Question

Find the particular solution that satisfies the differential equation and the initial condition. f'(s) = 145 – 453, f(3) = 3 f(s) =

Answer 1

Gi'new f'(s) = 145=459 , f(3) = 3 1.0.B. 5, we get Sf'cs) ds = $(145 45°) es = f(s) = 14. -4. TC 7 f(s) = 7s"_5"+c ->() I put s=3 f(3) = 7x 9 - 81+ c Given f(3) = 3 => 3 = 6 3 – 81+ c

= 3 = -18+ c => C= 3+18 =Y C = 21 sub. 'c' value in (1), we get fis) = 75?_sty PI -