# Using the Laplace Transform to solve Integrodifferential Equations (example #1)

### Given the following differential equation:

$$v''(t) + 2v'(t) + 10v(t) = 3\cos(2t)$$ ...where: $$v(0) = 1 \;,\;\;\; v'(0) = -2$$

### ...determine v(t) using the Laplace transform.

We begin by taking the Laplace transform of both sides of the equation: $$\mathcal{L}[v''(t)] + 2\mathcal{L}[v'(t)] + 10\mathcal{L}[v(t)] = 3\mathcal{L}[\cos(2t)]$$ For the left side of the equation we use the time-differentiation properties (properties 35 and 36 in the Laplace transform table). For the right side of the equation, we use property # 8 in the table). Doing all of this gives us the following: $$s^2V(s) - sv(0) - v'(0) + 2[sV(s)-v(0)] + 10V(s) = 3\Big(\frac{s}{s^2+4}\Big)$$ Now substitute in for v(0) and v'(0): $$s^2V(s) - s + 2 + 2[sV(s)-1] + 10V(s) = \frac{3s}{s^2+4}$$ $$s^2V(s) - s + 2 + 2sV(s) - 2 + 10V(s) = \frac{3s}{s^2+4}$$ $$V(s)[s^2 + 2s + 10] - s = \frac{3s}{s^2+4}$$

\begin{align} V(s) &= \Big[ \frac{3s}{s^2+4}+s \Big] \Big[ \frac{1}{s^2+2s+10} \Big] \\\\ &= \Big[ \frac{3s+s^3+4s}{s^2+4} \Big] \Big[ \frac{1}{s^2+2s+10} \Big] \\\\ &= \Big[ \frac{s^3+7s}{s^2+4} \Big] \Big[ \frac{1}{s^2+2s+10} \Big] \\\\ V(s) &= \frac{s^3+7s}{(s^2+4)(s^2+2s+10)} \qquad,\;(Eqn\;1) \end{align}

## (Sidework: Partial Fraction Decomposition)

In order to determine the Laplace transform of equation #1, we need to expand it out using partial fraction decomposition. $$\frac{s^3+7s}{(s^2+4)(s^2+2s+10)} = \frac{As+B}{s^2+4} + \frac{Cs+D}{s^2+2s+10}$$ $$s^3+7s = (s^2+2s+10)(As+B)+(s^2+4)(Cs+D)$$ $$s^3+7s = As^2+2As^2+10As+Bs^2+2Bs+10B+Cs^2+Ds^2+4Cs+4D$$ $$s^3+7s = s^3(A+C) + s^2(2A+B+D) + s(10A+2B+4C) + 10B+4D$$ Now, equate coefficients: $$For\;s^3: \; A+C=1$$ $$For\;s^2: \; 2A+B+D=0$$ $$For\;s: \; 10A+2B+4C=7$$ $$constants: \; 10B+4D = 0$$ I won't go through the process of solving the system of equations here. You could use row operations, Cramer's Rule, etc... In the end however, you should arrive at the following values: $$A = \frac{9}{26}$$ $$B = \frac{6}{13}$$ $$C = \frac{17}{26}$$ $$D = \frac{-15}{13}$$ Therefore, we arrive at the following expansion: $$\frac{s^3+7s}{(s^2+4)(s^2+2s+10)} = \frac{\frac{9s}{26}+\frac{6}{13}}{s^2+4} + \frac{\frac{17s}{26}-\frac{15}{13}}{s^2+2s+10} \qquad,\;(Eqn\;2)$$

Picking up where we left off, we substitute equation #2 into equation #1 and get the following: $$V(s) = \frac{\frac{9s}{26}+\frac{6}{13}}{s^2+4} + \frac{\frac{17s}{26}-\frac{15}{13}}{s^2+2s+10}$$ \begin{align} V(s) &= \frac{9}{26}\Big( \frac{s}{s^2+4} \Big) + \frac{6}{13}\Big( \frac{1}{s^2+4} \Big) + \frac{17}{26}\Big(\frac{s}{s^2+2s+10}\Big) - \frac{15}{13}\Big( \frac{1}{s^2+2s+10} \Big)\\\\ \end{align} We will now rewrite the terms in the above equation so that they are in forms which allow us to utilize the properties of the Laplace transform and determine the inverse. Specifically, we will be looking to use the following properties in our Laplace transform table: Property #8 for the 1st term, protperty #7 for the 2nd term and property #19 for the 3rd/4th terms. In the case of the 3r and 4th terms, we will need to "complete the square" in order to accomplish this: \begin{align} V(s) &= \frac{9}{26}\Big( \frac{s}{s^2+(2)^2} \Big) + \frac{6}{13}\Big(\frac{1}{2}\Big)\Big( \frac{2}{s^2+(2)^2} \Big) + \frac{17}{26}\Big(\frac{s}{s^2+2s+(\frac{2}{2})^2+10-(\frac{2}{2})^2}\Big) - \frac{15}{13}\Big( \frac{1}{s^2+2s+(\frac{2}{2})^2+10-(\frac{2}{2})^2} \Big)\\\\ \end{align} \begin{align} V(s) &= \frac{9}{26}\Big( \frac{s}{s^2+(2)^2} \Big) + \frac{3}{13}\Big( \frac{2}{s^2+(2)^2} \Big) + \frac{17}{26}\Big(\frac{s}{(s^2+2s+1)+9}\Big) - \frac{15}{13}\Big( \frac{1}{(s^2+2s+1)+9} \Big)\\\\ &= \frac{9}{26}\Big( \frac{s}{s^2+(2)^2} \Big) + \frac{3}{13}\Big( \frac{2}{s^2+(2)^2} \Big) + \frac{17}{26}\Big(\frac{s-1+1}{(s+1)^2+(3)^2}\Big) - \frac{15}{13}\Big(\frac{1}{3}\Big)\Big( \frac{3}{(s+1)^2+(3)^2} \Big)\\\\ \end{align} $$= \frac{9}{26}\Big( \frac{s}{s^2+(2)^2} \Big) + \frac{3}{13}\Big( \frac{2}{s^2+(2)^2} \Big) + \frac{17}{26}\Big(\frac{s+1}{(s+1)^2+(3)^2}\Big)- \frac{17}{26}\Big(\frac{1}{(s+1)^2+(3)^2}\Big) - \frac{5}{13}\Big( \frac{3}{(s+1)^2+(3)^2}\Big)$$ $$= \frac{9}{26}\Big( \frac{s}{s^2+(2)^2} \Big) + \frac{3}{13}\Big( \frac{2}{s^2+(2)^2} \Big) + \frac{17}{26}\Big(\frac{s+1}{(s+1)^2+(3)^2}\Big)- \frac{17}{26}\Big(\frac{1}{3}\Big)\Big(\frac{3}{(s+1)^2+(3)^2}\Big) - \frac{5}{13}\Big( \frac{3}{(s+1)^2+(3)^2}\Big)$$ $$V(s) = \frac{9}{26}\Big(\frac{s}{s^2+(2)^2}\Big) + \frac{3}{13}\Big(\frac{2}{s^2+(2)^2}\Big) + \frac{17}{26}\Big(\frac{s+1}{(s+1)^2+(3)^2}\Big)- \frac{17}{78}\Big(\frac{3}{(s+1)^2+(3)^2}\Big) - \frac{5}{13}\Big( \frac{3}{(s+1)^2+(3)^2}\Big)$$ We now take the inverse Laplace transform of the above equation to get v(t): $$v(t) = \frac{9}{26}\mathcal{L}^{-1}\Big(\frac{s}{s^2+(2)^2}\Big) + \frac{3}{13}\mathcal{L}^{-1}\Big(\frac{2}{s^2+(2)^2}\Big) + \frac{17}{26}\mathcal{L}^{-1}\Big(\frac{s+1}{(s+1)^2+(3)^2}\Big)- \frac{17}{78}\mathcal{L}^{-1}\Big(\frac{3}{(s+1)^2+(3)^2}\Big) - \frac{5}{13}\mathcal{L}^{-1}\Big( \frac{3}{(s+1)^2+(3)^2}\Big)$$ $$v(t) = \frac{9}{26}\cos(2t)+\frac{3}{13}\sin(2t)+\frac{17}{26}e^{-t}\cos(3t)-\frac{17}{78}e^{-t}\sin(3t)-\frac{5}{13}e^{-t}\sin(3t)$$

$$v(t) = \frac{9}{26}\cos(2t)+\frac{3}{13}\sin(2t)+\frac{17}{26}e^{-t}\cos(3t)-\frac{47}{78}e^{-t}\sin(3t)$$

Continue on to example problem #2 (involving integrodifferential equations)...