Question

logy=sin inversex show (1-x2)d2/dx2=xdy/dx+y

Answer 1

Given   Log y = Sin⁻¹ x  ---(1)
To show (1 - x²) d²y/dx² = x dy/dx + y

Differentiating given equation (1)  wrt x.

   1/y * y' = 1/√[1 - x²] 
   √[1 - x² ]  * y' = y       ---- (2)

Differentiate again wrt x
   √[1 - x²] * y'' + y' * (1/2) * 1/√(1-x²) ] (-2x) = y'
   √[1-x²] * y" = y' + x y'/√[1-x²]
   (1-x²) y'' = √(1-x²) * y' + x y'
                 = y + x y'                   using (2).

Proved.

Answer 2

ANSWER:-----

Log y = Sin⁻¹ x  ---(1)

To show (1 - x²) d²y/dx² = x dy/dx + y

Differentiating given equation (1)  wrt x.

   1/y * y' = 1/√[1 - x²] 

   √[1 - x² ]  * y' = y       ---- (2)

Differentiate withx

   √[1 - x²] * y'' + y' * (1/2) * 1/√(1-x²) ] (-2x) = y'

   √[1-x²] * y" = y' + x y'/√[1-x²]

   (1-x²) y'' = √(1-x²) * y' + x y'

                 = y + x y'                   using (2).

HENCE PROVED:----

hope it helps:----