Let 0x 1 - (y'(t))2 dt = 0x y(t) dt, 0 x 3, y 0, y(0) = 0. Then at x = 2, y'' + y + 1 is equal to
- 2
- 2
- 1/2
- 1
Answer
(4) 1
Solution
Differentiating both sides w.r.t. x (Leibnitz rule): 1 - (y')2 = y
(y')2 = 1 - y2, so dy 1 - y2 = dx
Integrating, -1y = x + c, and y(0) = 0 gives c = 0.
Since y 0 for 0 x 3, we must take -1y = x, i.e. y = x.
Hence y' = x and y'' = - x, so y'' + y + 1 = - x + x + 1 = 1 for every x, and in particular at x = 2.
Answer: (4)