Solve the following equation x ^2y dx=y ^2x dx - Vidyadip

Question

Solve the following equation
$\text{x}\cos^2\text{y dx}=\text{y}\cos^2\text{x dx}$

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Answer

We have,$\text{x}\cos^2\text{y dx}=\text{y}\cos^2\text{x dx}$
$\Rightarrow\frac{\text{x}}{\cos^2\text{x}}\ \text{dx}=\frac{\text{y}}{\cos^2\text{y}}\ \text{dy}$
$\Rightarrow\text{x}\sec^2\text{x dx}=\text{y}\sec^2\text{y dy}$
Integrating both sides, we get
$\int\text{x}\sec^2\text{x dx}=\int\text{y}\sec^2\text{y dy}$
$\Rightarrow\text{x}\int\sec\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sec^2\text{x dx}\Big\}\text{dx}\\=\text{y}\int\sec^2\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}(\text{y})\int\sec^2\text{y dy}\Big\}\text{dy}$
$\Rightarrow\text{x}\tan\text{x}-\int\int\tan\text{x dx}=\text{y}\tan\text{y}-\int\tan\text{y dy}$
$\Rightarrow\text{x}\tan\text{x}-\log|\sec\text{x}|=\text{y}\tan\text{y}-\log|\sec\text{y}|+\text{C}$
$\Rightarrow\text{x}\tan\text{x}-\text{y}\tan\text{y}=\log|\sec\text{x}|-\log|\sec\text{y}|+\text{C}$
Hence, $\text{x}\tan\text{x}-\text{y}\tan\text{y}=\log|\sec\text{x}|-\log|\sec\text{y}|+\text{C}$ is the required solution.

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