$(x^2+y^2)^2$ = xy ... (i) Differentiating with respect to x, we have, 2 $(x^2+y^2)\left(2x + 2y\cdot\frac{dy}{dx}\right)$ = y + $\frac{xdy}{dx}$ ⇒ 4x $(x^2+y^2)$ + 4y $(x^2+y^2)\cdot\frac{dy}{dx}$ = y + $\frac{xdy}{dx}$ ⇒ $\frac{dy}{dx} (4x^2y+4y^3-x)$ = y - $4x^3-4xy^2$ ⇒ $\frac{dy}{dx}$ = $\frac{y-4x^3-4xy^2}{4x^2y+4y^3-x}$ y 3cos(logx) 4sin(logx) Differentiating the above function with respect to x, we have, $\frac{dy}{dx}$ = $\frac{-3\sin(\log x)}{x}$ + $\frac{4\cos(\log x)}{x}$ x $\frac{dy}{dx}$ = - 3 sin (logx) + 4 cos (log x) Again differentiating with respect to x, we have, x $\frac{d^2y}{dx^2} + \frac{dy}{dx}$ = $\frac{-3\cos(\log x)}{x}$ - $\frac{4\sin(\log x)}{x}$ ⇒ $x^2 \frac{d^2y}{dx^2} + \frac{dy}{dx}$ + $x\frac{dy}{dx}$ = - (3 cos (log x) + 4 sin (log x)) ⇒ $x^2 \frac{d^2y}{dx^2} + \frac{dy}{dx}$ + $x\frac{dy}{dx}$ = - y ⇒ $x^2 \frac{d^2y}{dx^2} + \frac{dy}{dx}$ + $x\frac{dy}{dx}$ + y = 0