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Question
The solution of the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\] is
Options
\[\phi\left( \frac{y}{x} \right) = kx\]
\[x\phi\left( \frac{y}{x} \right) = k\]
\[\phi\left( \frac{y}{x} \right) = ky\]
\[y\phi\left( \frac{y}{x} \right) = k\]
Solution
\[\phi\left( \frac{y}{x} \right) = kx\]
We have,
\[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\]
Let y = vx
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[ \therefore v + x\frac{dv}{dx} = v + \frac{\phi\left( v \right)}{\phi'\left( v \right)}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{\phi\left( v \right)}{\phi'\left( v \right)}\]
\[ \Rightarrow \frac{\phi\left( v \right)}{\phi'\left( v \right)}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[\int\frac{\phi'\left( v \right)}{\phi\left( v \right)}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| \phi\left( v \right) \right| = \log \left| x \right| + \log k\]
\[ \Rightarrow \log \left| \phi\left( \frac{y}{x} \right) \right| - \log \left| x \right| = \log k\]
\[ \Rightarrow \log\left| \frac{\phi\left( \frac{y}{x} \right)}{x} \right| = \log k\]
\[ \Rightarrow \frac{\phi\left( \frac{y}{x} \right)}{x} = k\]
\[ \Rightarrow \phi\left( \frac{y}{x} \right) = kx\]
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