Respuesta :
Answer:
[tex]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 )[/tex]
Step-by-step explanation:
Given: [tex]y''-y=\frac{1}{e^x+e^{-x}}[/tex]
First we will find homogeneous solution:
Let [tex]y=e^{rx}[/tex] be the solution of equation [tex]y''-y=0[/tex]
we get,
[tex]\left ( r^2-1 \right )e^{rx}=0[/tex]. Since [tex]e^{rx}\neq 0[/tex], we will solve equation: [tex]r^2-1=0[/tex]
[tex]r^2-1=0\Rightarrow \left ( r-1 \right )\left ( r+1 \right )=0\rightarrow r=-1 , 1[/tex]
We get homogeneous solution as [tex]y_c=c_1e^{x}+c_2e^{-x}[/tex]
For particular solution:
On comparing [tex]y''-y=\frac{1}{e^x+e^{-x}}[/tex] with [tex]y''+q(x)y'+r(x)y=g(x)[/tex], we get [tex]g(x)=\frac{1}{e^x+e^{-x}}[/tex]
Particular solution is of form [tex]Y_p(x)=y_1u_1+y_2u_2[/tex].
[tex]u_1= - \int \frac{y_2g(x)}{W(y_1,y_2)},\,\,\,\,u_2=\int \frac{y_1g(x)}{W(y_1,y_2)}[/tex]
Here, [tex]y_1=e^x\,\,,\,\,y_2=e^{-x}[/tex]
[tex]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[/tex]
[tex]u_1= - \int \frac{y_2g(x)}{W(y_1,y_2)}\\=\frac{1}{2}\int \frac{e^{-x}}{e^x+e^{-x}}\,dx\\[/tex]
Let [tex]e^{-x}=t\Rightarrow -e^{-x}\,dx=dt[/tex]
we get,
[tex]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 )[/tex]
[tex]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 )[/tex]
So, we get particular solution as:
[tex]Y_p(x)= \frac{-e^x}{4}\ln \left ( e^{-2x}+1 \right)+\frac{-e^{-x}}{4}\ln \left ( e^{2x}+1 \right )[/tex]
Therefore, solution is
[tex]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 )[/tex]