Question

How do you find `(d^2y)/(dx^2)` for `x^3=2y^2+5`?

Answer 1

Taken you to a point where you can take over. Hopefully my workings are correct.

Explanation:

`2y^2=x^3-5`

`y^2=1/2x^3-5/2`

`=>y=+-(1/2x^3-5/2)^(1/2)`

Let `u= 1/2x^3-5/2" " =>" " (du)/(dx)=3/2x^2`

Consider:

`y=+-u^(1/2)" " =>" " dy/(du)=1/2u^(-1/2)`

But `dy/dx=(du)/dxxxdy/(du)`

`=>dy/dx=1/2u^(-1/2)xx3/2x^2`

`=>dy/dx=1/2(1/2x^3-5/2)^(-1/2)xx3/2x^2`

`=>dy/dx=3/4x^3(1/2x^3-5/2)^(-1/2)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Second differential

Let `u=3/4x^3" " =>" " (du)/dx=9/4x^2`

Let `v=(1/2x^3-5/2)^(-1/2)`
`=>" "(dv)/dx=-1/2(1/2x^3-5/2)^(-3/2)(3/2x^2)`
`=>" "(dv)/dx=-3/4x^2(1/2x^3-5/2)^(-3/2)`

So for the second differential we have:

`(d^2y)/(dx^2) -> v (du)/dx+u(dv)/(dx)`

`color(white)(.)`

`(d^2y)/(dx^2) ->[(1/2x^3-5/2)^(-1/2)xx9/4x^2] + [3/4x^3xx(-3/4x^2(1/2x^3-5/2)^(-3/2)]`

`color(white)(.)`

`(d^2y)/(dx^2) ->(9x^2)/(4xxsqrt(1/2x^3-5/2))- (9x^5)/(16xx(sqrt(1/2x^3-5/2))^3)`

I will let you take it from this point