Question

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

Answer 1

The 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 of this equation, we get:

$∫dy=∫[\frac {e^x-e^{-x}}{e^x+e^{-x}}]dx +C$

⇒$$y=∫[\frac {e^x-e^{-x}}{e^x+e^{-x}}]dx +C ...(1)

$Let \ (e^x+e^{-x})=t$

Differentiating both sides with respect to x, we get:

$\frac {d}{dx}(e^x+e^{-x})$ = $\frac {dt}{dx}$

⇒$$e^x+e^{-x} = $\frac {dt}{dx}$

$⇒(e^x-e^{-x})dx = dt$

Substituting this value in equation (1), we get:

$y=∫\frac {1}{t}dt+C$

$⇒y=log\ (t)+C$

$⇒y=log\ (e^x+e^{-x})+C$

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