Let f be a differentiable function in the interval (0, ) such that f(1) = 1 and t x t2 f(x) - x2 f(t)t - x = 1 for each x > 0. Then 2f(2) + 3f(3) is equal to .
Answer
24
Solution
00 form; L'Hôpital's rule (w.r.t. t): t x [2t f(x) - x2 f'(t) ] = 1
2x f(x) - x2 f'(x) = 1 f'(x) - 2xf(x) = - 1x2
I.F. = e-2 x = 1x2 f(x)x2 = - 1x4 dx = 13x3 + C
f(1) = 1 C = 23 f(x) = 13x + 2x23
2f(2) + 3f(3) = 2 ( 16 + 83 ) + 3 ( 19 + 6 ) = 173 + 553 = 24
Answer: 24