I was wondering if someone could help me with this problem find the d^2y/dx^2 of x^2(y^2)-2x=3 in terms of x&y

Answer 1

x^2(y^2)-2x=3 differentiate

2 x^2 y dy/dx + 2x y^2 = 2

dy/dx = {1 - xy^2} / y x^2

x^2 y dy/dx + x y^2 = 1 differ
2x y dy/dx + x^2 d^2y/dx^2 + y^2 + 2xdy/dx = 0

dy/dx [ 2xy+2x] + y^2+ x^2 d^2y/dx^2 = 0

y^2 + x^2 d^2y/dx^2 = - dy/dx [ 2xy+2x] put in dy/dx

y^2 + x^2 d^2y/dx^2 = - [ 2xy+2x]{1 - xy^2} / y x^2

y^2 + x^2 d^2y/dx^2 = - 2 [y+1]{1 - xy^2} / y x

xy^3 + yx^3 d^2y/dx^2 = - 2 [y+1]{1 - xy^2}

d^2y/dx^2 = - {2 [y+1]{1 - xy^2} + x y^3 } / yx^3

d^2y/dx^2 = - {2 [y+1 - xy^3-xy^2] + x y^3 } / yx^3

d^2y/dx^2 = - {2y+2 -xy^3- 2xy^2} / yx^3

Answer 2

(x^2)(y^2) -2x =3
differentiate with respect to x(wrtx) :
2x(y^2) + (x^2)(2y)(dy/dx) -2 = 0
differentiate wrtx: 2(y^2) + 2x(2y)(dy/dx) + 2x(2y)(dy/dx) + (x^2)(2)(dy/dx)(dy/dx) + (x^2)(2y)(d^2y/dx^2) = 0

d^2y/dx^2 = -{8xy(dy/dx) + 2y^2 + 2(x^2)[(dy/dx)^2]}/(x^2)(2y)
= -{4xy(dy/dx) + y^2 + x^2[(dy/dx)^2)]}/(yx^2)

You can then subt. in dy/dx = (1-xy^2)/(yx^2) to obtain the eqn in terms of x&y...