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

Question

Solve the following differential equations : $x ^2 \frac{d y}{d x}= x ^2+ xy - y ^2$

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Answer

$
\begin{aligned}
& x^2 \frac{d y}{d x}=x^2+x y-y^2 \\
& \therefore \frac{d y}{d x}=1+\frac{y}{x}-\frac{y^2}{x^2}
\end{aligned}
$
Put $y=v x \quad$ i.e. $\frac{y}{x}=v$
$
\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}
$
$\therefore$ (1) becomes
$
\begin{aligned}
& v+x \frac{d v}{d x}=1+v-v^2 \\
& \therefore x \frac{d v}{d x}=1-v^2 \\
& \therefore \frac{d v}{1-v^2}=\frac{d x}{x}
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{d v}{1-v^2}=\int \frac{d x}{x} \\
& \therefore \frac{1}{2} \log \left|\frac{1+v}{1-v}\right|=\log x+\log c_1
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{1}{2} \log \left|\frac{1+\frac{y}{x}}{1-\frac{y}{x}}\right|=\log \left(x c_1\right) & \\
\therefore \log \left|\frac{x+y}{x-y}\right| & =2 \log \left(x c_1\right) \\
& =\log \left(x^2 c_1{ }^2\right)
\end{aligned}
$
$\therefore \frac{x+y}{x-y}=c x^2, c=c_1{ }^2$, is the required solution.

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