Solve the following differential equations: dx+ ^2y dy=0 - Vidyadip

Question

Solve the following differential equations:

$\tan\text{y dx}+\sec^2\text{y}\tan\text{x dy}=0$

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Answer

We have,
$\tan\text{y dx}+\sec^2\text{y}\tan\text{x dy}=0$
$\Rightarrow\sec^2\text{y}\tan\text{x dy}=-\tan\text{y dx}$
$\Rightarrow\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}=-\frac{1}{\tan\text{x}}\text{dx}$
$\Rightarrow\frac{1}{\cos^2\text{y}}\times\frac{\cos\text{y}}{\sin\text{y}}\text{dy}=-\cot\text{x dx}$
$\Rightarrow\frac{1}{\sin\text{y}\cos\text{y}}\text{dy}=-\cot\text{x dx}$
$\Rightarrow\frac{2}{\sin2\text{y}}\text{dy}=-\cot\text{x dx}$
$\Rightarrow2\text{ cosec }2\text{y dy}=-\cot\text{x dx}$
Integrating both sides, we get
$2\int\text{cosec}\text{ 2y dy}=-\int\cot\text{x dx}$
$\Rightarrow\log\tan\text{x}=-\log\sin\text{x}=\log\text{C}$
$\Rightarrow\log \tan\text{x}+\log\sin\text{x}=\log\text{C}$
$\Rightarrow\log(\tan\text{ x}\times\sin\text{x})=\log\text{C}$
$\Rightarrow\tan\text{x}\times\sin\text{x}=\text{C}$

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