#### Question

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

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

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

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

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

#### Solution

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

We have,

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,

\[\int\frac{1}{1 + v^2}dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \tan^{- 1} v = \log x + C\]

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

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,

\[\int\frac{1}{1 + v^2}dv = \int\frac{1}{x}dx\]

\[ \Rightarrow \tan^{- 1} v = \log x + C\]

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

Is there an error in this question or solution?

Solution The Solution of the Differential Equation D Y D X = X 2 + X Y + Y 2 X 2 , is (A) Tan − 1 ( X Y ) = Log Y + C (B) Tan − 1 ( Y X ) = Log X + C (C) Tan − 1 ( X Y ) = Log X + C Concept: General and Particular Solutions of a Differential Equation.