Question

For the differential equation find the general solution of
$(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

Answer 1

Given differential equation is

$(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

$(e^x+e^{-x})dy=(e^x-e^{-x})dx$

$dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx$

Integrating both sides
We get,

$\int dy=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$ (i)

Let $u=e^x+e^{-x}$

Differentiate $u$ w.r.t $u$
we get,

$\frac{du}{dx}=(e^x-e^{-x})$

$dx=\frac{du}{e^x-e^{-x}}\left(a\right)$

Putting value of $\left(a\right)$ in $\left(i\right)$ then we get,

$\int dy=\int \frac{e^x-e^{-x}}{u}\frac{du}{e^x-e^{-x}}$

$\int dy=\int \frac{du}{u}$

$y=\log |u|+c$

Now, putting $u=e^x-e^{-x}$ then we get,

$y=\log |e^x-e^{-x}|+c$

Final Answer:
Hence, the required general solution is

$y=\log |e^x-e^{-x}|+c$