Question

As in Examples 3 and 4 , use Laplace transform techniques to solve the initial value problem. \(y^{\prime \prime}-4 y=e^{3 t}, \quad y(0)=0, \quad y^{\prime}(0)=0\)

Short answer

## Question Solve the following initial value problem using Laplace transforms: \(y^{\prime \prime} - 4y = e^{3 t}\), with initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 0\). ## Answer \(y(t) = -\frac{1}{5} e^{3t} - \frac{1}{10} e^{2t} + \frac{1}{10} e^{-2t}\).

Step-by-step solution

Write down the given ODE and initial conditions

The given initial value problem is: \(y^{\prime \prime} - 4y = e^{3 t}\), with initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 0\).

Apply Laplace transform to the given ODE

Taking the Laplace transform of the given ODE, we have: \(\mathcal{L}\{y^{\prime \prime} - 4y\} = \mathcal{L}\{e^{3t}\}\). Using the Laplace transform properties, we get: \(s^2Y(s) - sy(0) - y^{\prime}(0) - 4Y(s) = \frac{1}{s - 3}\), where \(Y(s) = \mathcal{L}\{y(t)\}\). Since we are given initial conditions \(y(0) = 0\) and \(y^{\prime}(0) = 0\), the equation simplifies to: \(s^2 Y(s) - 4 Y(s) = \frac{1}{s - 3}\).

Solve for Y(s)

Rearrange the equation to solve for \(Y(s)\): \(Y(s) = \frac{1}{(s - 3)(s^2 - 4)}\).

Perform partial fraction decomposition on Y(s)

We perform partial fraction decomposition on \(Y(s)\): \(\frac{1}{(s - 3)(s^2 - 4)} = \frac{A}{s - 3} + \frac{B}{s - 2} + \frac{C}{s + 2}\), where we need to find the values of constants A, B, and C. Clearing the denominator and setting the numerators equal, we obtain: \(1 = A(s^2 - 4) + (s - 3) B(s + 2) + (s - 3) C(s - 2)\). Solving for A, B, and C: \(A = -\frac{1}{5}\), \(B = -\frac{1}{10}\), and \(C = \frac{1}{10}\). Now we can rewrite the \(Y(s)\) as: \(Y(s) = -\frac{1}{5} \cdot \frac{1}{s - 3} - \frac{1}{10} \cdot \frac{1}{s - 2} + \frac{1}{10} \cdot \frac{1}{s + 2}\).

Calculate the inverse Laplace transform of Y(s)

Now, we find the inverse Laplace transform of each term in \(Y(s)\) to obtain the solution y(t): \(y(t) = \mathcal{L}^{-1}\{Y(s)\}\) \(= -\frac{1}{5} \mathcal{L}^{-1}\left\{\frac{1}{s - 3}\right\} - \frac{1}{10} \mathcal{L}^{-1}\left\{\frac{1}{s - 2}\right\} + \frac{1}{10} \mathcal{L}^{-1}\left\{\frac{1}{s + 2}\right\}\). Using the inverse Laplace transform properties, we get: \(y(t) = -\frac{1}{5} e^{3t} - \frac{1}{10} e^{2t} + \frac{1}{10} e^{-2t}\).

Write down the final solution

The solution to the given initial value problem is: \(y(t) = -\frac{1}{5} e^{3t} - \frac{1}{10} e^{2t} + \frac{1}{10} e^{-2t}\).

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation along with specific values provided for the function and its derivatives at a given point, often times represented as t=0. These given conditions allow us to tailor the general solution of the differential equation to a particular scenario.

For instance, in our exercise, we are given the second-order linear ordinary differential equation (ODE) alongside initial values at t=0: \(y(0)=0\) and \(y'\)(0)=0. These values are essential for determining the unique solution that corresponds to the physical or geometrical situation in question. Without initial conditions, we would have a family of solutions rather than a single, specific response to the problem.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler, more manageable ones. This technique becomes particularly useful when solving an equation using the Laplace transform, as we aim to invert the transformed function back into the time domain.

In our exercise, we've used partial fraction decomposition to split the function \(Y(s)\) into a sum of simpler terms, facilitating the computation of the inverse Laplace transform. By finding constants A, B, and C that satisfy the equation, we obtained three simpler fractions that can be easily inverted. This step is vital for maneuvering through the complexity of inverse transformations.

Inverse Laplace Transform

The inverse Laplace transform is a powerful tool that allows us to convert a function from the s-domain back into the time domain, t. It is especially useful when analyzing systems described by differential equations.

In our case, after applying the Laplace transform to solve the given ODE and performing partial fraction decomposition, we arrive at \(Y(s)\) in a form that can be conveniently inverted. We then use the properties of the inverse Laplace transform to find the corresponding time-domain functions for each term in \(Y(s)\). This process ultimately provides us with the specific solution \(y(t)\) to our initial value problem.

Ordinary Differential Equations (ODE)

Ordinary differential equations (ODEs) are equations involving derivatives of an unknown function with respect to a single variable. These mathematical constructs are pivotal in describing various phenomena, from physics to finance.

The equation from the exercise, \(y'' - 4y = e^{3t}\), is a second-order linear ODE with constant coefficients. ODEs can be intimidating, but the Laplace transform is one technique that simplifies the solving process by converting the ODE into an algebraic equation which is often easier to manipulate and solve. The exercise showcases how, with an initial value problem setup, one can skillfully handle ODEs using the Laplace transform and its inverse.