Mathematics - Differential Equation Question with Solution | TestHub

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

$\Rightarrow \frac{dy}{dx}+\frac{2}{x}y=x$.  It is in linear form.

I.F.$=e^{\int \frac{2}{x}dx}=e^{2lnx}=x^2$

$\therefore y.x^2=\int x.x^2 dx=\frac{x^4}{4}+c$

$\because$At $x=1, y=-2$,

$\therefore -2\times 1^2=\frac{{\left(1\right)}^4}{4}+c$

$\Rightarrow c=-2-\frac{1}{4}=-\frac{9}{4}$

$\therefore$curve is $y.x^2=\frac{x^4}{4}-\frac{9}{4}$

$\therefore$It passes through $\left(\sqrt{3}, 0\right)$.