Question

Use variation of parameters to find a particular solution of y" - y = 1/e^x + e^-x.

Answer 1

Answer:

$y(x)=c_1e^{x}+c_2e^{-x}+\frac{-e^x}{4}\ln \left ( e^{-2x}+1 \right)+\frac{-e^{-x}}{4}\ln \left ( e^{2x}+1 \right )$

Step-by-step explanation:

Given: $y''-y=\frac{1}{e^x+e^{-x}}$

First we will find homogeneous solution:

Let $y=e^{rx}$ be the solution of equation $y''-y=0$

we get,

$\left ( r^2-1 \right )e^{rx}=0$. Since $e^{rx}\neq 0$, we will solve equation: $r^2-1=0$

$r^2-1=0\Rightarrow \left ( r-1 \right )\left ( r+1 \right )=0\rightarrow r=-1 , 1$

We get homogeneous solution as $y_c=c_1e^{x}+c_2e^{-x}$

For particular solution:

On comparing $y''-y=\frac{1}{e^x+e^{-x}}$ with $y''+q(x)y'+r(x)y=g(x)$, we get $g(x)=\frac{1}{e^x+e^{-x}}$

Particular solution is of form $Y_p(x)=y_1u_1+y_2u_2$.

$u_1= - \int \frac{y_2g(x)}{W(y_1,y_2)},\,\,\,\,u_2=\int \frac{y_1g(x)}{W(y_1,y_2)}$

Here, $y_1=e^x\,\,,\,\,y_2=e^{-x}$

$W(y_1,y_2)=\left | \begin{matrix} y_1&y_2\\y_1'&y_2\end{matrix} \right |\\=\left | \begin{matrix} e^x& e^{-x}\\e^x&-e^{-x}\end{matrix} \right |\\=-1-1\\=-2$

$u_1= - \int \frac{y_2g(x)}{W(y_1,y_2)}\\=\frac{1}{2}\int \frac{e^{-x}}{e^x+e^{-x}}\,dx$

Let $e^{-x}=t\Rightarrow -e^{-x}\,dx=dt$

we get,

$u_1=\frac{-1}{2}\int \frac{t\,dt}{t^2+1}\\=\frac{-1}{4}\int \frac{2t\,dt}{t^2+1}\\=\frac{-1}{4}\ln \left ( t^2+1 \right )\\=\frac{-1}{4}\ln \left ( e^{-2x}+1 \right )$

$u_2=\frac{-1}{2}\int \frac{e^x\,dx}{e^x+e^{-x}}\\=\frac{-1}{4}\int \frac{2t}{t^2+1}\\=\frac{-1}{4}\ln \left ( t^2+1 \right )\\=\frac{-1}{4}\ln \left ( e^{2x}+1 \right )$

So, we get particular solution as:

$Y_p(x)= \frac{-e^x}{4}\ln \left ( e^{-2x}+1 \right)+\frac{-e^{-x}}{4}\ln \left ( e^{2x}+1 \right )$

Therefore, solution is

$y(x)=c_1e^{x}+c_2e^{-x}+\frac{-e^x}{4}\ln \left ( e^{-2x}+1 \right)+\frac{-e^{-x}}{4}\ln \left ( e^{2x}+1 \right )$