Differential Equations - Laplace Transforms: IVP's With Step Functions - i

In this section we will use Laplace transforms to solve IVP’s which contain Heaviside functions in the forcing function. This is where Laplace transform really starts to come into its own as a solution method.

To work these problems we’ll just need to remember the following two formulas,

$$\begin{array}{cccc}L\left\{u_c\left(t\right)f\left(t-c\right)\right\}& =e^{-cs}F\left(s\right)& & \text{where}F\left(s\right)=L\left\{f\left(t\right)\right\}\\ L^{-1}\left\{e^{-cs}F\left(s\right)\right\}& =u_c\left(t\right)f\left(t-c\right)& & \text{where}f\left(t\right)=L^{-1}\left\{F\left(s\right)\right\}\end{array}$$

In other words, we will always need to remember that in order to take the transform of a function that involves a Heaviside we’ve got to make sure the function has been properly shifted.

Let’s work an example.

Example 1 Solve the following IVP.$$y^{\prime \prime }-y^{\prime }+5y=4+u_2\left(t\right)e^{4-2t},y\left(0\right)=2y^{\prime }\left(0\right)=-1$$

First let’s rewrite the forcing function to make sure that it’s being shifted correctly and to identify the function that is actually being shifted.
$$y^{\prime \prime }-y^{\prime }+5y=4+u_2\left(t\right)e^{-2\left(t-2\right)}$$So, it is being shifted correctly and the function that is being shifted is $e^{-2t}$. Taking the Laplace transform of everything and plugging in the initial conditions gives,
$$\begin{array}{cc}{s}^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)-\left(sY\left(s\right)-y\left(0\right)\right)+5Y\left(s\right)& =\frac{4}{s}+\frac{e^{-2s}}{s+2}\\ \left(s^2-s+5\right)Y\left(s\right)-2s+3& =\frac{4}{s}+\frac{e^{-2s}}{s+2}\end{array}$$Now solve for $Y\left(s\right)$.
$$\begin{array}{cc}\left(s^2-s+5\right)Y\left(s\right)& =\frac{4}{s}+\frac{e^{-2s}}{s+2}+2s-3\\ \left(s^2-s+5\right)Y\left(s\right)& =\frac{2s^2-3s+4}{s}+\frac{e^{-2s}}{s+2}\\ Y\left(s\right)& =\frac{2s^2-3s+4}{s\left(s^2-s+5\right)}+e^{-2s}\frac{1}{\left(s+2\right)\left(s^2-s+5\right)}\\ Y\left(s\right)& =F\left(s\right)+e^{-2s}G\left(s\right)\end{array}$$Notice that we combined a couple of terms to simplify things a little. Now we need to partial fraction $F\left(s\right)$ and $G\left(s\right)$. We’ll leave it to you to check the details of the partial fractions.
$$\begin{array}{cc}F\left(s\right)& =\frac{2s^2-3s+4}{s\left(s^2-s+5\right)}=\frac{1}{5}\left(\frac{4}{s}+\frac{6s-11}{s^2-s+5}\right)\\ G\left(s\right)& =\frac{1}{\left(s+2\right)\left(s^2-s+5\right)}=\frac{1}{11}\left(\frac{1}{s+2}-\frac{s-3}{s^2-s+5}\right)\end{array}$$We now need to do the inverse transforms on each of these. We’ll start with $F\left(s\right)$.
$$\begin{array}{cc}F\left(s\right)& =\frac{1}{5}\left(\frac{4}{s}+\frac{6\left(s-\frac{1}{2}+\frac{1}{2}\right)-11}{{\left(s-\frac{1}{2}\right)}^2+\frac{19}{4}}\right)\\ & =\frac{1}{5}\left(\frac{4}{s}+\frac{6\left(s-\frac{1}{2}\right)}{{\left(s-\frac{1}{2}\right)}^2+\frac{19}{4}}-\frac{8\frac{\sqrt{19}}{2}\frac{2}{\sqrt{19}}}{{\left(s-\frac{1}{2}\right)}^2+\frac{19}{4}}\right)\\ f\left(t\right)& =\frac{1}{5}\left(4+6e^{\frac{t}{2}}\cos \left(\frac{\sqrt{19}}{2}t\right)-\frac{16}{\sqrt{19}}{e}^{\frac{t}{2}}\sin \left(\frac{\sqrt{19}}{2}t\right)\right)\end{array}$$Now $G\left(s\right)$.
$$\begin{array}{cc}G\left(s\right)& =\frac{1}{11}\left(\frac{1}{s+2}-\frac{s-\frac{1}{2}+\frac{1}{2}-3}{{\left(s-\frac{1}{2}\right)}^2+\frac{19}{4}}\right)\\ & =\frac{1}{11}\left(\frac{1}{s+2}-\frac{s-\frac{1}{2}}{{\left(s-\frac{1}{2}\right)}^2+\frac{19}{4}}+\frac{\frac{5}{2}\frac{\sqrt{19}}{\sqrt{19}}}{{\left(s-\frac{1}{2}\right)}^2+\frac{19}{4}}\right)\\ g\left(t\right)& =\frac{1}{11}\left(e^{-2t}-e^{\frac{t}{2}}\cos \left(\frac{\sqrt{19}}{2}t\right)+\frac{5}{\sqrt{19}}{e}^{\frac{t}{2}}\sin \left(\frac{\sqrt{19}}{2}t\right)\right)\end{array}$$Okay, we can now get the solution to the differential equation. Starting with the transform we get,
$$\begin{array}{cc}Y\left(s\right)& =F\left(s\right)+e^{-2s}G\left(s\right)\\ y\left(t\right)& =f\left(t\right)+u_2\left(t\right)g\left(t-2\right)\end{array}$$where $f\left(t\right)$ and $g\left(t\right)$ are the functions shown above.

There can be a fair amount of work involved in solving differential equations that involve Heaviside functions.

Let’s take a look at another example or two.

Example 2 Solve the following IVP.$$y^{\prime \prime }-y^{\prime }=\cos \left(2t\right)+\cos \left(2t-12\right)u_6\left(t\right)y\left(0\right)=-4,y^{\prime }\left(0\right)=0$$

Let’s rewrite the differential equation so we can identify the function that is actually being shifted.
$$y^{\prime \prime }-y^{\prime }=\cos \left(2t\right)+\cos \left(2\left(t-6\right)\right)u_6\left(t\right)$$So, the function that is being shifted is $\cos \left(2t\right)$ and it is being shifted correctly. Taking the Laplace transform of everything and plugging in the initial conditions gives,
$$\begin{array}{cc}{s}^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)-\left(sY\left(s\right)-y\left(0\right)\right)& =\frac{s}{s^2+4}+\frac{s{e}^{-6s}}{s^2+4}\\ \left(s^2-s\right)Y\left(s\right)+4s-4& =\frac{s}{s^2+4}+\frac{s{e}^{-6s}}{s^2+4}\end{array}$$Now solve for $Y\left(s\right)$.
$$\begin{array}{cc}\left(s^2-s\right)Y\left(s\right)& =\frac{s+s{e}^{-6s}}{s^2+4}-4s+4\\ Y\left(s\right)& =\frac{s\left(1+e^{-6s}\right)}{s\left(s-1\right)\left(s^2+4\right)}-4\frac{s-1}{s\left(s-1\right)}\\ & =\frac{1+e^{-6s}}{\left(s-1\right)\left(s^2+4\right)}-\frac{4}{s}\\ Y\left(s\right)& =\left(1+e^{-6s}\right)F\left(s\right)-\frac{4}{s}\end{array}$$Notice that we combined the first two terms to simplify things a little. Also, there was some canceling going on in this one. Do not expect that to happen on a regular basis. We now need to partial fraction $F\left(s\right)$. We’ll leave the details to you to check.
$$\begin{array}{cc}F\left(s\right)& =\frac{1}{\left(s-1\right)\left(s^2+4\right)}=\frac{1}{5}\left(\frac{1}{s-1}-\frac{s+1}{s^2+4}\right)\\ f\left(t\right)& =\frac{1}{5}\left(e^t-\cos \left(2t\right)-\frac{1}{2}\sin \left(2t\right)\right)\end{array}$$Okay, we can now get the solution to the differential equation. Starting with the transform we get,
$$\begin{array}{cc}Y\left(s\right)& =F\left(s\right)+F\left(s\right)e^{-6s}-\frac{4}{s}\\ y\left(t\right)& =f\left(t\right)+u_6\left(t\right)f\left(t-6\right)-4\end{array}$$where $f\left(t\right)$ is given above.