Question

The solution of the differential equation \[x\frac{dy}{dx} = y + x \tan\frac{y}{x}\], is

विकल्प

• \[\sin\frac{x}{y} = x + C\]

• \[\sin\frac{y}{x} = Cx\]

• \[\sin\frac{x}{y} = Cy\]

• \[\sin\frac{y}{x} = Cy\]

Answer 1

\[\sin\frac{y}{x} = Cx\]

 

We have,
\[x\frac{dy}{dx} = y + x \tan\frac{y}{x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} + \tan\frac{y}{x} . . . . . \left( 1 \right)\]
\[\text{ Let }y = vx\]
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ Putting the above value in }\left( 1 \right),\text{ we get}\]
\[v + x\frac{dv}{dx} = v + \tan v\]
\[ \Rightarrow x\frac{dv}{dx} = \tan v\]
\[ \Rightarrow \frac{dv}{\tan v} = \frac{dx}{x}\]
Integrating both sides, we get
\[\log \sin v = \log x + \log C\]
\[ \Rightarrow \log \sin v - \log x = \log C\]
\[ \Rightarrow \log\frac{\sin v}{x} = \log C\]
\[ \Rightarrow \frac{\sin v}{x} = C\]
\[ \Rightarrow \sin v = Cx\]
\[ \Rightarrow \sin\left( \frac{y}{x} \right) = Cx .........\left[\because y = vx \right]\]