Solve the following differential equations : (x^2-y^2 ) d x+2 x y… - Vidyadip

Question

Solve the following differential equations : $\left(x^2-y^2\right) d x+2 x y d y=0$

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Answer

$
\begin{aligned}
& \left(x^2-y^2\right) d x+2 x y d y=0 \\
& \therefore 2 x y d y=-\left(x^2-y^2\right) d x=\left(y^2-x^2\right) d x
\end{aligned}
$
$
\therefore \frac{d y}{d x}=\frac{y^2-x^2}{2 x y}
$
Put $y=v x \quad \therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 x \cdot v x}$
$\therefore v+x \frac{d v}{d x}=\frac{v^2-1}{2 v}$
$\therefore x \frac{d v}{d x}=\frac{v^2-1}{2 v}-v=\frac{v^2-1-2 v^2}{2 v}$
$\therefore x \frac{d v}{d x}=\frac{-1-v^2}{2 v}=-\left(\frac{1+v^2}{2 v}\right)$
$
\therefore \frac{2 v}{1+v^2} d v=-\frac{1}{x} d x
$
Integrating, we get
$
\begin{aligned}
& \int \frac{2 v}{1+v^2} d v=-\int \frac{1}{x} d x \\
\therefore & \log \left|1+v^2\right|=-\log x+\log c \\
\therefore & {\left[\because \frac{d}{d v}\left(1+v^2\right)=2 v \text { and } \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right] } \\
\therefore & \log \left|1+\frac{y^2}{x^2}\right|=-\log x+\log c \\
\therefore & \log \left|\frac{x^2+y^2}{}\right|=\log \left|\frac{c}{-c}\right|
\end{aligned}
$
$
\begin{aligned}
& \therefore \frac{x^2+y^2}{x^2}=\frac{c}{x} \\
& \therefore x^2+y^2=c x
\end{aligned}
$
This is the general solution.

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