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The solution of the differential equation \(x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)\) is (Here, \(k\) is an arbitrary constant)
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Verified Answer
The correct answer is:
\(y=x \sin ^{-1}\left(\frac{k}{x}\right)\)
Given differential equation is
\(\begin{aligned}
\quad x \frac{d y}{d x} & =y-x \tan \frac{y}{x} \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right)
\end{aligned}\)
Let \(\quad y=v \cdot x\)
\(\Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}\)
So, \(\quad v+x \frac{d v}{d x}=v-\tan v \Rightarrow \frac{d v}{\tan v}=-\frac{d x}{x}\)
\(\Rightarrow \int \cot v d v=\int\left(-\frac{1}{x}\right) d x\)
\(\Rightarrow \quad \log |\sin v|=-\log |x|+\log k\)
\(\Rightarrow \quad \sin v=\frac{k}{x}\)
\(\Rightarrow \quad v=\sin ^{-1}\left(\frac{k}{x}\right) \Rightarrow y=x \sin ^{-1}\left(\frac{k}{x}\right)\)
Hence, option (b) is correct.
\(\begin{aligned}
\quad x \frac{d y}{d x} & =y-x \tan \frac{y}{x} \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right)
\end{aligned}\)
Let \(\quad y=v \cdot x\)
\(\Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}\)
So, \(\quad v+x \frac{d v}{d x}=v-\tan v \Rightarrow \frac{d v}{\tan v}=-\frac{d x}{x}\)
\(\Rightarrow \int \cot v d v=\int\left(-\frac{1}{x}\right) d x\)
\(\Rightarrow \quad \log |\sin v|=-\log |x|+\log k\)
\(\Rightarrow \quad \sin v=\frac{k}{x}\)
\(\Rightarrow \quad v=\sin ^{-1}\left(\frac{k}{x}\right) \Rightarrow y=x \sin ^{-1}\left(\frac{k}{x}\right)\)
Hence, option (b) is correct.
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