Find the general solution of y^2dx+(x^2-xy+y^2)dy=0. - Vidyadip

Question

Find the general solution of $\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0.$

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Answer

Given, differential equation is
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
$\Rightarrow\text{y}^2\text{dx}=-(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}$
$\Rightarrow\text{y}^2\frac{\text{dy}}{\text{dx}}=-(\text{x}^2-\text{xy}+\text{y}^2)$
Dividing both sides by y²,we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{\text{x}^2}{\text{y}^2}-\frac{\text{x}}{\text{y}}+1\Big)\ ....(\text{i})$
Which is a homogeneous differential equation.
Put $\frac{\text{x}}{\text{y}}=\text{v}$ or $\text{x}=\text{vy}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v}+\text{y}\frac{\text{dy}}{\text{dx}}$
On substituting these values in Eq. (i), we get
$\text{v}+\text{y}\frac{\text{dy}}{\text{dx}}=-[\text{v}^2-\text{v}+1]$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=\text{v}^2+\text{v}-1-\text{v}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=-\text{v}^2-1$
$\Rightarrow\frac{\text{dv}}{\text{v}^2+1}=-\frac{\text{dy}}{\text{y}}$
On integrating both sides, we get
$\frac{\text{dv}}{\text{v}^2+1}=-\frac{\text{dy}}{\text{y}}$
$\tan^{-1}(\text{v})=-\log\text{y}+\text{C}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)+\log\text{y}=\text{C}$ $\Big[\because\text{v}=\frac{\text{x}}{\text{y}}\Big]$

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