Question

Find nth derivative of x^2y2 + xy1 + y = 0

(a) x^2yn+2 + (2n+1)xyn+1 + (n^2+1)yn = 0

(b) x^2yn+2 + nxyn+1 + (n^2+1)yn = 0

(c) x^2yn+2 + (2n+1)xyn+1 + n^2yn = 0

(d) x^2yn+2 + 2nxyn+1 + n^2yn = 0

This question was addressed to me during an interview for a job.

Question is taken from Leibniz Rule topic in division Differential Calculus of Engineering Mathematics

Answer 1

Right option is (a) x^2yn+2 + (2n+1)xyn+1 + (n^2+1)yn = 0

To explain I would say: x^2y2 + xy1 + y = 0

By Leibniz theorem

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

x^2yn+2 + xyn+1(2n+1) + yn(n^2+1) = 0.