Question

For the question given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$x^2=2y^{2}logy:(x^2+y^2)\frac{dy}{dx}-xy=0.$

Answer 1

Given, $x^2=2y^{2}logy$ ....(i)
On differentiating both sides w.r.t. x, we get
$2x=2[y^2\times \frac{1}{y}\frac{dy}{dx}+logy\times 2y\frac{dy}{dx}]\Rightarrow 2x=2(y+2ylogy)\frac{dy}{dx}\Rightarrow x=(y+2ylogy)\frac{dy}{dx}$
On multiplying both sides by y, we get
$xy=(y^2+2y^{2}logy)\frac{dy}{dx}\Rightarrow xy=(y^2+x^2)\frac{dy}{dx}$ [Using Eq. (i),~ $x^2=2y^{2}logy$]
$\Rightarrow (x^2+y^2)\frac{dy}{dx}-xy=0$