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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 72

If y=[x+x2−1]15+[x−x2−1]15y=\left[x+\sqrt{x^2-1}\right]^{15}+\left[x-\sqrt{x^2-1}\right]^{15}, then (x2−1)d2ydx2+xdydx(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx} is equal to :

Choose one option
Answer and worked solution

Official answer: D

y=(x+x2−1)n+(x−x2−1)ny=(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n satisfies x2−1 y′=n[(x+x2−1)n−(x−x2−1)n]\sqrt{x^2-1}\,y'=n\left[(x+\sqrt{x^2-1})^n-(x-\sqrt{x^2-1})^n\right]. Differentiating again gives
(x2−1)y′′+xy′=n2y=225y.(x^2-1)y''+xy'=n^2y=225y.

Topics:Higher-order derivatives
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