If ∫ √1 - (y′(t))^2 dt; t ∈ [0, x] = ∫ y'(t) dt; t ∈ [0, x] and 0 ≤ x ≤ 3, y ≥ 0, y(0) = 0

Question

If \(\int \limits_{0}^x \sqrt{1 - (y'(t))^2} dt = \int \limits_{0}^xy'(t) dt\) and \(0 \le x \le 3, y \ge 0, y(0) =0\), then find \(y'' + 1 + y\).

(1) \(\frac x{\sqrt 2} - 1\)

(2) \(\frac x{\sqrt 2}+ 1\)

(3) \(\frac x{ 2}+ 1\)

(4) \(\frac x{ 2} -1\)

Answer 1

Correct option is (2) \(\frac x{\sqrt 2}+ 1\)