Question
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
✓
Answer
We have
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
$=\frac{2\sin^2\text{y}}{2\cos^2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\tan^2\text{y}$
$\frac{\text{dy}}{\tan^2\text{y}}=\text{dx}$
$\int\cot^2\text{y dy}=\int\text{dx}$
$\int(\text{cosec}^2\text{y}-1)\text{dy}=\int\text{dx}$
$-\cot\text{y}-\text{y}+\text{C}=\text{x}$
$\text{C}=\text{x}+\text{y}+\cot\text{y}$
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
$=\frac{2\sin^2\text{y}}{2\cos^2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\tan^2\text{y}$
$\frac{\text{dy}}{\tan^2\text{y}}=\text{dx}$
$\int\cot^2\text{y dy}=\int\text{dx}$
$\int(\text{cosec}^2\text{y}-1)\text{dy}=\int\text{dx}$
$-\cot\text{y}-\text{y}+\text{C}=\text{x}$
$\text{C}=\text{x}+\text{y}+\cot\text{y}$
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