Question

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

Answer 1

`y(x)=c_1e^(x)+c_2e^(-x)+(-e^x)/(4)\ln \left ( e^(-2x)+1 \right)+(-e^(-x))/(4)\ln \left ( e^(2x)+1 \right )`

Explanation:

Given:
`y''-y=(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)\\eq 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=(1)/(e^x+e^(-x))` with
`y''+q(x)y'+r(x)y=g(x)`, we get
`g(x)=(1)/(e^x+e^(-x))`

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

`u_1= - \int (y_2g(x))/(W(y_1,y_2)),\,\,\,\,u_2=\int (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 (y_2g(x))/(W(y_1,y_2))\\=(1)/(2)\int (e^(-x))/(e^x+e^(-x))\,dx\\`

Let
`e^(-x)=t\Rightarrow -e^(-x)\,dx=dt`

we get,

`u_1=(-1)/(2)\int (t\,dt)/(t^2+1)\\=(-1)/(4)\int (2t\,dt)/(t^2+1)\\=(-1)/(4)\ln \left ( t^2+1 \right )\\=(-1)/(4)\ln \left ( e^(-2x)+1 \right )`

`u_2=(-1)/(2)\int (e^x\,dx)/(e^x+e^(-x))\\=(-1)/(4)\int (2t)/(t^2+1)\\=(-1)/(4)\ln \left ( t^2+1 \right )\\=(-1)/(4)\ln \left ( e^(2x)+1 \right )`

So, we get particular solution as:

`Y_p(x)= (-e^x)/(4)\ln \left ( e^(-2x)+1 \right)+(-e^(-x))/(4)\ln \left ( e^(2x)+1 \right )`

Therefore, solution is

`y(x)=c_1e^(x)+c_2e^(-x)+(-e^x)/(4)\ln \left ( e^(-2x)+1 \right)+(-e^(-x))/(4)\ln \left ( e^(2x)+1 \right )`