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.