Let f : R R be a twice differentiable function such that f''(x) = -f(x) . Let h : R R be a differentiable function such that h'(x) = f(x)^2 + (f'(x))^2 for all x R . If h(0) = 1 and h ( 1 3 ) = 5 3 , then the value of h(100) is

Step-by-step solution

Given f''(x) = -f(x) Multiplying both sides by 2f'(x) : 2f'(x)f''(x) + 2f(x)f'(x) = 0 Integrating with respect to x : (f'(x))^2 + (f(x))^2 = C , where C is a constant. Given h'(x) = (f(x))^2 + (f'(x))^2 h'(x) = C Integrating with respect to x : h(x) = Cx + D , where D is a constant. Given h(0) = 1 , we get D = 1 . Given h ( 1 3 ) = 5 3 , we get: C ( 1 3 ) + 1 = 5 3 C = 2 Thus, h(x) = 2x + 1 . For x = 100 : h(100) = 2(100) + 1 = 201 Answer: 201