Problem 19 - Differential Equations and Linear Algebra

Question

Determine the inverse Laplace transform of the given function. $$F(s)=\frac{s+4}{(s-1)(s+2)(s-3)}.$$

Step-by-Step Solution

Answer

The inverse Laplace transform of the given function \(F(s) = \frac{s+4}{(s-1)(s+2)(s-3)}\) is: \[f(t) = -\frac{5}{4} e^{t} + \frac{2}{5} e^{-2t} + \frac{7}{10} e^{3t}\]

Step 1: Perform partial fraction decomposition

In order to perform partial fraction decomposition on \(F(s)\), we will first set up the associated equation: \[\frac{s+4}{(s-1)(s+2)(s-3)} = \frac{A}{s-1} + \frac{B}{s+2} + \frac{C}{s-3}\] Now, we will need to multiply both sides of the equation by the common denominator to clear the fractions and express the function in terms of A, B, and C.

Step 2: Clear the fractions by multiplying both sides by the common denominator

Multiply both sides by \((s-1)(s+2)(s-3)\): \[(s+4) = A(s+2)(s-3) + B(s-1)(s-3) + C(s-1)(s+2)\]

Step 3: Solve for A, B, and C using either the simultaneous equations method or by plugging in suitable values of s

In this case, we can use the method of plugging in suitable values of \(s\): Choose \(s = 1\) to get A: \[1 + 4 = A(1 + 2)(1 - 3)\] \[A = -\frac{5}{4}\] Choose \(s = -2\) to get B: \[-2 + 4 = B(-2 - 1)(-2 - 3)\] \[B = \frac{2}{5}\] Choose \(s = 3\) to get C: \[3 + 4 = C(3 - 1)(3 + 2)\] \[C = \frac{7}{10}\]

Step 4: Write the partial fractions with the found coefficients

We can now replace A, B, and C by the values found in the previous step: \[F(s) = \frac{-\frac{5}{4}}{s-1} + \frac{\frac{2}{5}}{s+2} + \frac{\frac{7}{10}}{s-3}\] Now, we will use the inverse Laplace transform rules to find the inverse transform of each of the fractions.

Step 5: Apply the inverse Laplace transform rules to each of the fractions

Recall that the inverse Laplace transform of \(\frac{k}{s-a}\) is \(k e^{at}\). So, we can find the inverse Laplace transform of each of the fractions: \[\mathcal{L}^{-1}\{F(s)\}(t) =\mathcal{L}^{-1}\left\{-\frac{5}{4} \cdot \frac{1}{s-1} + \frac{2}{5} \cdot \frac{1}{s+2} + \frac{7}{10} \cdot \frac{1}{s-3}\right\}(t)\] \[f(t) = -\frac{5}{4} e^{t} + \frac{2}{5} e^{-2t} + \frac{7}{10} e^{3t}\] The inverse Laplace transform of the given function \(F(s)\) is: \[f(t) = -\frac{5}{4} e^{t} + \frac{2}{5} e^{-2t} + \frac{7}{10} e^{3t}\]

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a useful technique in calculus and engineering, particularly when dealing with complex fractions. It allows us to break down a complex rational expression into simpler fractions, which are easier to work with, especially when taking inverse Laplace transforms.
To start with partial fraction decomposition, we express the given function, in this case \(F(s) = \frac{s+4}{(s-1)(s+2)(s-3)}\), as a sum of simpler fractions. We set it equal to a sum of fractions with unknown coefficients:

• \(\frac{A}{s-1}\)
• \(\frac{B}{s+2}\)
• \(\frac{C}{s-3}\)

Next, we clear the fractions by multiplying through by the common denominator, leading to an expression we can solve for \(A\), \(B\), and \(C\). We do this by substituting strategic values of \(s\) to isolate each variable. This results in:

• \(A = -\frac{5}{4}\)
• \(B = \frac{2}{5}\)
• \(C = \frac{7}{10}\)

Using these values, the decomposed function becomes \(\frac{-\frac{5}{4}}{s-1} + \frac{\frac{2}{5}}{s+2} + \frac{\frac{7}{10}}{s-3}\). This makes finding inverse transforms more manageable.

Laplace Transform Rules

Laplace Transform Rules are essential tools for solving differential equations and transforming functions from the time domain into the frequency domain, and vice versa. When dealing with inverse Laplace transforms, rules help us move from a function like \(F(s)\) back to \(f(t)\) with relative ease.
One of the key rules for inverse Laplace transforms is:
\[ \mathcal{L}^{-1}\{\frac{k}{s-a}\} = k e^{at} \]
This rule tells us how to transform basic fractions of the form \(\frac{k}{s-a}\) into exponential functions in time-domain \(t\). By applying this rule to each term of the decomposed function in our example, we can obtain the inverse transform simply by adjusting \(a\) and \(k\) in our exponential functions.

• For \(\frac{-\frac{5}{4}}{s-1}\), the transform is \(-\frac{5}{4} e^{t}\)
• For \(\frac{\frac{2}{5}}{s+2}\), the transform is \(\frac{2}{5} e^{-2t}\)
• For \(\frac{\frac{7}{10}}{s-3}\), the transform is \(\frac{7}{10} e^{3t}\)

These calculations provide the components needed to construct the complete inverse Laplace transform, combining into \(f(t) = -\frac{5}{4} e^{t} + \frac{2}{5} e^{-2t} + \frac{7}{10} e^{3t}\).

Exponential Functions

Exponential Functions are a fundamental part of the inverse Laplace transform, allowing us to convert frequency-domain representations back into the time domain. The behavior of exponential functions, \(e^{at}\), is dictated by the power \(a\), which indicates the rate of growth or decay.
When \(a > 0\), the exponential function \(e^{at}\) represents continuous growth over time. Conversely, \(a < 0\) reflects decay, as seen in the function \(e^{-2t}\) from our solution.
The scaling coefficients, such as \(-\frac{5}{4}\), \(\frac{2}{5}\), and \(\frac{7}{10}\), determine how sharply or gently each component influences the system's overall response.

• In \(-\frac{5}{4} e^{t}\), the function grows, moving negatively due to the constant.
• \(\frac{2}{5} e^{-2t}\) reflects a slower decay, plateauing as time increases.
• \(\frac{7}{10} e^{3t}\) grows rapidly, influenced strongly by the larger coefficient.

Exponential functions thus allow engineers and scientists to model real-world phenomena such as population growth, radioactive decay, and voltage across capacitors. Understanding their dynamics is key to interpreting the result of an inverse Laplace transform effectively.