Question

\[\cos^2 \left( x - 2y \right) = 1 - 2\frac{dy}{dx}\]

Answer 1

We have, 

\[ \cos^2 \left( x - 2y \right) = 1 - 2\frac{dy}{dx}\]

\[ \Rightarrow 2\frac{dy}{dx} = 1 - \cos^2 \left( x - 2y \right)\]

\[\text{Let }x - 2y = v\]

\[ \Rightarrow 1 - 2\frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow 2\frac{dy}{dx} = 1 - \frac{dv}{dx}\]

\[ \therefore 1 - \frac{dv}{dx} = 1 - \cos^2 v\]

\[ \Rightarrow \frac{dv}{dx} = \cos^2 v\]

\[ \Rightarrow \sec^2 v dv = dx\]

Integrating both sides, we get

\[\int \sec^2 v dv = \int dx\]

\[ \Rightarrow \tan v = x - C\]

\[ \Rightarrow \tan\left( x - 2y \right) = x - C\]

\[ \Rightarrow x = \tan\left( x - 2y \right) + C\]