Question

If x = at² and y = 2at, then show that `xy(d^2y)/(dx^2) + a` = 0.

Answer 1

x = at² and y = 2at     ...(1)

Differentiating x and y w.r.t. t, we get

`(dx)/(dt) = (d)/(dt)(at^2)`

= `a d/(dt)(t^2)`

= a × 2t

= 2at                          ...(2)

and

`(dy)/(dt) = d/(dt)(2at)`

= `2a d/(dt)(t)`

= 2a × 1

= 2a

∴ `(dy)/(dx) = ((dy/dt))/((dx/dt)`

= `(2a)/(2at) = (1)/t`

and

`(d^2y)/(dx^2) = d/(dx)(1/t)`

= `d/(dt)(t^-1).(dt)/(dx)`

= `(-1)t^-2.(1)/((dx/dt)`

= `(-1)/t^2 xx (1)/(2at)`           ...[By (2)]

= `(-1)/(2at^3)`

∴ `(d^2y)/(dx^2) = (-1)/(2at^3)`

= `(-1)/(yt^2) = (-1)/(y xx x/a)`  ...[Using (1)]

∴ `xy(d^2y)/(dx^2)` = – a

∴ `xy(d^2y)/(dx^2) + a` = 0.