Differential Equations - Laplace Transforms: Step Functions

Differential Equations - Laplace Transforms: Step Functions - iii

F (s) = \frac{4 s + e^{- s}}{(s - 1) (s + 2)}

In this case, unlike the previous part, we will need to break up the transform since one term has a constant in it and the other has an

s

. Note as well that we don’t consider the exponential in this, only its coefficient. Breaking up the transform gives,

F (s) = \frac{4 s}{(s - 1) (s + 2)} + e^{- s} \frac{1}{(s - 1) (s + 2)} = G (s) + e^{- s} H (s)

We will need to partial fraction both of these terms up. We’ll start with

G (s)

G (s) = \frac{A}{s - 1} + \frac{B}{s + 2}

Setting numerators equal gives,

4 s = A (s + 2) + B (s - 1)

Now, pick values of

s

to find the constants.

\begin{aligned} s = - 2 & - 8 & = - 3 B & \Rightarrow B & = \frac{8}{3} \\ s = 1 & 4 & = 3 A & \Rightarrow A & = \frac{4}{3} \end{aligned}

G (s)

and its inverse transform is,

\begin{aligned} G (s) & = \frac{\frac{4}{3}}{s - 1} + \frac{\frac{8}{3}}{s + 2} \\ g (t) & = \frac{4}{3} e^{t} + \frac{8}{3} e^{- 2 t} \end{aligned}

Now, repeat the process for

H (s)

H (s) = \frac{A}{s - 1} + \frac{B}{s + 2}

Setting numerators equal gives,

1 = A (s + 2) + B (s - 1)

Now, pick values of

s

to find the constants.

\begin{aligned} s = - 2 & 1 & = - 3 B & \Rightarrow B & = - \frac{1}{3} \\ s = 1 & 1 & = 3 A & \Rightarrow A & = \frac{1}{3} \end{aligned}

H (s)

and its inverse transform is,

\begin{aligned} H (s) & = \frac{\frac{1}{3}}{s - 1} - \frac{\frac{1}{3}}{s + 2} \\ h (t) & = \frac{1}{3} e^{t} - \frac{1}{3} e^{- 2 t} \end{aligned}

Putting all of this together gives the following,

\begin{aligned} F (s) & = G (s) + e^{- s} H (s) \\ f (t) & = g (t) + u_{1} (t) h (t - 1) \end{aligned}

where,

g (t) = \frac{4}{3} e^{t} + \frac{8}{3} e^{- 2 t} and h (t) = \frac{1}{3} e^{t} - \frac{1}{3} e^{- 2 t}

G (s) = \frac{3 s + 8 e^{- 20 s} - 2 s e^{- 3 s} + 6 e^{- 7 s}}{s^{2} (s + 3)}

This one looks messier than it actually is. Let’s first rearrange the numerator a little.

G (s) = \frac{s (3 - 2 e^{- 3 s}) + (8 e^{- 20 s} + 6 e^{- 7 s})}{s^{2} (s + 3)}

In this form it looks like we can break this up into two pieces that will require partial fractions. When we break these up we should always try and break things up into as few pieces as possible for the partial fractioning. Doing this can save you a great deal of unnecessary work. Breaking up the transform as suggested above gives,

\begin{aligned} G (s) & = (3 - 2 e^{- 3 s}) \frac{1}{s (s + 3)} + (8 e^{- 20 s} + 6 e^{- 7 s}) \frac{1}{s^{2} (s + 3)} \\ = (3 - 2 e^{- 3 s}) F (s) + (8 e^{- 20 s} + 6 e^{- 7 s}) H (s) \end{aligned}

Note that we canceled an

s

F (s)

. You should always simplify as much a possible before doing the partial fractions.
Let’s partial fraction up

F (s)

first.

F (s) = \frac{A}{s} + \frac{B}{s + 3}

Setting numerators equal gives,

1 = A (s + 3) + B s

Now, pick values of

s

to find the constants.

\begin{aligned} s = - 3 & 1 & = - 3 B & \Rightarrow B & = - \frac{1}{3} \\ s = 0 & 1 & = 3 A & \Rightarrow A & = \frac{1}{3} \end{aligned}

F (s)

and its inverse transform is,

\begin{aligned} F (s) & = \frac{\frac{1}{3}}{s} - \frac{\frac{1}{3}}{s + 3} \\ f (t) & = \frac{1}{3} - \frac{1}{3} e^{- 3 t} \end{aligned}

Now partial fraction

H (s)

H (s) = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s + 3}

Setting numerators equal gives,

1 = A s (s + 3) + B (s + 3) + C s^{2}

Pick values of

s

to find the constants.

\begin{aligned} s = - 3 & 1 & = 9 C & \Rightarrow C & = \frac{1}{9} \\ s = 0 & 1 & = 3 B & \Rightarrow B & = \frac{1}{3} \\ s = 1 & 1 & = 4 A + 4 B + C = 4 A + \frac{13}{9} & \Rightarrow A & = - \frac{1}{9} \end{aligned}

So,

H (s)

and its inverse transform is,

\begin{aligned} H (s) & = - \frac{\frac{1}{9}}{s} + \frac{\frac{1}{3}}{s^{2}} + \frac{\frac{1}{9}}{s + 3} \\ h (t) & = - \frac{1}{9} + \frac{1}{3} t + \frac{1}{9} e^{- 3 t} \end{aligned}

Now, let’s go back to the original problem, remembering to multiply the transform through the parenthesis.

G (s) = 3 F (s) - 2 e^{- 3 s} F (s) + 8 e^{- 20 s} H (s) + 6 e^{- 7 s} H (s)

Taking the inverse transform gives,

g (t) = 3 f (t) - 2 u_{3} (t) f (t - 3) + 8 u_{20} (t) h (t - 20) + 6 u_{7} (t) h (t - 7)

So, as this example has shown, these can be a somewhat messy. However, the mess is really only that of notation and amount of work. The actual partial fraction work was identical to the previous sections work. The main difference in this section is we had to do more of it. As far as the inverse transform process goes. Again, the vast majority of that was identical to the previous section as well.
So, don’t let the apparent messiness of these problems get you to decide that you can’t do them. Generally, they aren’t as bad as they seem initially.

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Calculus III - 3-Dimensional Space: Equations of Lines