Question

Prove that x+y=xy , if d^2y/dx^2=2 (y/x)^3?

Answer 1

We cannot prove that:

`(d^2y)/(dx^2)=2 (y/x)^3 => x+y=xy`

Which given the wording is the suggested outcome, due to the fact that a second order ODE will introduce two constants of integration, and no initial condition are given. We can however prove the converse. That is:

`x+y=xy => (d^2y)/(dx^2)=2 (y/x)^3`

Let us start with `x+y=xy` and re-arrange:

`x + y = xy => xy - y = x`
`:. (x-1)y = x`
`:. y = x/(x-1)`, and also `x-1 = x/y`

Now we differentiate using the quotient rule:

`dy/dx = ( (x-1)(1) -(x)(1) ) / (x-1)^2`
`\ \ \ \ \ \ = ( x-1 -x ) / (x-1)^2`
`\ \ \ \ \ \ = ( -1 ) / (x-1)^2`

Then differentiating again:

`(d^2y)/(dx^2) = (-1)(-2)(x-1)^(-3)(1)`
`\ \ \ \ \ \ \ = 2 / (x-1)^3`
`\ \ \ \ \ \ \ = 2 / (x/y)^3`
`\ \ \ \ \ \ \ = 2 (y/x)^3 \ \ \ \` QED