# Show that

Question:

$\left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0$

Solution:

The given differential equation is:

$\left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0$

$\Rightarrow\left(e^{x}+e^{-x}\right) d y=\left(e^{x}-e^{-x}\right) d x$

$\Rightarrow d y=\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x$

Integrating both sides of this equation, we get:

$\int d y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}$

$\Rightarrow y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}$      ...(1)

Differentiating both sides with respect to x, we get:

$y=\int \frac{1}{t} d t+\mathrm{C}$

$\Rightarrow y=\log (t)+\mathrm{C}$

$\Rightarrow y=\log \left(e^{x}+e^{-x}\right)+\mathrm{C}$

This is the required general solution of the given differential equation.