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Question
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Solution
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
\[\int \sec^2 v dv = \int dx\]
\[ \Rightarrow \tan v = x - C\]
\[ \Rightarrow \tan\left( x - 2y \right) = x - C\]
\[ \Rightarrow x = \tan\left( x - 2y \right) + C\]
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