Differential Equations - Laplace Transforms: Inverse - i

Finding the Laplace transform of a function is not terribly difficult if we’ve got a table of transforms in front of us to use as we saw in the last section. What we would like to do now is go the other way.

We are going to be given a transform, $F\left(s\right)$, and ask what function (or functions) did we have originally. As you will see this can be a more complicated and lengthy process than taking transforms. In these cases we say that we are finding the Inverse Laplace Transform of $F\left(s\right)$ and use the following notation.

$$f\left(t\right)=L^{-1}\left\{F\left(s\right)\right\}$$

As with Laplace transforms, we’ve got the following fact to help us take the inverse transform.

Fact

Given the two Laplace transforms $F\left(s\right)$ and $G\left(s\right)$ then
$$L^{-1}\left\{aF\left(s\right)+bG\left(s\right)\right\}=a{L}^{-1}\left\{F\left(s\right)\right\}+b{L}^{-1}\left\{G\left(s\right)\right\}$$for any constants $a$ and $b$.

So, we take the inverse transform of the individual transforms, put any constants back in and then add or subtract the results back up.

Let’s take a look at a couple of fairly simple inverse transforms.

Example 1 Find the inverse transform of each of the following.

1. $F\left(s\right)=\frac{6}{s}-\frac{1}{s-8}+\frac{4}{s-3}$
2. $H\left(s\right)=\frac{19}{s+2}-\frac{1}{3s-5}+\frac{7}{s^5}$
3. $F\left(s\right)=\frac{6s}{s^2+25}+\frac{3}{s^2+25}$
4. $G\left(s\right)=\frac{8}{3s^2+12}+\frac{3}{s^2-49}$

We’ve always felt that the key to doing inverse transforms is to look at the denominator and try to identify what you’ve got based on that. If there is only one entry in the table that has that particular denominator, the next step is to make sure the numerator is correctly set up for the inverse transform process. If it isn’t, correct it (this is always easy to do) and then take the inverse transform.
If there is more than one entry in the table that has a particular denominator, then the numerators of each will be different, so go up to the numerator and see which one you’ve got. If you need to correct the numerator to get it into the correct form and then take the inverse transform.
So, with this advice in mind let’s see if we can take some inverse transforms.

a) $F\left(s\right)=\frac{6}{s}-\frac{1}{s-8}+\frac{4}{s-3}$ 

From the denominator of the first term it looks like the first term is just a constant. The correct numerator for this term is a “1” so we’ll just factor the 6 out before taking the inverse transform. The second term appears to be an exponential with $a=8$ and the numerator is exactly what it needs to be. The third term also appears to be an exponential, only this time $a=3$ and we’ll need to factor the 4 out before taking the inverse transforms.
So, with a little more detail than we’ll usually put into these,
$$\begin{array}{cc}F\left(s\right)& =6\frac{1}{s}-\frac{1}{s-8}+4\frac{1}{s-3}\\ & f\left(t\right)=6\left(1\right)-e^{8t}+4\left(e^{3t}\right)\\ & =6-e^{8t}+4e^{3t}\end{array}$$

b) $H\left(s\right)=\frac{19}{s+2}-\frac{1}{3s-5}+\frac{7}{s^5}$ 

The first term in this case looks like an exponential with $a=-2$ and we’ll need to factor out the 19. Be careful with negative signs in these problems, it’s very easy to lose track of them.
The second term almost looks like an exponential, except that it’s got a $3s$ instead of just an $s$ in the denominator. It is an exponential, but in this case, we’ll need to factor a 3 out of the denominator before taking the inverse transform.
The denominator of the third term appears to be #3 in the table with $n=4$. The numerator however, is not correct for this. There is currently a 7 in the numerator and we need a $4!=24$ in the numerator. This is very easy to fix. Whenever a numerator is off by a multiplicative constant, as in this case, all we need to do is put the constant that we need in the numerator. We will just need to remember to take it back out by dividing by the same constant.
So, let’s first rewrite the transform.
$$\begin{array}{cc}H\left(s\right)& =\frac{19}{s-\left(-2\right)}-\frac{1}{3\left(s-\frac{5}{3}\right)}+\frac{7\frac{4!}{4!}}{s^{4+1}}\\ & =19\frac{1}{s-\left(-2\right)}-\frac{1}{3}\frac{1}{s-\frac{5}{3}}+\frac{7}{4!}\frac{4!}{s^{4+1}}\end{array}$$So, what did we do here? We factored the 19 out of the first term. We factored the 3 out of the denominator of the second term since it can’t be there for the inverse transform and in the third term we factored everything out of the numerator except the 4! since that is the portion that we need in the numerator for the inverse transform process.
Let’s now take the inverse transform.
$$h\left(t\right)=19e^{-2t}-\frac{1}{3}{e}^{\frac{5t}{3}}+\frac{7}{24}{t}^4$$

c) $F\left(s\right)=\frac{6s}{s^2+25}+\frac{3}{s^2+25}$ In this part we’ve got the same denominator in both terms and our table tells us that we’ve either got #7 or #8. The numerators will tell us which we’ve actually got. The first one has an 

$s$ in the numerator and so this means that the first term must be #8 and we’ll need to factor the 6 out of the numerator in this case. The second term has only a constant in the numerator and so this term must be #7, however, in order for this to be exactly #7 we’ll need to multiply/divide a 5 in the numerator to get it correct for the table.
The transform becomes,
$$\begin{array}{cc}F\left(s\right)& =6\frac{s}{s^2+{\left(5\right)}^2}+\frac{3\frac{5}{5}}{s^2+{\left(5\right)}^2}\\ & =6\frac{s}{s^2+{\left(5\right)}^2}+\frac{3}{5}\frac{5}{s^2+{\left(5\right)}^2}\end{array}$$Taking the inverse transform gives,
$$f\left(t\right)=6\cos \left(5t\right)+\frac{3}{5}\sin \left(5t\right)$$

d) $G\left(s\right)=\frac{8}{3s^2+12}+\frac{3}{s^2-49}$ 

In this case the first term will be a sine once we factor a 3 out of the denominator, while the second term appears to be a hyperbolic sine (#17). Again, be careful with the difference between these two. Both of the terms will also need to have their numerators fixed up. Here is the transform once we’re done rewriting it.
$$\begin{array}{cc}G\left(s\right)& =\frac{1}{3}\frac{8}{s^2+4}+\frac{3}{s^2-49}\\ & =\frac{1}{3}\frac{\left(4\right)\left(2\right)}{s^2+{\left(2\right)}^2}+\frac{3\frac{7}{7}}{s^2-{\left(7\right)}^2}\end{array}$$Notice that in the first term we took advantage of the fact that we could get the 2 in the numerator that we needed by factoring the 8. The inverse transform is then,
$$g\left(t\right)=\frac{4}{3}\sin \left(2t\right)+\frac{3}{7}\sinh \left(7t\right)$$

So, probably the best way to identify the transform is by looking at the denominator. If there is more than one possibility use the numerator to identify the correct one. Fix up the numerator if needed to get it into the form needed for the inverse transform process. Finally, take the inverse transform.

Let’s do some slightly harder problems. These are a little more involved than the first set.

Example 2 Find the inverse transform of each of the following.

1. $F\left(s\right)=\frac{6s-5}{s^2+7}$
2. $F\left(s\right)=\frac{1-3s}{s^2+8s+21}$
3. $G\left(s\right)=\frac{3s-2}{2s^2-6s-2}$
4. $H\left(s\right)=\frac{s+7}{s^2-3s-10}$

a) $F\left(s\right)=\frac{6s-5}{s^2+7}$ 

From the denominator of this one it appears that it is either a sine or a cosine. However, the numerator doesn’t match up to either of these in the table. A cosine wants just an $s$ in the numerator with at most a multiplicative constant, while a sine wants only a constant and no $s$ in the numerator.
We’ve got both in the numerator. This is easy to fix however. We will just split up the transform into two terms and then do inverse transforms.
$$\begin{array}{cc}F\left(s\right)& =\frac{6s}{s^2+7}-\frac{5\frac{\sqrt{7}}{\sqrt{7}}}{s^2+7}\\ f\left(t\right)& =6\cos \left(\sqrt{7}t\right)-\frac{5}{\sqrt{7}}\sin \left(\sqrt{7}t\right)\end{array}$$Do not get too used to always getting the perfect squares in sines and cosines that we saw in the first set of examples. More often than not (at least in my class) they won’t be perfect squares!

b) $F\left(s\right)=\frac{1-3s}{s^2+8s+21}$ 

In this case there are no denominators in our table that look like this. We can however make the denominator look like one of the denominators in the table by completing the square on the denominator. So, let’s do that first.
$$\begin{array}{cc}{s}^2+8s+21& =s^2+8s+16-16+21\\ & =s^2+8s+16+5\\ & ={\left(s+4\right)}^2+5\end{array}$$Recall that in completing the square you take half the coefficient of the $s$, square this, and then add and subtract the result to the polynomial. After doing this the first three terms should factor as a perfect square.
So, the transform can be written as the following.
$$F\left(s\right)=\frac{1-3s}{{\left(s+4\right)}^2+5}$$Okay, with this rewrite it looks like we’ve got #19 and/or #20’s from our table of transforms. However, note that in order for it to be a #19 we want just a constant in the numerator and in order to be a #20 we need an $s–a$ in the numerator. We’ve got neither of these, so we’ll have to correct the numerator to get it into proper form.
In correcting the numerator always get the $s–a$ first. This is the important part. We will also need to be careful of the 3 that sits in front of the $s$. One way to take care of this is to break the term into two pieces, factor the 3 out of the second and then fix up the numerator of this term. This will work; however, it will put three terms into our answer and there are really only two terms.
So, we will leave the transform as a single term and correct it as follows,
$$\begin{array}{cc}F\left(s\right)& =\frac{1-3\left(s+4-4\right)}{{\left(s+4\right)}^2+5}\\ & =\frac{1-3\left(s+4\right)+12}{{\left(s+4\right)}^2+5}\\ & =\frac{-3\left(s+4\right)+13}{{\left(s+4\right)}^2+5}\end{array}$$We needed an $s+4$ in the numerator, so we put that in. We just needed to make sure and take the 4 back out by subtracting it back out. Also, because of the 3 multiplying the $s$ we needed to do all this inside a set of parenthesis. Then we partially multiplied the 3 through the second term and combined the constants. With the transform in this form, we can break it up into two transforms each of which are in the tables and so we can do inverse transforms on them,
$$\begin{array}{cc}F\left(s\right)& =-3\frac{s+4}{{\left(s+4\right)}^2+5}+\frac{13\frac{\sqrt{5}}{\sqrt{5}}}{{\left(s+4\right)}^2+5}\\ f\left(t\right)& =-3e^{-4t}\cos \left(\sqrt{5}t\right)+\frac{13}{\sqrt{5}}{e}^{-4t}\sin \left(\sqrt{5}t\right)\end{array}$$

c) $G\left(s\right)=\frac{3s-2}{2s^2-6s-2}$ 

This one is similar to the last one. We just need to be careful with the completing the square however. The first thing that we should do is factor a 2 out of the denominator, then complete the square. Remember that when completing the square a coefficient of 1 on the $s^2$ term is needed! So, here’s the work for this transform.
$$\begin{array}{cc}G\left(s\right)& =\frac{3s-2}{2\left(s^2-3s-1\right)}\\ & =\frac{1}{2}\frac{3s-2}{s^2-3s+\frac{9}{4}-\frac{9}{4}-1}\\ & =\frac{1}{2}\frac{3s-2}{{\left(s-\frac{3}{2}\right)}^2-\frac{13}{4}}\end{array}$$So, it looks like we’ve got #21 and #22 with a corrected numerator. Here’s the work for that and the inverse transform.
$$\begin{array}{cc}G\left(s\right)& =\frac{1}{2}\frac{3\left(s-\frac{3}{2}+\frac{3}{2}\right)-2}{{\left(s-\frac{3}{2}\right)}^2-\frac{13}{4}}\\ & =\frac{1}{2}\frac{3\left(s-\frac{3}{2}\right)+\frac{5}{2}}{{\left(s-\frac{3}{2}\right)}^2-\frac{13}{4}}\\ & =\frac{1}{2}\left(\frac{3\left(s-\frac{3}{2}\right)}{{\left(s-\frac{3}{2}\right)}^2-\frac{13}{4}}+\frac{\frac{5}{2}\frac{\sqrt{13}}{\sqrt{13}}}{{\left(s-\frac{3}{2}\right)}^2-\frac{13}{4}}\right)\\ g\left(t\right)& =\frac{1}{2}\left(3e^{\frac{3t}{2}}\cosh \left(\frac{\sqrt{13}}{2}t\right)+\frac{5}{\sqrt{13}}{e}^{\frac{3t}{2}}\sinh \left(\frac{\sqrt{13}}{2}t\right)\right)\end{array}$$In correcting the numerator of the second term, notice that I only put in the square root since we already had the “over 2” part of the fraction that we needed in the numerator.

d) $H\left(s\right)=\frac{s+7}{s^2-3s-10}$ 

This one appears to be similar to the previous two, but it actually isn’t. The denominators in the previous two couldn’t be easily factored. In this case the denominator does factor and so we need to deal with it differently. Here is the transform with the factored denominator.
$$H\left(s\right)=\frac{s+7}{\left(s+2\right)\left(s-5\right)}$$The denominator of this transform seems to suggest that we’ve got a couple of exponentials, however in order to be exponentials there can only be a single term in the denominator and no $s$’s in the numerator.
To fix this we will need to do partial fractions on this transform. In this case the partial fraction decomposition will be
$$H\left(s\right)=\frac{A}{s+2}+\frac{B}{s-5}$$Don’t remember how to do partial fractions? In this example we’ll show you one way of getting the values of the constants and after this example we’ll review how to get the correct form of the partial fraction decomposition.
Okay, so let’s get the constants. There is a method for finding the constants that will always work, however it can lead to more work than is sometimes required. Eventually, we will need that method, however in this case there is an easier way to find the constants.
Regardless of the method used, the first step is to actually add the two terms back up. This gives the following.
$$\frac{s+7}{\left(s+2\right)\left(s-5\right)}=\frac{A\left(s-5\right)+B\left(s+2\right)}{\left(s+2\right)\left(s-5\right)}$$Now, this needs to be true for any $s$ that we should choose to put in. So, since the denominators are the same we just need to get the numerators equal. Therefore, set the numerators equal.
$$s+7=A\left(s-5\right)+B\left(s+2\right)$$Again, this must be true for ANY value of $s$ that we want to put in. So, let’s take advantage of that. If it must be true for any value of $s$ then it must be true for $s=-2$, to pick a value at random. In this case we get,
$$5=A\left(-7\right)+B\left(0\right)\Rightarrow A=-\frac{5}{7}$$We found $A$ by appropriately picking $s$. We can $B$ in the same way if we chose $s=5$.
$$12=A\left(0\right)+B\left(7\right)\Rightarrow B=\frac{12}{7}$$This will not always work, but when it does it will usually simplify the work considerably.
So, with these constants the transform becomes,
$$H\left(s\right)=\frac{-\frac{5}{7}}{s+2}+\frac{\frac{12}{7}}{s-5}$$We can now easily do the inverse transform to get,
$$h\left(t\right)=-\frac{5}{7}{e}^{-2t}+\frac{12}{7}{e}^{5t}$$