Differential Equations - Laplace Transforms: Inverse - ii

The last part of this example needed partial fractions to get the inverse transform. When we finally get back to differential equations and we start using Laplace transforms to solve them, you will quickly come to understand that partial fractions are a fact of life in these problems. Almost every problem will require partial fractions to one degree or another.

Note that we could have done the last part of this example as we had done the previous two parts. If we had we would have gotten hyperbolic functions. However, recalling the definition of the hyperbolic functions we could have written the result in the form we got from the way we worked our problem. However, most students have a better feel for exponentials than they do for hyperbolic functions and so it’s usually best to just use partial fractions and get the answer in terms of exponentials. It may be a little more work, but it will give a nicer (and easier to work with) form of the answer.

Be warned that in my class I’ve got a rule that if the denominator can be factored with integer coefficients then it must be.

So, let’s remind you how to get the correct partial fraction decomposition. The first step is to factor the denominator as much as possible. Then for each term in the denominator we will use the following table to get a term or terms for our partial fraction decomposition.

Factor in denominator	Term in partial fraction decomposition
$ax+b$	$\frac{A}{ax+b}$
${\left(ax+b\right)}^k$	$\frac{A_1}{ax+b}+\frac{A_2}{{\left(ax+b\right)}^2}+\cdots +\frac{A_k}{{\left(ax+b\right)}^k}$
$a{x}^2+bx+c$	$\frac{Ax+B}{a{x}^2+bx+c}$
${\left(a{x}^2+bx+c\right)}^k$	$\frac{A_{1}x+B_1}{a{x}^2+bx+c}+\frac{A_{2}x+B_2}{{\left(a{x}^2+bx+c\right)}^2}+\cdots +\frac{A_{k}x+B_k}{{\left(a{x}^2+bx+c\right)}^k}$

Notice that the first and third cases are really special cases of the second and fourth cases respectively.

So, let’s do a couple more examples to remind you how to do partial fractions.

Example 3 Find the inverse transform of each of the following.

1. $G\left(s\right)=\frac{86s-78}{\left(s+3\right)\left(s-4\right)\left(5s-1\right)}$
2. $F\left(s\right)=\frac{2-5s}{\left(s-6\right)\left(s^2+11\right)}$
3. $G\left(s\right)=\frac{25}{s^3\left(s^2+4s+5\right)}$

a) $G\left(s\right)=\frac{86s-78}{\left(s+3\right)\left(s-4\right)\left(5s-1\right)}$ 

Here’s the partial fraction decomposition for this part.
$$G\left(s\right)=\frac{A}{s+3}+\frac{B}{s-4}+\frac{C}{5s-1}$$Now, this time we won’t go into quite the detail as we did in the last example. We are after the numerator of the partial fraction decomposition and this is usually easy enough to do in our heads. Therefore, we will go straight to setting numerators equal.
$$86s-78=A\left(s-4\right)\left(5s-1\right)+B\left(s+3\right)\left(5s-1\right)+C\left(s+3\right)\left(s-4\right)$$As with the last example, we can easily get the constants by correctly picking values of $s$.
$$\begin{array}{cccccc}& s=-3& -336& =A\left(-7\right)\left(-16\right)& \Rightarrow A& =-3\\ & s=\frac{1}{5}& -\frac{304}{5}& =C\left(\frac{16}{5}\right)\left(-\frac{19}{5}\right)& \Rightarrow C& =5\\ & s=4& 266& =B\left(7\right)\left(19\right)& \Rightarrow B& =2\end{array}$$So, the partial fraction decomposition for this transform is,
$$G\left(s\right)=-\frac{3}{s+3}+\frac{2}{s-4}+\frac{5}{5s-1}$$Now, in order to actually take the inverse transform we will need to factor a 5 out of the denominator of the last term. The corrected transform as well as its inverse transform is.
$$\begin{array}{cc}G\left(s\right)& =-\frac{3}{s+3}+\frac{2}{s-4}+\frac{1}{s-\frac{1}{5}}\\ g\left(t\right)& =-3e^{-3t}+2e^{4t}+e^{\frac{t}{5}}\end{array}$$

b) $F\left(s\right)=\frac{2-5s}{\left(s-6\right)\left(s^2+11\right)}$ 

So, for the first time we’ve got a quadratic in the denominator. Here’s the decomposition for this part.
$$F\left(s\right)=\frac{A}{s-6}+\frac{Bs+C}{s^2+11}$$Setting numerators equal gives,
$$2-5s=A\left(s^2+11\right)+\left(Bs+C\right)\left(s-6\right)$$Okay, in this case we could use $s=6$ to quickly find $A$, but that’s all it would give. In this case we will need to go the “long” way around to getting the constants. Note that this way will always work but is sometimes more work than is required.
The “long” way is to completely multiply out the right side and collect like terms.
$$\begin{array}{cc}2-5s=A\left(s^2+11\right)+\left(Bs+C\right)\left(s-6\right)& \\ & =A{s}^2+11A+B{s}^2-6B+Cs-6C\\ & =\left(A+B\right)s^2+\left(-6B+C\right)s+11A-6C\end{array}$$In order for these two to be equal the coefficients of the $s^2$, $s$ and the constants must all be equal. So, setting coefficients equal gives the following system of equations that can be solved.
$$\begin{array}{cccc}& s^2:& A+B& =0\\ & s^1:& -6B+C& =-5\\ & s^0:& 11A-6C& =2\end{array}\}\Rightarrow A=-\frac{28}{47},B=\frac{28}{47},C=-\frac{67}{47}$$Notice that we used $s^0$ to denote the constants. This is habit on my part and isn’t really required, it’s just what I’m used to doing. Also, the coefficients are fairly messy fractions in this case. Get used to that. They will often be like this when we get back into solving differential equations.
There is a way to make our life a little easier as well with this. Since all of the fractions have a denominator of 47 we’ll factor that out as we plug them back into the decomposition. This will make dealing with them much easier. The partial fraction decomposition is then,
$$\begin{array}{cc}F\left(s\right)& =\frac{1}{47}\left(-\frac{28}{s-6}+\frac{28s-67}{s^2+11}\right)\\ & =\frac{1}{47}\left(-\frac{28}{s-6}+\frac{28s}{s^2+11}-\frac{67\frac{\sqrt{11}}{\sqrt{11}}}{s^2+11}\right)\end{array}$$The inverse transform is then.
$$f\left(t\right)=\frac{1}{47}\left(-28e^{6t}+28\cos \left(\sqrt{11}t\right)-\frac{67}{\sqrt{11}}\sin \left(\sqrt{11}t\right)\right)$$

c) $G\left(s\right)=\frac{25}{s^3\left(s^2+4s+5\right)}$ 

With this last part do not get excited about the $s^3$. We can think of this term as
$$s^3={\left(s-0\right)}^3$$and it becomes a linear term to a power. So, the partial fraction decomposition is
$$G\left(s\right)=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{Ds+E}{s^2+4s+5}$$Setting numerators equal and multiplying out gives.
$$\begin{array}{cc}25& =A{s}^2\left(s^2+4s+5\right)+Bs\left(s^2+4s+5\right)+C\left(s^2+4s+5\right)+\left(Ds+E\right)s^3\\ & =\left(A+D\right)s^4+\left(4A+B+E\right)s^3+\left(5A+4B+C\right)s^2+\left(5B+4C\right)s+5C\end{array}$$Setting coefficients equal gives the following system.
$$\begin{array}{cccc}& s^4:& A+D& =0\\ & s^3:& 4A+B+E& =0\\ & s^2:& 5A+4B+C& =0\\ & s^1:& 5B+4C& =0\\ & s^0:& 5C& =25\end{array}\}\Rightarrow A=\frac{11}{5},B=-4,C=5,D=-\frac{11}{5},E=-\frac{24}{5}$$This system looks messy, but it’s easier to solve than it might look. First, we get $C$ for free from the last equation. We can then use the fourth equation to find $B$. The third equation will then give $A$, etc.
When plugging into the decomposition we’ll get everything with a denominator of 5, then factor that out as we did in the previous part in order to make things easier to deal with.
$$G\left(s\right)=\frac{1}{5}\left(\frac{11}{s}-\frac{20}{s^2}+\frac{25}{s^3}-\frac{11s+24}{s^2+4s+5}\right)$$Note that we also factored a minus sign out of the last two terms. To complete this part we’ll need to complete the square on the later term and fix up a couple of numerators. Here’s that work.
$$\begin{array}{cc}G\left(s\right)& =\frac{1}{5}\left(\frac{11}{s}-\frac{20}{s^2}+\frac{25}{s^3}-\frac{11s+24}{s^2+4s+5}\right)\\ & =\frac{1}{5}\left(\frac{11}{s}-\frac{20}{s^2}+\frac{25}{s^3}-\frac{11\left(s+2-2\right)+24}{{\left(s+2\right)}^2+1}\right)\\ & =\frac{1}{5}\left(\frac{11}{s}-\frac{20}{s^2}+\frac{25\frac{2!}{2!}}{s^3}-\frac{11\left(s+2\right)}{{\left(s+2\right)}^2+1}-\frac{2}{{\left(s+2\right)}^2+1}\right)\end{array}$$The inverse transform is then.
$$g\left(t\right)=\frac{1}{5}\left(11-20t+\frac{25}{2}{t}^2-11e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)\right)$$

So, one final time. Partial fractions are a fact of life when using Laplace transforms to solve differential equations. Make sure that you can deal with them.