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