Question

If y = sin^-1x, then

(a) (1-x^2)yn+2 – xyn+1(2n-1) = nyn(2n-1)

(b) x^2yn+2 – xyn+1(2n-1) = nyn(2n-1)

(c) (1-x^2)yn+2 – 2nxyn+1 = nyn(2n-1)

(d) (1-x^2)yn+2 – xyn+1(2n-1) +nyn(2n-1)=0

The question was asked in an interview.

My question comes from Leibniz Rule in portion Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) (1-x^2)yn+2 – xyn+1(2n-1) = nyn(2n-1)

To explain I would say: Y = sin^-1x

Differentiating it

Y1 = \(\frac{1}{\sqrt{1-x^2}}\)

(1-x^2)(y1)^2= 1

Again Differentiating we get

(1-x^2)2y1y2 – 2x(y1)^2 = 0

(1-x^2)y2 = xy1

By Leibniz Rule, Diff it n times,

(1-x^2)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn

(1-x^2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)

(1-x^2) yn+2 – xyn+1(2n-1) = nyn(2n-1).