Question

Find one-parameter families of solution curves of the following differential equation:-

\[e^{- y} \sec^2 y dy = dx + x dy\]

Solve the following differential equation:-

\[e^{- y} \sec^2 y dy = dx + x dy\]

Answer 1

We have,
\[ e^{- y} \sec^2 y dy = dx + x dy\]
\[ \Rightarrow dx = e^{- y} \sec^2 y dy - x dy\]
\[ \Rightarrow \frac{dx}{dy} = e^{- y} \sec^2 y - x\]
\[ \Rightarrow \frac{dx}{dy} + x = e^{- y} \sec^2 y . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form
\[\frac{dx}{dy} + Px = Q\]
where
\[P = 1\]
\[Q = e^{- y} \sec^2 y\]
\[ \therefore I . F . = e^{\int P\ dy} \]
\[ = e^{\int dy} \]
\[ = e^y \]
\[\text{ Multiplying both sides of }\left( 1 \right)\text{ by }e^y , \text{ we get }\]
\[ e^y \left( \frac{dx}{dy} + x \right) = e^y e^{- y} \sec^2 y\]
\[ \Rightarrow e^y \frac{dx}{dy} + e^y x = \sec^2 y\]
Integrating both sides with respect to y, we get
\[ e^y x = \int \sec^2 y\ dy + C\]
\[ \Rightarrow e^y x = \tan y + C\]
\[ \Rightarrow x = \left( \tan y + C \right) e^{- y} \]
\[\text{ Hence, }x = \left( \tan y + C \right) e^{- y}\text{ is the required solution.}\]