Question

if y = ( sin inverse of x )² + ksin inverse of x, then
(1-x²)d²y/dx² - xdy/dx is equal to :
a) 0
b) 1
c) 2
d) y

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Answer 1

We seek to show that:

(1 − x2) d2ydx2 − xdydx − 2 = 0 where y = (sin − 1x) 2

Using the result:

ddx (arcsinx) = 1√1 − x2

In conjunction with the chain rule, then differentiating y = (sin − 1x) 2 wrt x we have:

dydx = 2arcsinx√1 − x2

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

d2ydx2 = (√1 − x2) (2√1 − x2) - (12 (−2x) √1 − x2) (2arcsinx) (√1 − x2) 2

= 2 + 2x arcsinx√1 − x21 − x2

And so, considering the LHS of the given expression:

LHS = (1 − x2) d2ydx2 − xdydx − 2

= (1 − x2) ⎧⎪⎨⎪⎩2 + 2x arcsinx√1 − x21 − x2⎫⎪⎬⎪⎭ − x {2arcsinx√1 − x2} −2

= 2 + 2x arcsinx√1 − x2−2xarcsinx√1 − x2−2

= 0 QED