Differential Equations - Laplace Transforms: Solving Initial Value Problems - ii

Example 3 Solve the following IVP.$$y^{\prime \prime }-6y^{\prime }+15y=2\sin \left(3t\right),y\left(0\right)=-1y^{\prime }\left(0\right)=-4$$

Take the Laplace transform of everything and plug in the initial conditions.
$$\begin{array}{cc}{s}^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)-6\left(sY\left(s\right)-y\left(0\right)\right)+15Y\left(s\right)& =2\frac{3}{s^2+9}\\ \left(s^2-6s+15\right)Y\left(s\right)+s-2& =\frac{6}{s^2+9}\end{array}$$Now solve for $Y\left(s\right)$ and combine into a single term as we did in the previous two examples.
$$Y\left(s\right)=\frac{-s^3+2s^2-9s+24}{\left(s^2+9\right)\left(s^2-6s+15\right)}$$Now, do the partial fractions on this. First let’s get the partial fraction decomposition.
$$Y\left(s\right)=\frac{As+B}{s^2+9}+\frac{Cs+D}{s^2-6s+15}$$Now, setting numerators equal gives,
$$\begin{array}{cc}-s^3+2s^2-9s+24& =\left(As+B\right)\left(s^2-6s+15\right)+\left(Cs+D\right)\left(s^2+9\right)\\ & =\left(A+C\right)s^3+\left(-6A+B+D\right)s^2+\left(15A-6B+9C\right)s+15B+9D\end{array}$$Setting coefficients equal and solving for the constants gives,
$$\begin{array}{cccc}& s^3:& A+C& =-1\\ & s^2:& -6A+B+D& =2\\ & s^1:& 15A-6B+9C& =-9\\ & s^0:& 15B+9D& =24\end{array}\}\Rightarrow \begin{array}{cccc}A& =\frac{1}{10}& B& =\frac{1}{10}\\ C& =-\frac{11}{10}& D& =\frac{5}{2}\end{array}$$Now, plug these into the decomposition, complete the square on the denominator of the second term and then fix up the numerators for the inverse transform process.
$$\begin{array}{cc}Y\left(s\right)& =\frac{1}{10}\left(\frac{s+1}{s^2+9}+\frac{-11s+25}{s^2-6s+15}\right)\\ & =\frac{1}{10}\left(\frac{s+1}{s^2+9}+\frac{-11\left(s-3+3\right)+25}{{\left(s-3\right)}^2+6}\right)\\ & =\frac{1}{10}\left(\frac{s}{s^2+9}+\frac{1\frac{3}{3}}{s^2+9}-\frac{11\left(s-3\right)}{{\left(s-3\right)}^2+6}-\frac{8\frac{\sqrt{6}}{\sqrt{6}}}{{\left(s-3\right)}^2+6}\right)\end{array}$$Finally, take the inverse transform.
$$y\left(t\right)=\frac{1}{10}\left(\cos \left(3t\right)+\frac{1}{3}\sin \left(3t\right)-11e^{3t}\cos \left(\sqrt{6}t\right)-\frac{8}{\sqrt{6}}{e}^{3t}\sin \left(\sqrt{6}t\right)\right)$$

To this point we’ve only looked at IVP’s in which the initial values were at $t=0$. This is because we need the initial values to be at this point in order to take the Laplace transform of the derivatives. The problem with all of this is that there are IVP’s out there in the world that have initial values at places other than $t=0$. Laplace transforms would not be as useful as it is if we couldn’t use it on these types of IVP’s. So, we need to take a look at an example in which the initial conditions are not at $t=0$ in order to see how to handle these kinds of problems.

Example 4 Solve the following IVP.$$y^{\prime \prime }+4y^{\prime }=\cos \left(t-3\right)+4t,y\left(3\right)=0y^{\prime }\left(3\right)=7$$

The first thing that we will need to do here is to take care of the fact that initial conditions are not at $t=0$. The only way that we can take the Laplace transform of the derivatives is to have the initial conditions at $t=0$.
This means that we will need to formulate the IVP in such a way that the initial conditions are at $t=0$. This is actually fairly simple to do, however we will need to do a change of variable to make it work. We are going to define
$$\eta =t-3\Rightarrow t=\eta +3$$Let’s start with the original differential equation.
$$y^{\prime \prime }\left(t\right)+4y^{\prime }\left(t\right)=\cos \left(t-3\right)+4t$$Notice that we put in the $\left(t\right)$ part on the derivatives to make sure that we get things correct here. We will next substitute in for $t$.
$$y^{\prime \prime }\left(\eta +3\right)+4y^{\prime }\left(\eta +3\right)=\cos \left(\eta \right)+4\left(\eta +3\right)$$Now, to simplify life a little let’s define,
$$u\left(\eta \right)=y\left(\eta +3\right)$$Then, by the chain rule, we get the following for the first derivative.
$$u^{\prime }\left(\eta \right)=\frac{du}{d\eta }=\frac{dy}{dt}\frac{dt}{d\eta }=y^{\prime }\left(\eta +3\right)$$By a similar argument we get the following for the second derivative.
$$u^{\prime \prime }\left(\eta \right)=y^{\prime \prime }\left(\eta +3\right)$$The initial conditions for $u\left(\eta \right)$ are,
$$\begin{array}{cc}u\left(0\right)& =y\left(0+3\right)=y\left(3\right)=0\\ u^{\prime }\left(0\right)& =y^{\prime }\left(0+3\right)=y^{\prime }\left(3\right)=7\end{array}$$The IVP under these new variables is then,
$$u^{\prime \prime }+4u^{\prime }=\cos \left(\eta \right)+4\eta +12,u\left(0\right)=0u^{\prime }\left(0\right)=7$$This is an IVP that we can use Laplace transforms on provided we replace all the $t$’s in our table with $\eta$’s. So, taking the Laplace transform of this new differential equation and plugging in the new initial conditions gives,
$$\begin{array}{cc}{s}^{2}U\left(s\right)-su\left(0\right)-u^{\prime }\left(0\right)+4\left(sU\left(s\right)-u\left(0\right)\right)& =\frac{s}{s^2+1}+\frac{4}{s^2}+\frac{12}{s}\\ \left(s^2+4s\right)U\left(s\right)-7& =\frac{s}{s^2+1}+\frac{4+12s}{s^2}\end{array}$$Solving for $U\left(s\right)$ gives,
$$\begin{array}{cc}\left(s^2+4s\right)U\left(s\right)& =\frac{s}{s^2+1}+\frac{4+12s+7s^2}{s^2}\\ U\left(s\right)& =\frac{1}{\left(s+4\right)\left(s^2+1\right)}+\frac{4+12s+7s^2}{s^3\left(s+4\right)}\end{array}$$Note that unlike the previous examples we did not completely combine all the terms this time. In all the previous examples we did this because the denominator of one of the terms was the common denominator for all the terms. Therefore, upon combining, all we did was make the numerator a little messier and reduced the number of partial fractions required down from two to one. Note that all the terms in this transform that had only powers of $s$ in the denominator were combined for exactly this reason.
In this transform however, if we combined both of the remaining terms into a single term we would be left with a fairly involved partial fraction problem. Therefore, in this case, it would probably be easier to just do partial fractions twice. We’ve done several partial fractions problems in this section and many partial fraction problems in the previous couple of sections so we’re going to leave the details of the partial fractioning to you to check. Partial fractioning each of the terms in our transform gives us the following.
$$\begin{array}{cc}\frac{1}{\left(s+4\right)\left(s^2+1\right)}& =\frac{\frac{1}{17}}{s+4}+\frac{1}{17}\left(\frac{-s+4}{s^2+1}\right)\\ \frac{4+12s+7s^2}{s^3\left(s+4\right)}& =\frac{1}{s^3}+\frac{\frac{11}{4}}{s^2}+\frac{\frac{17}{16}}{s}-\frac{\frac{17}{16}}{s+4}\end{array}$$Plugging these into our transform and combining like terms gives us
$$\begin{array}{cc}U\left(s\right)& =\frac{1}{s^3}+\frac{\frac{11}{4}}{s^2}+\frac{\frac{17}{16}}{s}-\frac{\frac{273}{272}}{s+4}+\frac{1}{17}\left(\frac{-s+4}{s^2+1}\right)\\ & =\frac{1\frac{2!}{2!}}{s^3}+\frac{\frac{11}{4}}{s^2}+\frac{\frac{17}{16}}{s}-\frac{\frac{273}{272}}{s+4}+\frac{1}{17}\left(\frac{-s}{s^2+1}+\frac{4}{s^2+1}\right)\end{array}$$Now, taking the inverse transform will give the solution to our new IVP. Don’t forget to use $\eta$’s instead of $t$’s!
$$u\left(\eta \right)=\frac{1}{2}{\eta }^2+\frac{11}{4}\eta +\frac{17}{16}-\frac{273}{272}{e}^{-4\eta }+\frac{1}{17}\left(4\sin \left(\eta \right)-\cos \left(\eta \right)\right)$$This is not the solution that we are after of course. We are after $y\left(t\right)$. However, we can get this by noticing that
$$y\left(t\right)=y\left(\eta +3\right)=u\left(\eta \right)=u\left(t-3\right)$$So, the solution to the original IVP is,
$$\begin{array}{cc}y\left(t\right)& =\frac{1}{2}{\left(t-3\right)}^2+\frac{11}{4}\left(t-3\right)+\frac{17}{16}-\frac{273}{272}{e}^{-4\left(t-3\right)}+\frac{1}{17}\left(4\sin \left(t-3\right)-\cos \left(t-3\right)\right)\\ y\left(t\right)& =\frac{1}{2}{t}^2-\frac{1}{4}t-\frac{43}{16}-\frac{273}{272}{e}^{-4\left(t-3\right)}+\frac{1}{17}\left(4\sin \left(t-3\right)-\cos \left(t-3\right)\right)\end{array}$$

So, we can now do IVP’s that don’t have initial conditions that are at $t=0$. We also saw in the last example that it isn’t always the best to combine all the terms into a single partial fraction problem as we have been doing prior to this example.

The examples worked in this section would have been just as easy, if not easier, if we had used techniques from the previous chapter. They were worked here using Laplace transforms to illustrate the technique and method.