independent solution.
Since $y_1(x)=e^{x}+1 \Rightarrow y_1 '(x)=e^{x}$, and $y_1 ''(x)=e^{x}$
Plugging $y_1$, $y_1 '$, and $y_1 ''$ into the ODE, we have
$$\begin{align}-e^{x}(e^{x}+e^{-x}+2)+e^{x}(e^{x}-e^{-x})+2(e^{x}+1)&=0\\-2e^x-e^{2x}-2+2e^{x}+e^{2x}+2&=0\end{align}$$
$y_1(x)$ satisfies the ODE $\Rightarrow$ $y_1(x)$ is a solution.
Given $y_1(x)$, we can find another linearly independent solution.
We know from the definition of the Wronskian that
$$W(y_1,y_2)(x)=y_1y_2 '-y_1 'y_2=(e^x+1)y_2 '-y_2 e^x$$
Equating the two expressions for the Wronskian, we get
$$(e^x+1)y_2 '-y_2 e^x=e^x+e^{-x}+2$$
Dividing both sides by $e^x+1$, and multiplying by integrating factor $\mu=\frac{1}{e^x+1}$,
$$\frac{1}{e^x+1}y_2=\int{\frac{e^x+e^{-x}+2}{e^{2x}+2e^x+1}}dx+C$$
$$\frac{1}{e^x+1}y_2=-\frac{1}{e^x}$$
$$y_2=-\frac{e^x+1}{e^x}$$
$$y_2=-1-\frac{1}{e^{x}}$$
c) Write the general solution. Find solution such that $y(0)=0$, $y'(0)=2$
The general solution to the ODE is
$$y(x)=c_1 (e^x+1)+c_2(-1-\frac{1}{e^{x}}).$$
$\Rightarrow y'(x)=c_1e^x+c_2e^{-x}$
$$\cases{c_1-c_2=0\\c_1+c_2=2} \Rightarrow \cases{c_1=1\\c_2=1}$$
Thus, the solution that satisfies $y(0)=1$, $y'(0)=2$ is
$$y(x)=e^x-\frac{1}{e^{x}}.$$
