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

Explanation

`y=(x+sqrt(x^(2)-1))^(m)`
`thereforelogy=m.log(x+sqrt(x^(2)-1))`
Differentiating both sides w.r.t. x, we get,
`(1)/(y).(dy)/(dx)=m(d)/(dx)[log(x+sqrt(x^(2)-1))]`
`=(m)/(x+sqrt(x^(2)-1)).(d)/(dx)(x+sqrt(x^(2)-1))`
`=(m)/(x+sqrt(x^(2)-1)).[1+(1)/(2sqrt(x^(2)-1)).(d)/(dx)(x^(2)-1)]`
`=(m)/(x+sqrt(x^(2)-1))(1+(2x)/(2sqrt(x^(2)-1)))`
`=(m)/(x+sqrt(x^(2)-1)).(sqrt(x^(2)-1)+x)/(sqrt(x^(2)-1))=(m)/(sqrt(x^(2)-1))`
`thereforesqrt(x^(2)-1).(dy)/(dx)=my therefore(x^(2)-1)((dy)/(dx))^(2)=m^(2)y^(2)`
Differentiating both sides w.r.t. x, we get,
`(x^(2)-1)(d)/(dx)((dy)/(dx))^(2)+((dy)/(dx))^(2).(d)/(dx)(x^(2)-1)=m^(2)(d)/(dx)(y^(2))`
`(x^(2)-1).2(dy)/(dx)(d^(2)y)/(dx^(2))+2x((dy)/(dx))^(2)=2m^(2)y.(dy)/(dx)`
Cancelling `2(dy)/(dx)` throughout, we get, `(x^(2)-1)(d^(2)y)/(dx^(2))+x(dy)/(dx)=m^(2)y`