Question

If y^3+x^3-3axy=0,show that y"=-2a^3xy/(y^2-ax)^3

Answer 1

Given[math]~x^3+ y^3 - 3axy = 0\quad (1)[/math]Differentiate w.r.t.[math]~x.~[/math]We get,[math] 3x^2 + 3y^2(\frac{\text{d}y}{\text{d}x})- (3ay + 3ax(\frac{\text{d}y}{\text{d}x})) = 0[/math][math]\implies\frac{\text{d}y}{\text{d}x} =\frac{x^2 - ay}{ax - y^2} \quad (2)[/math]Differentiate again w.r.t.[math]~x.~[/math]We get,[math]\implies\frac{\text{d}^2y}{\text{d}x^2} =\frac{(ax - y^2)(2x - a(\frac{\text{d}y}{\text{d}x})) - (x^2 - ay)(a - 2y(\frac{\text{d}y}{\text{d}x}))}{(ax- y^2)^2} [/math][math]\implies\frac{\text{d}^2y}{\text{d}x^2} =\frac{(ax - y^2)(2x - a(\frac{x^2 - ay}{ax - y^2})) - (x^2 - ay)(a - 2y(\frac{x^2 - ay}{ax - y^2}))}{(ax - y^2)^2}\quad[/math][Using[math]~(2)[/math]][math] [/math][math]\implies\frac{\text{d}^2y}{\text{d}x^2} =\frac{(ax - y^2)(2x - a(\frac{x^2 - ay}{ax - y^2})) - (x^2 - ay)(a - 2y(\frac{x^2 - ay}{ax - y^2}))}{(ax - y^2)^2}[/math][math]= \frac{(ax - y^2)(2x(ax - y^2)- a(x^2 - ay)) - (x^2 - ay)(a(ax - y^2) - 2y(x^2 - ay))}{(ax - y^2)^3}[/math][math]= \frac{(ax - y^2)(2ax^2- 2xy^2 - ax^2 + a^2y) - (x^2 - ay)(a^2x - ay^2 - 2x^2y + 2ay^2)}{(ax - y^2)^3}[/math][math]= \frac{(ax - y^2)(ax^2- 2xy^2+ a^2y) - (x^2 - ay)(a^2x + ay^2 - 2x^2y)}{(ax - y^2)^3}[/math][math]=\frac{(a^2x^3 - 2ax^2y^2 + a^3yx- ax^2y^2 + 2xy^4 - a^2y^3) - (a^2x^3 + ay^2x^2 - 2x^4y - a^3xy - a^2y^3 + 2ax^2y^2)}{(ax - y^2)^3}[/math][math]=\frac{2a^3yx - 6ax^2y^2 + 2xy^4+ 2x^4y}{(ax - y^2)^3}[/math][math]=\frac{2a^3yx - 6ax^2y^2 + 2xy(x^3 + y^3)}{(ax - y^2)^3}[/math][math]=\frac{2a^3yx - 6ax^2y^2 + 2xy(3axy)}{(ax - y^2)^3}\quad [[/math]Using[math]~(1)] [/math][math]=\frac{2a^3yx}{(ax - y^2)^3}[/math][math]\therefore\boxed{\frac{\text{d}^2y}{\text{d}x^2} =\frac{2a^3xy}{(ax - y^2)^3}}[/math]