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

• \[\phi\left( \frac{y}{x} \right) = kx\]

• \[x\phi\left( \frac{y}{x} \right) = k\]

• \[\phi\left( \frac{y}{x} \right) = ky\]

• \[y\phi\left( \frac{y}{x} \right) = k\]

Answer 1

\[\phi\left( \frac{y}{x} \right) = kx\]

 

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

\[\int\frac{\phi'\left( v \right)}{\phi\left( v \right)}dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \log \left| \phi\left( v \right) \right| = \log \left| x \right| + \log k\]

\[ \Rightarrow \log \left| \phi\left( \frac{y}{x} \right) \right| - \log \left| x \right| = \log k\]

\[ \Rightarrow \log\left| \frac{\phi\left( \frac{y}{x} \right)}{x} \right| = \log k\]

\[ \Rightarrow \frac{\phi\left( \frac{y}{x} \right)}{x} = k\]

\[ \Rightarrow \phi\left( \frac{y}{x} \right) = kx\]