工程數學 initial value problem

(dy/dt) - y = 1 + 3sint

積分因式，exp[∫(-1)dt] = e-t

e-t(dy/dt - y) = e-t + 3e-tsint

d/dt (ye-t) = e-t + 3e-tsint

ye-t = -e-t + 3∫e-tsint dt

考慮∫e-tsint dt

= -∫e-t d(cost)

= -e-tcost + ∫cost d(e-t) (分部積分)

= -e-tcost - ∫e-td(sint)

= -e-tcost - e-tsint + ∫sintd(e-t)

= -e-tcost - e-tsint - ∫e-tsintdt

所以，2∫e-tsintdt = -e-tcost - e-tsint + c'

∫e-tsintdt = -e-t/2 (cost + sint) + c

所以

ye-t = -e-t + 3[-e-t/2 (cost + sint) + c]

y = -1 + 3/2 (cost + sint) + cet



2009-09-15 12:21:24 補充：
y(0) = y0
所以，c = y0 - 1/2
但要當t → 無限大，y是finite
唯有c = 0
所以，yo = 1/2

2009-09-15 12:21:36 補充：
y(0) = y0
所以，c = y0 - 1/2
但要當t → 無限大，y是finite
唯有c = 0
所以，yo = 1/2