Question

If $x^3+y^3-3axy=0$ then prove that $\frac{d^{2}y}{d{x}^2}=\frac{2a^{2}xy}{{(ax-y^2)}^3}$

Answer 1

$\begin{array}{c}wehave\\ x^3+y^3-3axy=0\\ 3x^2+3y^2\frac{dy}{dx}-3a\left(x\frac{dy}{dx}+y\right)=0\\ 3x^2+3y^2\frac{dy}{dx}-3ax\frac{dy}{dx}-3ay=0\\ 3x^2-3ay+\left(3y^2-3ax\right)\frac{dy}{dx}=0\\ \frac{dy}{dx}=\frac{ay-x^2}{y^2-ax}\\ \frac{dy}{dx}=\frac{x^2-ay}{ax-y^2}\\ Again,\\ \frac{d^{2}y}{d{x}^2}=\frac{ax-y^2\left(2x-a\frac{dy}{dx}\right)-\left(x^2-ay\right)\left(a-2y\frac{dy}{dx}\right)}{{\left(ax-y^2\right)}^2}\\ \frac{d^{2}y}{d{x}^2}=\frac{ax-y^2\left(2x-a\left(\frac{x^2-ay}{ax-y^2}\right)\right)-\left(x^2-ay\right)\left(a-2y\left(\frac{x^2-ay}{ax-y^2}\right)\right)}{{\left(ax-y^2\right)}^2}\\ \frac{d^{2}y}{d{x}^2}=\frac{ax-y^2\left[\frac{2a{x}^2-2x{y}^2-a{x}^2-a^{2}y}{ax-y^2}\right]-\left(x^2-ay\right)\left[\frac{a^{2}x-a{y}^2-2x^{2}y+2a{x}^2}{ax-y^2}\right]}{{\left(ax-y^2\right)}^2}\\ \frac{d^{2}y}{d{x}^2}=\frac{ax-y^2\left(a{x}^2-a^{2}y-2x{y}^2\right)-\left(x^2-ay\right)\left(a{x}^2+a{y}^2-2x^{2}y\right)}{{\left(ax-y^2\right)}^3}\\ \frac{d^{2}y}{d{x}^2}=\frac{2a^{2}xy}{{\left(ax-y^2\right)}^3}\\ proved.\end{array}$