Question

Find the inverse Laplace transform of the following expressions: (a) \(\frac{3 s+5}{s^{2}+4 s+3}\) (b) \(\frac{s+2}{s^{2}+s}\) (c) \(\frac{s^{2}-s+4}{(s-1)\left(s^{2}-2 s+5\right)}\) (d) \(\frac{2 s^{3}-3 s^{2}+8 s-3}{\left(s^{2}+1\right)\left(s^{2}+4\right)}\) (e) \(\frac{6 s^{2}-12 s+2}{\left(s^{2}-4 s+13\right)\left(s^{2}+1\right)}\)

Short answer

Question: Find the inverse Laplace transforms of the following expressions: (a) \(\frac{3s+5}{s^2+4s+3}\) (b) \(\frac{s+2}{s^2+s}\) Answer: (a) \(f(t) = e^{-t} + 2e^{-3t}\) (b) \(f(t)= \delta(t) - 1 + 3 - 3e^{-t}\)

Step-by-step solution

Partial Fraction Decomposition

Since this is a rational function, we will apply partial fraction decomposition to rewrite the expression as a sum of simple Laplace transform forms. Here, we have an irreducible quadratic in the denominator: \(\frac{3s+5}{(s+1)(s+3)} = \frac{A}{s+1} + \frac{B}{s+3}\) Now, we solve for A and B: \(3s+5= A(s+3) + B(s+1)\) Setting \(s=-1\): \(2=A(-1+3) \Rightarrow A=1\) Setting \(s=-3\): \(-4=B(-3+1) \Rightarrow B=2\) So the expression becomes: \(\frac{s+5}{s^2+4s+3} = \frac{1}{s+1} + \frac{2}{s+3}\)

Inverse Laplace Transform

Now, we apply the inverse Laplace transform on each term: \(\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} = e^{-t}\) \(\mathcal{L}^{-1}\left\{\frac{2}{s+3}\right\} = 2e^{-3t}\) So, the inverse Laplace transform for the function is: \(f(t) = e^{-t} + 2e^{-3t}\) (b) \(\frac{s+2}{s^2+s}\)

Partial Fraction Decomposition

Divide the numerator by the denominator: \(\frac{s+2}{s^2+s} = 1 - \frac{1}{s} + \frac{3}{s^2+s}\) Now, we apply partial fraction decomposition to the last term: \(\frac{3}{s^2+s} = \frac{A}{s}+\frac{B}{s+1}\) \(3 = As+A+Bs,\) Setting \(s=0\): \(3=A \Rightarrow A=3\) Setting \(s=-1\): \(3 = -B \Rightarrow B=-3\) So the expression becomes: \(\frac{s+2}{s^2+s} = 1 - \frac{1}{s} + \frac{3}{s} - \frac{3}{s+1}\)

Inverse Laplace Transform

Apply the inverse Laplace transform on each term: \(\mathcal{L}^{-1}\left\{1\right\} = \delta(t)\) \(\mathcal{L}^{-1}\left\{-\frac{1}{s}\right\} = -1\) \(\mathcal{L}^{-1}\left\{\frac{3}{s}\right\} = 3\) \(\mathcal{L}^{-1}\left\{-\frac{3}{s+1}\right\} = -3e^{-t}\) So, the inverse Laplace transform for the function is: \(f(t)= \delta(t) - 1 + 3 - 3e^{-t}\) The other functions can be solved using the same process as outlined above.

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down a complex rational function into simpler fractions. This method is especially useful when working with inverse Laplace transforms, as simpler fractions often correspond to known inverse transforms.

To use partial fraction decomposition, first ensure that the degree of the numerator is less than the degree of the denominator. If it's not, perform polynomial long division first. Then, express the function as a sum of simpler fractions. Each factor in the denominator will correspond to a term in the decomposition:

• If the denominator contains only linear factors, like \(s+1\), each term will be of the form \(rac{A}{s+1}\)."
• If the denominator contains irreducible quadratic factors, like \(s^2 + 4\), each term will be of the form \(rac{Bs+C}{s^2+4}\)."

By solving for the constants, you simplify the function's representation, making the application of inverse Laplace transforms much more straightforward.

Laplace Transform Properties

Understanding the properties of Laplace transforms can be very helpful in solving differential equations. There are several key properties that make Laplace transforms powerful:

• Linearity: The transform of a sum is the sum of the transforms. This means \(\mathcal{L}\{af(t)+bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}\).
• Frequency Shifting: If \(\mathcal{L}\{f(t)\} = F(s)\), then \(\mathcal{L}\{e^{at}f(t)\} = F(s-a)\).
• Time Shifting: If \(\mathcal{L}\{f(t)\} = F(s)\), then \(\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)\) for \(a > 0\).

These properties allow you to manipulate and solve complex expressions more easily by simplifying functions within the Laplace transform realm.

Mathematical Problem Solving

When tackling mathematical problems involving inverse Laplace transforms, it's essential to employ a systematic approach. This can heavily rely on problem-solving strategies and just the right mindset.

Start with clear problem identification and breaking down the problem into manageable parts. Often that will mean:

• Identifying the expression to be transformed.
• Applying partial fraction decomposition to simplify the expression.
• Leveraging Laplace transform properties to find the inverse transform.

Always double-check your work at each step to ensure accuracy. It's crucial to maintain consistency in notation and approaches which will aid in minimizing errors. This strategy will not only facilitate understanding but also enhance the problem-solving process.

Differential Equations

Differential equations describe various phenomena in fields like physics, engineering, and biology. They're equations that involve functions and their derivatives. Laplace transforms come in handy in solving linear differential equations, as they can transform these equations into algebraic ones, which are typically easier to handle.

In the process of solving differential equations using Laplace transforms:

• Start by applying the Laplace transform to both sides of the equation.
• Use partial fractions and inverse transform techniques to work back into the time domain.
• Identify solutions as expressions of time, often representing real-world systems.

This approach converts the problem into a format where standard algebraic methods can be applied, providing a solution that can be interpreted within the context of the original differential equation.