Question

If $y^3-y=2x$, then prove that $\frac{d^{2}y}{d{x}^2}=\frac{-24y}{(3y^2-1{)}^3}$. Hence, show that $9(x^2-\frac{1}{27})\frac{d^{2}y}{d{x}^2}+9x\frac{dy}{dx}=y$.

Answer 1

$y^3-y=2x$

differentiate on both sides w.r.t x

$3y^2\frac{dy}{dx}-\frac{dy}{dx}=2$

$\frac{dy}{dx}=\frac{2}{(3y^2-1)}$

Again differentiate on both sides w.r.t x

$\frac{d^{2}y}{d^{2}x}=\frac{-2}{(3y^2-1{)}^2}\left(6y\right)\frac{dy}{dx}$

$\frac{d^{2}y}{d^{2}x}=\frac{-12}{(3y^2-1{)}^2}\times \frac{2}{(3y^2-1)}$ $(\because \frac{dy}{dx}=\frac{2}{(3y^2-1)})$

$\therefore \frac{d^{2}y}{d^{2}x}=\frac{-24y}{(3y^2-1{)}^3}$

$9(x^2-\frac{1}{27})\frac{d^{2}y}{d{x}^2}+9x\frac{dy}{dx}=[\frac{9(y^3-y{)}^2}{4}-\frac{1}{3}]\left(\frac{-24y}{(3y^2-1{)}^3}\right)+\frac{9(y^3-y)}{(3y^2-1)}$

$\left(\frac{27y^6-54y^4+27y^2-4}{12}\right)\left(\frac{-24y}{(3y^2-1{)}^2}\right)+\frac{9(y^3-y)}{(3y^2-1)}$

$=-2y+\frac{6y}{(3y^2-1)}+\frac{9y^3-9y}{(3y^2-1)}$

$=-2y+\frac{3y(3y^2-1)}{(3y^2-1)}$

$\therefore 9(x^2-\frac{1}{27})\frac{d^{2}y}{d^{2}x}+9x\frac{dy}{dx}=y$