Question

dx + xdy = e^−y sec² y dy

Answer 1

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