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Section 4-2 : Laplace Transforms

As we saw in the last section computing Laplace transforms directly can be fairly complicated. Usually we just use a table of transforms when actually computing Laplace transforms. The table that is provided here is not an all-inclusive table but does include most of the commonly used Laplace transforms and most of the commonly needed formulas pertaining to Laplace transforms.

Before doing a couple of examples to illustrate the use of the table let’s get a quick fact out of the way.

Fact

Given \(f(t)\) and \(g(t)\) then,

\[\mathcal{L}\left\{ {af\left( t \right) + bg\left( t \right)} \right\} = a\,F\left( s \right) + b\,G\left( s \right)\]

for any constants \(a\) and \(b\).

In other words, we don’t worry about constants and we don’t worry about sums or differences of functions in taking Laplace transforms. All that we need to do is take the transform of the individual functions, then put any constants back in and add or subtract the results back up.

So, let’s do a couple of quick examples.

Example 1 Find the Laplace transforms of the given functions.
  1. \(f\left( t \right) = 6{{\bf{e}}^{ - 5t}} + {{\bf{e}}^{3t}} + 5{t^3} - 9\)
  2. \(g\left( t \right) = 4\cos \left( {4t} \right) - 9\sin \left( {4t} \right) + 2\cos \left( {10t} \right)\)
  3. \(h\left( t \right) = 3\sinh \left( {2t} \right) + 3\sin \left( {2t} \right)\)
  4. \(g\left( t \right) = {{\bf{e}}^{3t}} + \cos \left( {6t} \right) - {{\bf{e}}^{3t}}\cos \left( {6t} \right)\)
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Okay, there’s not really a whole lot to do here other than go to the table, transform the individual functions up, put any constants back in and then add or subtract the results.

We’ll do these examples in a little more detail than is typically used since this is the first time we’re using the tables.


a \(f\left( t \right) = 6{{\bf{e}}^{ - 5t}} + {{\bf{e}}^{3t}} + 5{t^3} - 9\) Show Solution
\[\begin{align*}F\left( s \right) & = 6\frac{1}{{s - \left( { - 5} \right)}} + \frac{1}{{s - 3}} + 5\frac{{3!}}{{{s^{3 + 1}}}} - 9\frac{1}{s}\\ & = \frac{6}{{s + 5}} + \frac{1}{{s - 3}} + \frac{{30}}{{{s^4}}} - \frac{9}{s}\end{align*}\]

b \(g\left( t \right) = 4\cos \left( {4t} \right) - 9\sin \left( {4t} \right) + 2\cos \left( {10t} \right)\) Show Solution
\[\begin{align*}G\left( s \right) & = 4\frac{s}{{{s^2} + {{\left( 4 \right)}^2}}} - 9\frac{4}{{{s^2} + {{\left( 4 \right)}^2}}} + 2\frac{s}{{{s^2} + {{\left( {10} \right)}^2}}}\\ & = \frac{{4s}}{{{s^2} + 16}} - \frac{{36}}{{{s^2} + 16}} + \frac{{2s}}{{{s^2} + 100}}\end{align*}\]

c \(h\left( t \right) = 3\sinh \left( {2t} \right) + 3\sin \left( {2t} \right)\) Show Solution
\[\begin{align*}H\left( s \right) & = 3\frac{2}{{{s^2} - {{\left( 2 \right)}^2}}} + 3\frac{2}{{{s^2} + {{\left( 2 \right)}^2}}}\\ & = \frac{6}{{{s^2} - 4}} + \frac{6}{{{s^2} + 4}}\end{align*}\]

d \(g\left( t \right) = {{\bf{e}}^{3t}} + \cos \left( {6t} \right) - {{\bf{e}}^{3t}}\cos \left( {6t} \right)\) Show Solution
\[\begin{align*}G\left( s \right) & = \frac{1}{{s - 3}} + \frac{s}{{{s^2} + {{\left( 6 \right)}^2}}} - \frac{{s - 3}}{{{{\left( {s - 3} \right)}^2} + {{\left( 6 \right)}^2}}}\\ & = \frac{1}{{s - 3}} + \frac{s}{{{s^2} + 36}} - \frac{{s - 3}}{{{{\left( {s - 3} \right)}^2} + 36}}\end{align*}\]

Make sure that you pay attention to the difference between a “normal” trig function and hyperbolic functions. The only difference between them is the “\( + {a^2}\)” for the “normal” trig functions becomes a “\( - {a^2}\)” in the hyperbolic function! It’s very easy to get in a hurry and not pay attention and grab the wrong formula. If you don’t recall the definition of the hyperbolic functions see the notes for the table.

Let’s do one final set of examples.

Example 2 Find the transform of each of the following functions.
  1. \(f\left( t \right) = t\cosh \left( {3t} \right)\)
  2. \(h\left( t \right) = {t^2}\sin \left( {2t} \right)\)
  3. \(g\left( t \right) = {t^{\frac{3}{2}}}\)
  4. \(f\left( t \right) = {\left( {10t} \right)^{\frac{3}{2}}}\)
  5. \(f\left( t \right) = tg'\left( t \right)\)
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a \(f\left( t \right) = t\cosh \left( {3t} \right)\) Show Solution

This function is not in the table of Laplace transforms. However, we can use #30 in the table to compute its transform. This will correspond to #30 if we take n=1.

\[F\left( s \right) = \mathcal{L}\left\{ {tg\left( t \right)} \right\} = - G'\left( s \right),\hspace{0.25in}{\mbox{where }}g\left( t \right) = \cosh \left( {3t} \right)\]

So, we then have,

\[G\left( s \right) = \frac{s}{{{s^2} - 9}}\hspace{0.25in}G'\left( s \right) = - \frac{{{s^2} + 9}}{{{{\left( {{s^2} - 9} \right)}^2}}}\]

Using #30 we then have,

\[F\left( s \right) = \frac{{{s^2} + 9}}{{{{\left( {{s^2} - 9} \right)}^2}}}\]

b \(h\left( t \right) = {t^2}\sin \left( {2t} \right)\) Show Solution

This part will also use #30 in the table. In fact, we could use #30 in one of two ways. We could use it with \(n = 1\).

\[H\left( s \right) = \mathcal{L}\left\{ {tf\left( t \right)} \right\} = - F'\left( s \right),\hspace{0.25in}{\mbox{where }}f\left( t \right) = t\sin \left( {2t} \right)\]

Or we could use it with \(n = 2\).

\[H\left( s \right) = \mathcal{L}\left\{ {{t^2}f\left( t \right)} \right\} = F''\left( s \right),\hspace{0.25in}{\mbox{where }}f\left( t \right) = \sin \left( {2t} \right)\]

Since it’s less work to do one derivative, let’s do it the first way. So, using #9 we have,

\[F\left( s \right) = \frac{{4s}}{{{{\left( {{s^2} + 4} \right)}^2}}}\hspace{0.25in}F'\left( s \right) = - \frac{{12{s^2} - 16}}{{{{\left( {{s^2} + 4} \right)}^3}}}\]

The transform is then,

\[H\left( s \right) = \frac{{12{s^2} - 16}}{{{{\left( {{s^2} + 4} \right)}^3}}}\]

c \(g\left( t \right) = {t^{\frac{3}{2}}}\) Show Solution

This part can be done using either #6 (with \(n = 2\)) or #32 (along with #5). We will use #32 so we can see an example of this. In order to use #32 we’ll need to notice that

\[\int_{{\,0}}^{{\,t}}{{\sqrt v \,dv}} = \frac{2}{3}{t^{\frac{3}{2}}}\hspace{0.25in} \Rightarrow \hspace{0.25in}{t^{\frac{3}{2}}} = \frac{3}{2}\int_{{\,0}}^{{\,t}}{{\sqrt v \,dv}}\]

Now, using #5,

\[f\left( t \right) = \sqrt t \hspace{0.25in}F\left( s \right) = \frac{{\sqrt \pi }}{{2{s^{\frac{3}{2}}}}}\]

we get the following.

\[G\left( s \right) = \frac{3}{2}\left( {\frac{{\sqrt \pi }}{{2{s^{\frac{3}{2}}}}}} \right)\left( {\frac{1}{s}} \right) = \frac{{3\sqrt \pi }}{{4{s^{\frac{5}{2}}}}}\]

This is what we would have gotten had we used #6.


d \(f\left( t \right) = {\left( {10t} \right)^{\frac{3}{2}}}\) Show Solution

For this part we will use #24 along with the answer from the previous part. To see this note that if

\[g\left( t \right) = {t^{\frac{3}{2}}}\]

then

\[f\left( t \right) = g\left( {10t} \right)\]

Therefore, the transform is.

\[\begin{align*}F\left( s \right) & = \frac{1}{{10}}G\left( {\frac{s}{{10}}} \right)\\ & = \frac{1}{{10}}\left( {\frac{{3\sqrt \pi }}{{4{{\left( {\frac{s}{{10}}} \right)}^{\frac{5}{2}}}}}} \right)\\ & = {10^{\frac{3}{2}}}\frac{{3\sqrt \pi }}{{4{s^{\frac{5}{2}}}}}\end{align*}\]

e \(f\left( t \right) = tg'\left( t \right)\) Show Solution

This final part will again use #30 from the table as well as #35.

\[\begin{align*}\mathcal{L}\left\{ {tg'\left( t \right)} \right\} & = - \frac{d}{{ds}}\mathcal{L}\left\{ {g'} \right\}\\ & = - \frac{d}{{ds}}\left\{ {sG\left( s \right) - g\left( 0 \right)} \right\}\\ & = - \left( {G\left( s \right) + sG'\left( s \right) - 0} \right)\\ & = - G\left( s \right) - sG'\left( s \right)\end{align*}\]

Remember that \(g(0)\) is just a constant so when we differentiate it we will get zero!

As this set of examples has shown us we can’t forget to use some of the general formulas in the table to derive new Laplace transforms for functions that aren’t explicitly listed in the table!