Question
Solve the following differential equation:
$\cos^2(\text{x}-2\text{y}) = 1-2\frac{\text{dy}}{\text{dx}}$

Answer

We have,
$\cos^2(\text{x}-2\text{y}) = 1-2\frac{\text{dy}}{\text{dx}}$
$\Rightarrow 2\frac{\text{dy}}{\text{dx}} = 1 - \cos^2(\text{x}-2\text{y} )$
Let $\text{x}-2\text{y}=\text{v}$
$\Rightarrow1-2\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}$
$\Rightarrow 2\frac{\text{dy}}{\text{dx}} = 1 -\frac{\text{dv}}{\text{dx}}$
$\therefore 1 - \frac{\text{dv}}{\text{dx}} = 1 - \cos^2\text{v}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \cos^2\text{v}$
$\Rightarrow \sec^2 \text{v}\text{ dv} = \text{dx}$
Integrating both sides, we get
$\int\sec^2\text{v}\text{ dv} = \int \text{dx}$
$\Rightarrow \tan \text{v} = \text{x} - \text{C}$
$\Rightarrow \tan (\text{x}-2\text{y}) = \text{x}-\text{C}$
$\Rightarrow \text{x} = \tan (\text{x}-2\text{y})+\text{C}$

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