Question

In Problems 17-32 use the procedures developed in this chapter to find the general solution of each differential equation.

$\mathit{y}\mathbf{"}\mathbf{-}\mathit{y}\mathbf{=}\frac{\mathbf{2}{\mathbf{e}}^{\mathbf{x}}}{{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}}$

Short answer

The general solution is $y\left(x\right)=c_1{e}^{-x}+c_2{e}^x-1+e^x{\tan }^{-1}\left(e^x\right)+e^x{\tan }^{-1}\left(e^x\right).$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

$y"-y=\frac{2e^x}{e^x+e^{-x}}$

Find homogeneous equation

Given differential equation,

$y"-y=\frac{2e^x}{e^x+e^{-x}}$

The characteristic equation for the homogeneous differential equation is

$$\begin{gathered} r^2-1=0 \\ r=-1,1 \end{gathered}$$

C.F. is $y_c=c_1{e}^{-x}+c_2{e}^x$

Using the method of variation of parameters,

$y_p\left(x\right)=y_1\left(x\right)u\left(x\right)+y_2\left(x\right)v\left(x\right)$

$y_1\left(x\right)=e^{-x}and{y}_2\left(x\right)=e^x$

Obtain Wronskian’s

Wronskian,

$$\begin{gathered} W(y_1,y_2)=y_1\left(x\right)y_2^{\text{'}}\left(x\right)-y_2\left(x\right)y_1^{\text{'}}\left(x\right) \\ =e^{-x}{e}^x-e^x(-e^{-x}) \\ =2\ne 0 \end{gathered}$$

$u\left(x\right)=-\int \frac{y_2\left(x\right)g\left(x\right)}{W\left(y_1,y_2\right)}dx,whereg\left(x\right)=\frac{2e^x}{e^x+e^{-x}}$

$$\begin{gathered} =-\int \frac{e^x\frac{2e^x}{e^x+e^{-x}}}{2}dx \\ =-\int \frac{e^{3x}dx}{e^{2x}+1} \\ =-\int \frac{t^2}{t^2+1}dt,t=e^x \end{gathered}$$

$$\begin{gathered} =-t+{\tan }^{-1}\left(t\right) \\ =-e^x+{\tan }^{-1}\left(e^x\right) \end{gathered}$$

Simplify and find general solution

$$\begin{gathered} v\left(x\right)=\int \frac{y_1\left(x\right)g\left(x\right)}{W\left(y_1,y_2\right)}dt,whereg\left(x\right)=\frac{2e^x}{e^x+e^{-x}} \\ =\int \frac{e^{-x}\frac{2e^x}{e^x+e^{-x}}}{2}dx \\ =\int \frac{e^{x}dx}{e^{2x}+1} \\ =ta{n}^{-1}\left(e^x\right) \end{gathered}$$

So, $y_p\left(x\right)=e^{-x}\left(-e^x+{\tan }^{-1}\left(e^x\right)\right)+e^x{\tan }^{-1}\left(e^x\right)$

$=-1+e^{-x}ta{n}^{-1}\left(e^x\right)+e^{x}ta{n}^{-1}\left(e^x\right)$

Therefore, the general solution is $y=y_c\left(x\right)+y_p\left(x\right)$

$y\left(x\right)=c_1{e}^{-x}+c_2{e}^x-1+e^{-x}ta{n}^{-1}\left(e^x\right)+e^{x}ta{n}^{-1}\left(e^x\right)$