Problem 43 - Precalculus

Question

Refer to the following: Laplace transforms are used to solve differential equations. The Laplace transform of \(f(t)\) is denoted by \(L\\{f(t)\\} ;\) thus, \(L\left\\{e^{3 t}\right\\}\) is the Laplace transform of \(f(t)=e^{3 t} .\) It is known that \(L\left\\{e^{k t}\right\\}=\frac{1}{s-k}\) and \(L\left\\{e^{-k t}\right\\}=\frac{1}{s+k} .\) Then the inverse Laplace transform of \(g(s)=\frac{1}{s-k}\) is \(L^{-1}\left\\{\frac{1}{s-k}\right\\}=e^{k t} .\) Inverse Laplace transforms are linear: $$L^{-1}\\{f(t)+g(t)\\}=L^{-1}\\{f(t)\\}+L^{-1}\\{g(t)\\}$$ Use partial fractions to find the inverse Laplace transform of \(\frac{9+s}{4-s^{2}}\).

Step-by-Step Solution

Answer

The inverse Laplace transform is \( \frac{11}{4}e^{2t} - \frac{7}{4}e^{-2t} \).

Step 1: Factor the Denominator

First, recognize that the denominator can be factored. Since it is a difference of squares, we can write:\[ 4 - s^2 = (2 - s)(2 + s) \]

Step 2: Set Up Partial Fraction Decomposition

Express the function \( \frac{9+s}{4-s^2} \) as a sum of partial fractions. We write:\[ \frac{9+s}{(2-s)(2+s)} = \frac{A}{2-s} + \frac{B}{2+s} \]Identify the individual components, knowing each has a denominator corresponding to the factors we just determined.

Step 3: Solve for Coefficients

Multiply through by the denominator \((2-s)(2+s)\) to eliminate the denominators:\[ 9+s = A(2+s) + B(2-s) \]Expand and collect like terms:\[ 9+s = (A+B)s + (2A-2B) \]Equate coefficients:1. Coefficient of \(s\): \( A + B = 1 \)2. Constant term: \( 2A - 2B = 9 \)Solve this system of equations to find \(A\) and \(B\).

Step 4: Solving the System of Equations

Solve the equations:1. From \(A+B=1\), express \(A\) as \(A = 1-B\).2. Substitute \(A = 1-B\) into the second equation, \(2A - 2B = 9\): \[ 2(1-B) - 2B = 9 \] \[ 2 - 2B - 2B = 9 \] \[ 2 - 4B = 9 \] \[ -4B = 7 \] \[ B = -\frac{7}{4} \]3. Substitute \(B = -\frac{7}{4}\) back into \(A = 1-B\): \[ A = 1 + \frac{7}{4} = \frac{11}{4} \]

Step 5: Express the Original Function in Terms of Partial Fractions

Plug the values of \(A\) and \(B\) back into the partial fraction decomposition:\[ \frac{9+s}{(2-s)(2+s)} = \frac{11/4}{2-s} - \frac{7/4}{2+s} \]

Step 6: Convert Each Term to Inverse Laplace Transforms

Utilize the inverse Laplace transform formula:\[ L^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at} \]So,\[ L^{-1}\left\{ \frac{11/4}{2-s} \right\} = \frac{11}{4}e^{2t} \]\[ L^{-1}\left\{ \frac{7/4}{2+s} \right\} = \frac{7}{4}e^{-2t} \]The linear property of Laplace transforms gives:\[ L^{-1}\left\{ \frac{9+s}{4-s^2} \right\} = \frac{11}{4}e^{2t} - \frac{7}{4}e^{-2t} \]

Step 7: Conclusion

The inverse Laplace transform of \( \frac{9+s}{4-s^2} \) is:\[ \frac{11}{4}e^{2t} - \frac{7}{4}e^{-2t} \]

Key Concepts

Inverse Laplace Transform

The inverse Laplace transform allows us to transform a function from the Laplace domain back to the time domain. This is useful when dealing with differential equations, as it can simplify complex expressions. To apply the inverse Laplace transform, you need a good understanding of Laplace transform pairs, which relate a function in the time domain and its counterpart in the Laplace domain.

The foundational formula for the inverse Laplace transform is:

• \( L^{-1}igg\{ \rac{1}{s-a} \bigg\} = e^{at} \)

This means if you have a function like \( \rac{1}{s-a} \), its inverse Laplace transform is an exponential function \( e^{at} \).

Inverse Laplace transforms are linear. This means you can take the inverse of individual components and add them together. For instance,

• \( L^{-1}\{ f(t) + g(t) \} = L^{-1}\{ f(t) \} + L^{-1}\{ g(t) \} \)

This property is often used to break down complex fractions into simpler ones that match known transform pairs, simplifying the process of finding the inverse.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex algebraic fractions into simpler fractions. This is helpful when trying to find the inverse Laplace transform, especially when the denominator is a polynomial.

For example, if you have a fraction \( \rac{9+s}{4-s^2} \), the denominator can be factored into \( (2-s)(2+s) \). This allows us to express the fraction as a sum of two simpler fractions:

• \( \rac{9+s}{(2-s)(2+s)} = \rac{A}{2-s} + \rac{B}{2+s} \)

Where \( A \) and \( B \) are constants that we need to solve for.

To find these constants, multiply through by \( (2-s)(2+s) \) to clear the denominators, then solve the resulting equation by matching coefficients for corresponding powers of \( s \). This produces a system of equations that can be solved to find \( A \) and \( B \).

Once \( A \) and \( B \) are determined, the original function is written in terms of these simpler fractions, each of which can be easily transformed back to the time domain using known inverse Laplace transforms.

Differential Equations

Differential equations describe relationships involving rates of change, represented by derivatives. Solving these equations is crucial in fields like engineering, physics, and biology.

Laplace transforms are a powerful tool in solving differential equations. They transform a differential equation in the time domain into an algebraic equation in the Laplace domain, which is often much easier to solve. The Laplace transform turns derivatives into simple polynomial multiplications, making the problem more manageable.

Once you have an algebraic equation, you solve for the Laplace transform of the unknown function, and then use the inverse Laplace transform to convert the solution back to the time domain.

Because of this transformation process, Laplace transforms are especially beneficial for solving linear differential equations with constant coefficients. They simplify initial value problems by neatly incorporating initial conditions into the transformed equation.

In summary, the combination of Laplace transforms and inverse Laplace transforms streamlines the process of solving differential equations, turning complex calculus problems into simpler algebraic ones.