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प्रश्न
Solve the following initial value problem:
\[\left\{ x \sin^2 \left( \frac{y}{x} \right) - y \right\}dx + x dy = 0, y\left( 1 \right) = \frac{\pi}{4}\]
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उत्तर
\[{x \sin^2 \left( \frac{y}{x} \right) - y}dx + x dy = 0, y(1) = \frac{\pi}{4}\]
it is a homogeneous equation . so, we put y = vx
\[\frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ so,} v + x\frac{dv}{dx} = - \sin^2 \left( \frac{vx}{x} \right) + \frac{vx}{x}\]
\[x\frac{dv}{dx} = - si n^2 v\]
\[\frac{dv}{\sin^2 v} = - \frac{dx}{x}\]
integrating both sides, we get
\[cot\left( \frac{y}{x} \right) = \log\left| cx \right|\]
\[\text{ putting the values of }x = 1\text{ and }y = \frac{\pi}{4}\]
\[cot\left( \frac{\pi}{4} \right) = \log c\]
\[1 = \log c\]
\[c = e\]
\[\text{ Hence, }cot\left( \frac{y}{x} \right) = \log(ex)\]
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