Solve the following differential equation dy dx = 1- 1+ - Vidyadip

Question

Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos\text{x}}{1+\cos\text{x}}$

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Answer

We have,
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos\text{x}}{1+\cos\text{x}}$
$\Rightarrow\text{dy}=\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$
$\Rightarrow\text{dy}=\tan^2\frac{\text{x}}{2}$
$\Rightarrow\text{dy}=\Big(\tan^2\frac{\text{x}}{2}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\tan^2\frac{\text{x}}{2}\Big)\text{dx}$
$\Rightarrow\int\text{dy}=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$\Rightarrow\text{y}=2\tan\frac{\text{x}}{2}-\text{x}+\text{C}$
so, $\Rightarrow\text{y}=2\tan\frac{\text{x}}{2}-\text{x}+\text{C}$ is defined for all $\text{x}\in\text{R}$
Hence, $\Rightarrow\text{y}=\tan^{-1}\text{x}+\text{C}$, where $\text{x}\in\text{R}$ is the solution o the given differential equation.

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