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

Step $1$ : Differentiate with respect to $x$,

Given :$x^3+y^3–3axy=0$.

$\Rightarrow$ $\frac{d}{dx}{x}^3+\frac{d}{dx}{y}^3-\frac{d}{dx}\left(3axy\right)=\frac{d}{dx}\left(0\right)$

$\Rightarrow$$3x^2+3y^2\frac{dy}{dx}-3a\left(1.y+x\frac{dy}{dx}\right)=0$

$\Rightarrow$ $3x^2+3y^2\frac{dy}{dx}-3ay-3ax\frac{dy}{dx}=0$

$\Rightarrow$ $x^2+y^2\frac{dy}{dx}-ay-ax\frac{dy}{dx}=0$

$\Rightarrow$ $\frac{dy}{dx}\left(ax-y^2\right)=x^2-ay$

$\Rightarrow$ $\frac{dy}{dx}=\frac{(x^2-ay)}{(ax-y^2)}$

Step $2$: Find double differentiation of $y$ with respect to $x$.

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left[\frac{(x^2-ay)}{(ax-y^2)}\right]$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{(ax-y^2){\displaystyle \frac{d}{dx}}(x^2-ay)-(x^2-ay){\displaystyle \frac{d}{dx}}(ax-y^2)}{{(ax-y^2)}^2}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{(ax-y^2)(2x-a{\displaystyle \frac{dy}{dx}})-(x^2-ay)(a-2y{\displaystyle \frac{dy}{dx}})}{{(ax-y^2)}^2}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{(ax-y^2)(2x-a{\displaystyle \left(\frac{x^2-ay}{ax-y^2}\right)})-(x^2-ay)(a-2y{\displaystyle \left(\frac{{\displaystyle x^2-ay}}{{\displaystyle ax-y^2}}\right)})}{{(ax-y^2)}^2}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{(ax-y^2)\left[{\displaystyle \frac{2a{x}^2-2x{y}^2-a{x}^2-a^{2}y}{ax-y^2}}\right]-(x^2-ay)\left[{\displaystyle \frac{a^{2}x-a{y}^2-2x^{2}y+2a{x}^2}{ax-y^2}}\right]}{{(ax-y^2)}^2}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{(ax-y^2)\left(a{x}^2-a^{2}y-2x{y}^2\right)-(x^2-ay)\left(a{x}^2+a{y}^2-2x^{2}y\right)}{{(ax-y^2)}^3}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{2a^{2}xy}{{(ax-y^2)}^3}$

Hence proved, $\frac{d^{2}y}{d{x}^2}=\frac{2a^{2}xy}{{(ax-y^2)}^3}$