# Y2 Dx + (X2 − Xy + Y2) Dy = 0 - Mathematics

Sum

y2 dx + (x2 − xy + y2) dy = 0

#### Solution

We have,

$y^2 dx + \left( x^2 - xy + y^2 \right) dy = 0$

$\frac{dy}{dx} = \frac{- y^2}{x^2 - xy + y^2}$

This is a homogeneous differential equation.

$\text{Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx},\text{ we get}$

$v + x\frac{dv}{dx} = \frac{- v^2 x^2}{x^2 - v x^2 + v^2 x^2}$

$\Rightarrow v + x\frac{dv}{dx} = \frac{- v^2}{1 - v + v^2}$

$\Rightarrow x\frac{dv}{dx} = \frac{- v^2}{1 - v + v^2} - v$

$\Rightarrow x\frac{dv}{dx} = \frac{- v - v^3}{1 - v + v^2}$

$\Rightarrow \frac{1 - v + v^2}{v + v^3}dv = - \frac{1}{x}dx$

$\Rightarrow \frac{1 + v^2 - v}{v\left( 1 + v^2 \right)}dv = - \frac{1}{x}dx$

Integrating both sides, we get

$\int\frac{1 + v^2 - v}{v\left( 1 + v^2 \right)}dv = \int\frac{1}{x}dx$

$\Rightarrow \int\frac{1 + v^2}{v\left( 1 + v^2 \right)}dv - \int\frac{v}{v\left( 1 + v^2 \right)}dv = - \int\frac{1}{x}dx$

$\Rightarrow \int\frac{1}{v}dv - \int\frac{1}{1 + v^2}dv = - \int\frac{1}{x}dx$

$\Rightarrow \log \left| v \right| - \tan {}^{- 1} \left| v \right| = - \log \left| x \right| + \log C$

$\Rightarrow \log \left| \frac{vx}{C} \right| = \tan^{- 1} v$

$\Rightarrow \left| \frac{vx}{C} \right| = e^{\tan^{- 1} v}$

$\text{Putting }v = \frac{y}{x},\text{ we get}$

$\Rightarrow \left| y \right| = C e^{\tan^{- 1} v}$
$\text{Hence, }\left| y \right| = C e^{\tan^{- 1} v}\text{ is the required solution.}$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Exercise 22.9 | Q 20 | Page 83