# Show that if #y= (sin^-1 x)^2# then #(1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2 = 0# ?

##### 1 Answer

We seek to show that:

# (1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2 = 0 # where#y= (sin^-1 x)^2#

Using the result:

# d/dx(arcsin x) = 1/sqrt(1-x^2) #

In conjunction with the chain rule, then differentiating

# dy/dx = (2arcsinx)/sqrt(1-x^2) #

And differentiating a second time, in conjunction with the quotient rule, we have:

# (d^2y)/(dx^2) = { (sqrt(1-x^2))(2/sqrt(1-x^2)) - ((1/2 (-2x))/sqrt(1-x^2))(2arcsinx) } / (sqrt(1-x^2))^2 #

# \ \ \ \ \ \ \ = ( 2 + (2x \ arcsinx)/sqrt(1-x^2) ) / (1-x^2) #

And so, considering the LHS of the given expression:

# LHS = (1 - x^2) (d^2y)/(dx^2) - xdy/dx - 2 #

# \ \ \ \ \ \ \ \ = (1 - x^2) {( 2 + (2x \ arcsinx)/sqrt(1-x^2) ) / (1-x^2)} - x{(2arcsinx)/sqrt(1-x^2)} - 2 #

# \ \ \ \ \ \ \ \ = 2 + (2x \ arcsinx)/sqrt(1-x^2) - (2xarcsinx)/sqrt(1-x^2) - 2 #

# \ \ \ \ \ \ \ \ = 0 \ \ \ # QED