Application to Integrodifferential Equations

The Laplace transform is useful in solving linear integrodifferential equations. Using the differentiation and integration properties of Laplace transforms, each term in the integrodifferential equation is transformed. Initial conditions are automatically taken into account. We solve the resulting algebraic equation in the $ s $ domain. We then convert the solution back to the time domain by using the inverse transform. The following examples illustrate the process.
Example 1: Use the Laplace transform to solve the differential equation $$\frac{d^{2} v(t)}{d t^{2}}+6 \frac{d v(t)}{d t}+8 v(t)=2 u(t)$$ subject to $ v(0)=1, v^{\prime}(0)=-2 $.
Solution: We take the Laplace transform of each term in the given differential equation and obtain
$$\left[s^{2} V(s)-s v(0)-v^{\prime}(0)\right]+6[s V(s)-v(0)]+8 V(s)=\frac{2}{s}$$
Substituting $ v(0)=1, v^{\prime}(0)=-2 $,
$$s^{2} V(s)-s+2+6 s V(s)-6+8 V(s)=\frac{2}{s}$$
$$\left(s^{2}+6 s+8\right) V(s)=s+4+\frac{2}{s}=\frac{s^{2}+4 s+2}{s}$$
$$V(s)=\frac{s^{2}+4 s+2}{s(s+2)(s+4)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+4}$$
$$\begin{array}{c}A=\left.s V(s)\right|_{s=0}=\left.\frac{s^{2}+4 s+2}{(s+2)(s+4)}\right|_{s=0}=\frac{2}{(2)(4)}=\frac{1}{4} \\B=\left.(s+2) V(s)\right|_{s=-2}=\left.\frac{s^{2}+4 s+2}{s(s+4)}\right|_{s=-2}=\frac{-2}{(-2)(2)}=\frac{1}{2} \\C=\left.(s+4) V(s)\right|_{s=-4}=\left.\frac{s^{2}+4 s+2}{s(s+2)}\right|_{s=-4}=\frac{2}{(-4)(-2)}=\frac{1}{4}\end{array}$$
By the inverse Laplace transform,
$$v(t)=\frac{1}{4}\left(1+2 e^{-2 t}+e^{-4 t}\right) u(t)$$
Example 2: Solve for the response $ y(t) $ in the following integrodifferential equation.
$$\frac{d y}{d t}+5 y(t)+6 \int_{0}^{t} y(\tau) d \tau=u(t), \quad y(0)=2$$
Solution: Taking the Laplace transform of each term, we get
$$[s Y(s)-y(0)]+5 Y(s)+\frac{6}{s} Y(s)=\frac{1}{s}$$
Substituting $ y(0)=2 $ and multiplying through by $ s $,
$$Y(s)\left(s^{2}+5 s+6\right)=1+2 s$$
$$Y(s)=\frac{2 s+1}{(s+2)(s+3)}=\frac{A}{s+2}+\frac{B}{s+3}$$
$$\begin{array}{c}A=\left.(s+2) Y(s)\right|_{s=-2}=\left.\frac{2 s+1}{s+3}\right|_{s=-2}=\frac{-3}{1}=-3 \\B=\left.(s+3) Y(s)\right|_{s=-3}=\left.\frac{2 s+1}{s+2}\right|_{s=-3}=\frac{-5}{-1}=5\end{array}$$
Its inverse transform is
$$y(t)=\left(-3 e^{-2 t}+5 e^{-3 t}\right)$$