Solve yx d y d x = x ^2+2 y ^2 - Vidyadip

Question

Solve $yx \frac{d y}{d x}= x ^2+2 y ^2$

✓

Answer

$
\begin{aligned}
& x y \frac{d y}{d x}=x^2+2 y^2 \\
& \therefore \frac{d y}{d x}=x^2+\frac{2 y^2}{x y}
\end{aligned}
$
Put $y=t x \ldots$ (ii)
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=t+x \frac{d t}{d x}
$
Substituting (ii) and (iii) in (i), we get
$
\begin{aligned}
& t+x \frac{d t}{d x}=\frac{x^2+2 t^2 x^2}{x(t x)} \\
& \therefore t+x \frac{d t}{d x}=\frac{x^2\left(1+2 t^2\right)}{x^2 t} \\
& \therefore x \frac{d t}{d x} \frac{1+2 t^2}{t}-t=\frac{1+t^2}{t} \\
& \therefore \frac{t}{1+t^2} d t=\frac{1}{x} d x
\end{aligned}
$
Integrating on both sides, we get
$
\begin{aligned}
& \frac{1}{2} \int \frac{2 t}{1+t^2} d t=\int \frac{d x}{x} \\
& \therefore \frac{1}{2} \log \left|1+t^2\right|=\log |x|+\log |c|
\end{aligned}
$
$\begin{aligned} & \therefore \log \left|1+ t ^2\right|=2 \log | x |+2 \log | c | \\ & =\log \left| x ^2\right|+\log \left| c ^2\right| \\ & \therefore \log \left|1+ t ^2\right|=\log \left| c ^2 x ^2\right| \\ & \therefore 1+ t ^2= c ^2 x ^2 \\ & \therefore 1+\frac{y^2}{x^2}=c^2 x^2 \\ & \therefore x ^2+ y ^2= c ^2 x ^4\end{aligned}$

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