Question

The nth derivative of x^2y2 + (1-x^2)y1 + xy = 0 is,

(a) x^2yn+2 + yn(2n^2-2n-2nx+x) – yn-1(2n^2-3n)=0

(b) x^2yn+2 + yn+1(2nx-x^2) + yn(2n^2-2n-2nx+x) – yn-1(2n^2-3n)=0

(c) x^2yn+2 + yn+1(2nx+1-x^2) + yn(2n^2-2n-2nx+x)-yn – 1(2n^2-3n)=0

(d) x^2yn+2 + yn+1(2nx+1-x^2) + xyn2n^2-yn – 1(2n^2-3n)=0

The question was posed to me in an online quiz.

The query is from Leibniz Rule topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (c) x^2yn+2 + yn+1(2nx+1-x^2) + yn(2n^2-2n-2nx+x)-yn – 1(2n^2-3n)=0

The explanation is: x^2y2 + (1-x^2)y1 + xy = 0

Differentiating n times by Leibniz Rule

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

x^2yn+2 + yn+1(2nx+1-x^2) + yn(2n^2-2n-2nx+x) – yn-1(2n^2-3n)=0.