Question

Use the Laplace transform to solve the given initial value problem. $$ y^{\prime \prime}+2 y^{\prime}+5 y=0 ; \quad y(0)=2, \quad y^{\prime}(0)=-1 $$

Short answer

Question: Solve the initial value problem for the given second-order linear ordinary differential equation (ODE) with constant coefficients using the Laplace transform method: \(y^{\prime\prime} + 2y^{\prime} + 5y = 0\), with \(y(0) = 2\) and \(y^{\prime}(0) = -1\). Answer: The solution of the given initial value problem is: \(y(t) = \frac{3}{5}e^{-t}\cos(2t) + \frac{7}{5}e^{-t}\sin(2t)\).

Step-by-step solution

Apply Laplace transform to the ODE

Apply the Laplace transform to the given ODE: $$ \mathcal{L}\{y^{\prime\prime}\} + 2\mathcal{L}\{y^{\prime}\} + 5\mathcal{L}\{y\} = \mathcal{L}\{0\} $$

Use the rules of Laplace transform for derivatives

Using the Laplace transform rules for the first and second derivatives, and the initial conditions, we get: $$ s^2Y(s) - sy(0) - y^{\prime}(0) +2[sY(s) - y(0)] + 5Y(s) = 0 $$ Substitute the given initial conditions, \(y(0) = 2\) and \(y^{\prime}(0) = -1\): $$ s^2Y(s) - 2s +1 +2[sY(s) - 2] + 5Y(s) = 0 $$

Solve for Y(s)

Now, simplify the equation and isolate the function Y(s) in the expression: $$ \begin{aligned} (s^2+2s+5)Y(s) &= 2s -1 +4 \\ Y(s) &= \frac{2s + 3}{s^2+2s+5} \end{aligned} $$

Perform partial fraction decomposition

We'll now perform partial fraction decomposition on the Y(s) expression to obtain a form that is easier to find the inverse Laplace transform of. Using the partial fraction decomposition, we get: $$ Y(s) = \frac{As + B}{s^2+2s+5} $$ To find the constants A and B, we have: $$ A(s^2+2s+5)+Bs = 2s + 3 $$ Equating coefficients of "s" and the constant term, we get the two equations: $$ \begin{cases} A+B = 2 \\ 5A = 3 \end{cases} $$ Solving the system of equations, we get \(A = \frac{3}{5}\) and \(B = \frac{7}{5}\).

Apply inverse Laplace transform

Now we should apply the inverse Laplace transform to obtain the function y(t) back: $$ y(t) = \mathcal{L}^{-1}\{Y(s)\} = \mathcal{L}^{-1}\left\{\frac{3/5s + 7/5}{s^2+2s+5}\right\} $$

Find the solution in time domain

After computing the inverse Laplace transform, we get the final solution in the time domain as: $$ y(t) = \frac{3}{5}e^{-t}\cos(2t) + \frac{7}{5}e^{-t}\sin(2t) $$ This is the solution of the given initial value problem.

Key concepts

Inverse Laplace Transform

The Inverse Laplace Transform is a mathematical process used to convert a function in the complex frequency domain, commonly denoted as 'F(s)', back into the time domain as the original function, 'f(t)'. It essentially reverses the effects of the Laplace transform.

In solving differential equations, especially in engineering and physics, the Laplace transform is a powerful tool for handling initial value problems. It simplifies complex differential equations by converting them into algebraic equations which are easier to manage. Once we solve the algebraic equation, we then apply the inverse Laplace transform to retrieve the function in terms of time.

The inverse Laplace transform of a function is usually denoted as \( \mathcal{L}^{-1}\{F(s)\} = f(t) \). To find the inverse Laplace transform, one may use various techniques such as partial fraction decomposition, complex inversion formula, or consulting a table of Laplace transforms for known functions.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique commonly used to simplify complex rational expressions, especially when working with the Laplace transform. This method breaks down a complicated fraction into a sum of simpler ones, making them more amenable to the application of the inverse Laplace transform.

Usually, a rational function is expressed in a form of \( \frac{P(s)}{Q(s)} \), where \( P(s) \) and \( Q(s) \) are polynomials and the degree of \( P(s) \) is less than the degree of \( Q(s) \). The crucial step is to factor \( Q(s) \) into its irreducible parts, then for each factor, a term is assigned in the decomposition - this aids in isolating the constants that will later be determined.

For instance, suppose we have to decompose \( \frac{X(s)}{s^2+2s+5} \), we'll assign \( \frac{As+B}{s^2+2s+5} \). Then, equating it to the original rational expression and solving for 'A' and 'B' gives us simpler fractions which can be easily transformed back into the time domain via the inverse Laplace transform.

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function and provide a relationship between the function itself and its derivatives. They are essential tools in expressing physically observed phenomena such as heat conduction, wave motion, and population dynamics.

A typical form of a differential equation is \( f'(t) = g(t, f(t)) \), where \('f(t)'\) is an unknown function, and \( f'(t) \) is its derivative. The function \( g(t, f(t)) \) involves both the variable \( t \) and the function \( f(t) \).

For an initial value problem, the solution to the differential equation is sought given the initial conditions of the function at a specific time. The complexity of solving differential equations varies - some have straightforward analytical solutions, while others may require sophisticated numerical methods or approximations. The Laplace transform is particularly helpful in transforming second-order linear differential equations with constant coefficients into more manageable algebraic forms, facilitating the finding of solutions.