工程數學拉式轉換解

y" + 4y' + 3y = e^t, y(0) = 0, y'(0) = 2

L{y" + 4y' + 3y}(s) = L{e^t}(s)

Denote Y(s) = L{y(t)}(s)

So, [s^2Y(s) - sy(0) - y'(0)] + 4[sY(s) - y(0)] + 3Y(s) = 1/(s - 1)

(s^2 + 4s + 3)Y(s) - 2 = 1/(s - 1)

(s + 3)(s + 1)Y(s) = 1/(s - 1) + 2

Y(s) = 1/(s - 1)(s + 1)(s + 3) + 2/(s + 1)(s + 3)

Y(s) = [1/8(s - 1) - 1/4(s + 1) + 1/8(s + 3)] + [1/(s + 1) - 1/(s + 3)]

Y(s) = 1/8(s - 1) + 3/4(s + 1) - 7/8(s + 3)

y(t) = L^-1{Y(s)}(t)

= (1/8)e^t + (3/4)e^(-t) - (7/8)e^(-3t)