Question

If 2x = `y^(1/m) + y^(-1/m)`, then show that `(x^2 - 1) (dy/dx)^2` = m²y²

Answer 1

2x = `y^(1/m) + y^(-1/m)`

Differentiating w.r.t.x,

2 = `1/my^(1/m - 1).dy/dx + (-1/m)y^(1/m - 1).dy/dx`

∴ 2 = `1/my^-1(y^(1/m) - y^(-1/m)).dy/dx`

∴ 2 = `1/(my) dy/dx(y^(1/m) - y^(-1/m))`

∴ 2my = `dy/dx(y^(1/m) - y^(-1/m))`

∴ 2my = `dy/dx sqrt((y^(1/m) + y^(1/m))^2 - 4)`

∴ 2my = `dy/dx sqrt((2x)^2 - 4)`

Squaring both sides

4m²y² = `(dy/dx)^2 (4x^2 - 4)`

∴ 4m²y² = `(dy/dx)^2 . 4(x^2 - 1)`

∴ `(x^2 - 1)(dy/dx)^2` = m²y²

Hence proved.