Using the Laplace Transform to solve Integrodifferential Equations (example #2)

Given the following integrodifferential equation:

$$y'(t) + 4y(t) + 3\int_{0}^{t}y(\tau) \;d\tau = 18e^{-2t}$$ ...where: $$y(0) = -3$$

...determine y(t) using the Laplace transform.

Begin by taking the Laplace transform of both sides of the equation: $$\mathcal{L}[y'(t)] + 4\mathcal{L}[y(t)] + 3\mathcal{L}\Big[\int_{0}^{t}y(\tau) \;d\tau\Big] = 18\mathcal{L}[e^{-2t}]$$ For the left side of the equation we use the time-differentiation property (property 35 in the Laplace transform table) as well as the time-integration property (property #32 in the Laplace transform table). For the right side of the equation, we use property #2 in the table). $$sY(s)-y(0)+4Y(s)+\frac{3}{s}Y(s)=\frac{18}{s+2}$$ Now substitute in for y(0): $$sY(s)+3+4Y(s)+\frac{3}{s}Y(s)=\frac{18}{s+2}$$ $$Y(s)\Big[s+\frac{3}{s}+4\Big]+3=\frac{18}{s+2}$$ $$Y(s)\Big[\frac{s^2+4s+3}{s}\Big]+3=\frac{18}{s+2}$$ $$Y(s)=\Big(\frac{18}{s+2}-3\Big)\Big(\frac{s}{s^2+4s+3}\Big)$$ $$Y(s)=\Big(\frac{18-(3s+6)}{s+2}\Big)\Big(\frac{s}{s^2+4s+3}\Big)$$ $$Y(s)=\Big(\frac{-3s+12}{s+2}\Big)\Big(\frac{s}{(s+3)(s+1)}\Big)$$ $$Y(s) = \frac{-3s^2+12s}{(s+2)(s+3)(s+1)} \qquad,\;(Eqn\;1)$$

Partial Fraction Decomposition using the Residue Method:

We will use the Heaviside Theorem/Residue Method to expand equation #1: $$\frac{-3s^2+12s}{(s+2)(s+3)(s+1)} = \frac{A}{s+2}+\frac{B}{s+3}+\frac{C}{s+1}$$ To determine the expansion coefficients A, B and C, recall the definition of Heaviside Theorem: $$k_i = (s+p_i) F(s) \Big|_{s=-p_i}$$ Therefore: \begin{align} A &= (s+2) \; Y(s) \Big|_{s=-2} \\\\ &= \frac{-3s^2+12s}{(s+3)(s+1)}\Big|_{s=-2} \\\\ &= \frac{-3(4)-24}{1(-1)} \\\\ A &= 36 \\\\ B &= (s+3) \; Y(s) \Big|_{s=-3} \\\\ &= \frac{-3s^2+12s}{(s+2)(s+1)}\Big|_{s=-3} \\\\ &= \frac{-27-36}{-1(-2)} \\\\ B &= \frac{-63}{2} \\\\ C &= (s+1) \; Y(s) \Big|_{s=-1} \\\\ &= \frac{-3s^2+12s}{(s+2)(s+3)}\Big|_{s=-1} \\\\ &= \frac{-3-12}{1(2)} \\\\ C &= \frac{-15}{2} \\\\ \end{align} Therefore: $$\frac{-3s^2+12s}{(s+2)(s+3)(s+1)} = \frac{36}{s+2}-\frac{\frac{63}{2}}{s+3}-\frac{\frac{15}{2}}{s+1} \quad,\;(Eqn\;2)$$

Substitute equation #2 into equation #1:

$$Y(s) = 36\Big(\frac{1}{s+2}\Big)-\frac{63}{2}\Big(\frac{1}{s+3}\Big)-\frac{15}{2}\Big(\frac{1}{s+1}\Big)$$ Now, take the inverse Laplace Transform to get y(t): \begin{align} y(t) &= \mathcal{L}^{-1}[Y(s)] \\\\ &= 36\;\mathcal{L}^{-1}\Big[\frac{1}{s+2}\Big]-\frac{63}{2}\;\mathcal{L}^{-1}\Big[\frac{1}{s+3}\Big]-\frac{15}{2}\;\mathcal{L}^{-1}\Big[\frac{1}{s+1}\Big]\\\\ \end{align} For each of the three terms, we use property #2 from the Laplace transform table and arrive at our final answer:

$$y(t) = 36e^{-2t}-\frac{63}{2}e^{-3t}-\frac{15}{2}e^{-t}$$