Answer to Question #105505 in Differential Equations for khushi

Question

Solve, using the method of variation of parameters
d^2y/dx^2-y=2/1+e^x

Answer 1

Solution for the homogenous equation

$y''-y=0$

$D^2-1=0; D=\pm1$

$y_h=c_1e^x+c_2e^{-x}$

Using method of variations of parameters $c_1=c_1(x);c_2=c_2(x)$ will give us the system

$\begin{matrix} c_1'(x)e^x+c_2'(x)e^{-x}=0 \\ c_1'(x)e^x-c_2'(x)e^{-x}=\frac{2}{1+e^x} \end{matrix}\implies \begin{matrix} c_1'(x)e^x=\frac{1}{1+e^x} \\ c_2'(x)e^{-x}=-\frac{1}{1+e^x} \end{matrix}$

$c_1(x)=\int\frac{e^{-x}}{1+e^x}dx=\log(e^{-x}+1)-e^{-x}+c_1$

$c_2(x)=-\int\frac{e^{x}}{1+e^x}dx=-\log(e^{x}+1)+c_2$

Therefore

$y=c_1e^x+c_2e^{-x}+(\log(e^{-x}+1)-e^{-x})e^x-\log(e^{x}+1)e^{-x}$