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

\[\eta = t - 3\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,t = \eta + 3\]

Let’s start with the original differential equation.

\[y''\left( t \right) + 4y'\left( t \right) = \cos \left( {t - 3} \right) + 4t\]

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 \(t\).

\[y''\left( {\eta + 3} \right) + 4y'\left( {\eta + 3} \right) = \cos \left( \eta \right) + 4\left( {\eta + 3} \right)\]

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.

\[u'\left( \eta \right) = \frac{{du}}{{d\eta }} = \frac{{dy}}{{dt}}\frac{{dt}}{{d\eta }} = y'\left( {\eta + 3} \right)\]

By a similar argument we get the following for the second derivative.

\[u''\left( \eta \right) = y''\left( {\eta + 3} \right)\]

The initial conditions for \(u\left( \eta \right)\) are,

\[\begin{align*}u\left( 0 \right) & = y\left( {0 + 3} \right) = y\left( 3 \right) = 0\\ u'\left( 0 \right) & = y'\left( {0 + 3} \right) = y'\left( 3 \right) = 7\end{align*}\]

The IVP under these new variables is then,

\[u'' + 4u' = \cos \left( \eta \right) + 4\eta + 12,\hspace{0.25in}u\left( 0 \right) = 0\,\,\,\,\,\,\,\,u'\left( 0 \right) = 7\]

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,

\[\begin{align*}{s^2}U\left( s \right) - su\left( 0 \right) - u'\left( 0 \right) + 4\left( {sU\left( s \right) - u\left( 0 \right)} \right) & = \frac{s}{{{s^2} + 1}} + \frac{4}{{{s^2}}} + \frac{{12}}{s}\\ \left( {{s^2} + 4s} \right)U\left( s \right) - 7 & = \frac{s}{{{s^2} + 1}} + \frac{{4 + 12s}}{{{s^2}}}\end{align*}\]

Solving for \(U(s)\) gives,

\[\begin{align*}\left( {{s^2} + 4s} \right)U\left( s \right) & = \frac{s}{{{s^2} + 1}} + \frac{{4 + 12s + 7{s^2}}}{{{s^2}}}\\ U\left( s \right) & = \frac{1}{{\left( {s + 4} \right)\left( {{s^2} + 1} \right)}} + \frac{{4 + 12s + 7{s^2}}}{{{s^3}\left( {s + 4} \right)}}\end{align*}\]

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.

\[\begin{align*}\frac{1}{{\left( {s + 4} \right)\left( {{s^2} + 1} \right)}} & = \frac{{\frac{1}{{17}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s + 4}}{{{s^2} + 1}}} \right)\\ \frac{{4 + 12s + 7{s^2}}}{{{s^3}\left( {s + 4} \right)}} & = \frac{1}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \frac{{\frac{{17}}{{16}}}}{s} - \,\frac{{\frac{{17}}{{16}}}}{{s + 4}}\end{align*}\]

Plugging these into our transform and combining like terms gives us

\[\begin{align*}U\left( s \right) & = \frac{1}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \frac{{\frac{{17}}{{16}}}}{s} - \frac{{\frac{{273}}{{272}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s + 4}}{{{s^2} + 1}}} \right)\\ & = \frac{{1\frac{{2!}}{{2!}}}}{{{s^3}}} + \frac{{\frac{{11}}{4}}}{{{s^2}}} + \,\frac{{\frac{{17}}{{16}}}}{s} - \frac{{\frac{{273}}{{272}}}}{{s + 4}} + \frac{1}{{17}}\left( {\frac{{ - s}}{{{s^2} + 1}} + \frac{4}{{{s^2} + 1}}} \right)\end{align*}\]

Now, taking the inverse transform will give the solution to our new IVP. Don’t forget to use \(\eta \)’s instead of \(t\)’s!

\[u\left( \eta \right) = \frac{1}{2}{\eta ^2} + \frac{{11}}{4}\eta + \frac{{17}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\eta }} + \frac{1}{{17}}\left( {4\sin \left( \eta \right) - \cos \left( \eta \right)} \right)\]

This is not the solution that we are after of course. We are after \(y(t)\). 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,

\[\begin{align*}y\left( t \right) & = \frac{1}{2}{\left( {t - 3} \right)^2} + \frac{{11}}{4}\left( {t - 3} \right) + \frac{{17}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\left( {t - 3} \right)}} + \frac{1}{{17}}\left( {4\sin \left( {t - 3} \right) - \cos \left( {t - 3} \right)} \right)\\ y\left( t \right) & = \frac{1}{2}{t^2} - \frac{1}{4}t - \frac{{43}}{{16}} - \frac{{273}}{{272}}{{\bf{e}}^{ - 4\left( {t - 3} \right)}} + \frac{1}{{17}}\left( {4\sin \left( {t - 3} \right) - \cos \left( {t - 3} \right)} \right)\end{align*}\]