find d2y/dx2 in terms of x and y. x^2y^2-4x=8
The expression is: x^2.y^2 - 4x = 8
Applying thee product rule to x^2.y^2, we get
x^2[(2y).y'] + (2x).y^2 - 4 = 0
(Note that d(y^2) / dx = [d(y^2) / dy] [dy/dx] = 2y.y')
Continuing,
x^2.y.y' + x.y^2 - 2 = 0 ------------ (1)
x^2.y.y' = -x.y^2 + 2
y' = (-x.y^2 + 2) / (x^2.y) --------- (2)
Differentiating (1),
2x.(y.y') + x^2.(y.y')' + 1.y^2 + x.(2y.y') = 0
2x.(y.y') + x^2[(y')^2 + y.y''] + 1.y^2 + x.(2y.y') = 0
Substituting for y' = (2 - x.y^2) / (x^2.y) from (2)
2x.(y.[(2 - x.y^2) / (x^2.y)]) + x^2[[(2 - x.y^2) / (x^2.y)]^2 + y.y''] + y^2 + x.(2y.[(2 - x.y^2) / (x^2.y)]) = 0
2([(2 - x.y^2) / (x)]) + [(2 - x.y^2)^2 / (x^2.y^2)] + x^2.y.y'' + y^2 + 2.[(2 - x.y^2) / (x)] = 0
4([(2 - x.y^2) / (x)]) + (2 - x.y^2)^2 / (x^2.y^2) + y^2 = -x^2.y.y''
4x.y^2(2 - x.y^2) / (x^2.y^2) + (2 - x.y^2)^2 / (x^2.y^2) + x^2.y^4/(x^2.y^2) = -x^2.y.y''
[4x.y^2(2 - x.y^2) + (2 - x.y^2)^2 + x^2.y^4] / (x^2.y^2) = -x^2.y.y''
[(2 - x.y^2)[4x.y^2 + 2 - x.y^2] + x^2.y^4] / (x^2.y^2) = -x^2.y.y''
[(2 - x.y^2)[3x.y^2 + 2] + x^2.y^4] / (x^2.y^2) = -x^2.y.y''
[6x.y^2 + 4 - 3x^2.y^4 - 2x.y^2 + x^2.y^4] / (x^2.y^2) = -x^2.y.y''
[4x.y^2 + 4 - 2x^2.y^4] / (x^2.y^2) = -x^2.y.y''
y'' = 2(x^2.y^4 - 2x.y^2 - 2)/(x^4.y^2)