Question

Find the general solution of differential equation:

3 e^x tan y dx + (1 – e^x) sec² y dy = 0

Answer 1

3e^x tan y dx = (e^x - 1) sec²y dy

⇒ \(\frac{3e^x}{e^x - 1}dx\) = \(\frac{sec^2y}{tan y}dy\)

⇒ 3\(\int\frac{e^x}{e^x-1}dx\) = \(\int\frac{sec^2y}{tan y}dy\)

⇒ 3 log|e^x - 1| = log |tan y| + log c, where log c is an integral constant

⇒ (e^x - 1)³ = c tan y which is solution of given differential equation.