Question

The solution of the differential equation \[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}\], is

पर्याय

• \[\tan^{- 1} \left( \frac{x}{y} \right) = \log y + C\]

• \[\tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]

• \[\tan^{- 1} \left( \frac{x}{y} \right) = \log x + C\]

• \[\tan^{- 1} \left( \frac{y}{x} \right) = \log y + C\]

Answer 1

\[\tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]

 

We have,

\[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} \]  ................(1)
This is homogenous differential equation.
\[\text{ Let }y = vx\]
\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]
\[\text{ Now, putting }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ and }y = vx\text{ in }\left( 1 \right),\text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x^2 + x^2 v + x^2 v^2}{x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = 1 + v + v^2 \]
\[ \Rightarrow x\frac{dv}{dx} = 1 + v^2 \]
\[ \Rightarrow \left( \frac{1}{1 + v^2} \right)dv = \frac{1}{x}dx\]
Integrating both sides we get, 
\[\int\frac{1}{1 + v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow \tan^{- 1} v = \log x + C\]
\[ \Rightarrow \tan^{- 1} \left( \frac{y}{x} \right) = \log x + C\]