Question

Use the Laplace transform to solve the initial value problem. \(u^{\prime \prime}+u^{\prime}-2 u=e^{3 t}, \quad u(0)=1, \quad u^{\prime}(0)=-1\)

Short answer

Question: Use the Laplace transform to solve the initial value problem \(u^{\prime \prime} + u^{\prime} -2 u = e^{3 t}\) with initial conditions \(u(0) = 1\) and \(u^{\prime}(0) = -1\). Answer: After applying the Laplace transform and solving the given initial value problem, the solution is \(u(t) = e^{3t} - \left(\frac{2}{3}e^t-\frac{1}{3}e^{-2t}\right)\).

Step-by-step solution

Apply the Laplace transform to the entire equation

By applying the Laplace transform (\(\mathcal{L}\)) to each term, we will have: \(\mathcal{L}\{u^{\prime\prime}+u^{\prime}-2u\}=\mathcal{L}\{e^{3t}\}\) Now, we can apply the Laplace transform to each term separately: \(\mathcal{L}\{u^{\prime\prime}\}+\mathcal{L}\{u^{\prime}\}-2\mathcal{L}\{u\}=\frac{1}{s-3}\) Using the properties of the Laplace transform, we know that: \(\mathcal{L}\{u^{\prime\prime}\}=s^2U(s)-su(0)-u^{\prime}(0)\) \(\mathcal{L}\{u^{\prime}\}=sU(s)-u(0)\) \(\mathcal{L}\{u\}=U(s)\)

Apply the initial conditions

Now, we substitute the given initial conditions \(u(0) = 1\) and \(u^{\prime}(0)=-1\): \((s^2U(s)-s(1)-(-1))+ (sU(s)-(1))-2U(s)=\frac{1}{s-3}\)

Solve for U(s)

Simplify and solve for \(U(s)\): \(s^2U(s) - s + 1 + sU(s) - 1 - 2U(s) = \frac{1}{s-3}\) Combining terms, we get: \((s^2+s-2)U(s)=\frac{s-2}{s-3}\) Divide both sides by \((s^2+s-2)\): \(U(s) = \frac{s-2}{(s-3)(s^2+s-2)}\) To find the inverse Laplace transform, we need to perform partial fraction decomposition on \(U(s)\): \(U(s)=\frac{s-2}{(s-3)(s^2+s-2)}=\frac{A}{s-3}+\frac{Bs+C}{s^2+s-2}\)

Apply inverse Laplace transform and find the partial fraction decomposition

Multiply both sides by the common denominator to simplify: \((s-2)=(s-3)(A(s^2+s-2)+Bs+C)\) By equating the coefficients of like powers of \(s\), we obtain the following system of equations: \(A = 1\), \(B+C=-1A=-1\), \(-2A-2B=-2\) Solving this system gives \(A=1\), \(B=-1\), and \(C=0\). So, we have: \(U(s) = \frac{1}{s-3} - \frac{s}{s^2 + s - 2}\) Now, we can apply the inverse Laplace transform: \(u(t)=\mathcal{L}^{-1}\{U(s)\}=\mathcal{L}^{-1}\{\frac{1}{s-3}\}-\mathcal{L}^{-1}\{\frac{s}{s^2+s-2}\}\) Using properties of inverse Laplace transform, we get: \(u(t)=e^{3t}-\mathcal{L}^{-1}\{\frac{s}{(s-1)(s+2)}\}\) To find \(v(t)=\mathcal{L}^{-1}\{\frac{s}{(s-1)(s+2)}\}\), perform partial fraction decomposition: \(v(t) = \frac{s}{(s-1)(s+2)} = \frac{A}{s-1} + \frac{B}{s+2}\) Solving for A and B, we get \(A=\frac{2}{3}\) and \(B=-\frac{1}{3}\). Thus, \(v(t)=\frac{2}{3}e^t-\frac{1}{3}e^{-2t}\) Finally, combining the results, we have the solution: \(u(t)=e^{3t} - \left(\frac{2}{3}e^t-\frac{1}{3}e^{-2t}\right)\)

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with additional conditions specified at the outset (initial values). These conditions provide necessary information to determine a unique solution to the differential equation. In the given exercise, the initial conditions are that at time zero, the function \(u(t)\) equals 1, and its derivative \(u'(t)\) equals -1. These initial values allow us to apply techniques such as the Laplace transform effectively.

Initial value problems are vital in modeling real-world phenomena like motion or growth processes, where knowing initial states helps predict future behavior. These problems often show up in physics, engineering, and other sciences.

Differential Equations

Differential equations involve equations with functions and their derivatives. These equations can describe a wide range of phenomena in nature due to their ability to model change. The differential equation in this problem is \(u^{\prime \prime }+u^{\prime}-2 u=e^{3 t}\).

Solving differential equations often entails identifying a function, \(u(t)\), that satisfies the relationship given in the equation. Differential equations come in many forms, such as ordinary (ODEs) or partial (PDEs), and each form may require different techniques to solve.

In our exercise, we're dealing with an ordinary differential equation (ODE), solved using the Laplace transform, a powerful tool that converts differential equations into algebraic equations, simplifying the process of finding solutions.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions, making them easier to integrate or transform. For instance, the expression \(\frac{s-2}{(s-3)(s^2+s-2)}\) is decomposed into simpler fractions: \(\frac{A}{s-3}+\frac{Bs+C}{s^2+s-2}\).

This decomposition is crucial when dealing with the Laplace transforms, as it simplifies the inverse process by making expressions match more straightforward Laplace inverse formulas. By solving a system of equations derived from equating coefficients, specific constants \(A\), \(B\), and \(C\) are determined. This technique is essential for inverse Laplace transformations and finding solutions to initial value problems.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical technique used to convert a function from its frequency domain representation back to the time domain. After applying the Laplace transform and solving algebraically for \(U(s)\), the inverse Laplace transform helps find the original function \(u(t)\).

In the exercise, after decomposing \(U(s)\) into partial fractions, we use inverse Laplace tables or known transforms to revert \(U(s)\) into \(u(t)\). For instance, applying the inverse to \(\frac{1}{s-3}\) results in \(e^{3t}\), and similar steps are followed for more complicated expressions.

The inverse Laplace transform is an invaluable tool in solving initial value problems, allowing us to move from easier-to-handle algebraic equations in the frequency domain to interpretable functions in the time domain.