Question

If y^3 - 3x^2 y - 6x = 0, show that (y^2 - x^2)d^2 y/dx^2 + 2y(dy/dx)^2 - 4xdy/dx - 2y = 0 ?

Answer 1

`(y^2 - x^2)(d^2 y)/dx^2 + 2y(dy/dx)^2 - 4xdy/dx - 2y = 0`

Explanation:

Here,

`y^3 - 3x^2 y - 6x = 0`

Diff. w. r. t. `x` ,(using Product Rule)

`3y^2*y_1-3(x^2*y_1+y*2x)-6=0`

`=>3y^2y_1-3x^2y_1-6xy-6=0`

`=>y^2y_1-x^2y_1-2xy-2=0`

`=>(y^2-x^2)y_1-2xy-2=0`

Again diff, w, r, t, `x`,(using Product Rule)

`(y^2-x^2)y_2+y_1*(2yy_1-2x)-2[xy_1+y*1]=0`

`=>(y^2-x^2)y_2+2yy_1^2-2xy_1-2xy_1-2y=0`

`=>(y^2-x^2)y_2+2yy_1^2-4xy_1-2y=0`

`=>(y^2 - x^2)(d^2 y)/dx^2 + 2y(dy/dx)^2 - 4xdy/dx - 2y = 0`

Note:

`(1)d/(dx)(y^3)=3y^2d/(dx)(y)=3y^2(dy)/(dx)=3y^2y_1`

`(2)(dy)/(dx)=y_1 and (d^2y)/(dx^2)=y_2`