# For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation $( iv)\quad x^2=2y^2 \log y : \qquad (x^2+y^2)\large\frac{dy}{dx}$$-xy=0 This is fourth part of multipart q2 ## 1 Answer Toolbox: • \large\frac{d(e^x)}{dx }$$= e^x$; $\large\frac{d(x)}{dx}$$= 1; \large\frac{d(constant\; term )} {dx}=$$0$. The product rule of differentiation states $uv = uv' + vu'$
Step 1:
Given $x^2 = 2y^2\log y$
differentiating on both sides we get,
$2x = 4y\big(\frac{dy}{dx}\big)\log y + 2y^2(\large\frac{1}{y})$
dividing throughout by 2 we get,
$x = 2y(\large\frac{dy}{dx})$$\log y +y \large\frac{dy}{dx }=\large\frac{ x}{y(2\log y+1)} Step 2: Substituting for \large\frac{dy}{dx} in the given equation we get, (x^2+y^2) [\large\frac{x}{y(2\log y+1)}]$$ - xy$
subtituting for $x^2$ we get
$[2y^2\log y + y^2]\large\frac{x}{y(2\log y + 1)}$$- xy = y^2[2\log y +1]\large\frac{x}{2\log y+1}$$- xy$
$xy-xy=0$
Hence the given function is a solution to the corresponding differential equation.