Question

Solve the differential equation \(y^{\prime \prime}-a^{2} y=f(t),\) where \(f(t)=\left\\{\begin{array}{ll}0, & t<0 \\ 1, & t>0\end{array}\right.\) and \(y_{0}=y_{0}^{\prime}=0\). Hint: Use the convolution integral as in the example.

Short answer

The solution to the differential equation is \( y(t) = \frac{1}{a^2} + \frac{1}{a} \sinh(at) \).

Step-by-step solution

- Understand the problem

Given a differential equation of the form: \[ y'' - a^2 y = f(t), \] where the function \( f(t) \) is a piecewise function: \[ f(t) = \begin{cases} 0, & t < 0 \ 1, & t > 0 \end{cases} \] and the initial conditions are: \[ y(0) = 0 \ y'(0) = 0 \] The goal is to solve for \( y(t) \).

- Apply Laplace Transform

To simplify the differential equation, take the Laplace Transform of both sides. The Laplace Transform of \( y'' \) is \( s^2 Y(s) - sy(0) - y'(0) \), and for \( a^2 y \) it is \( a^2 Y(s) \). For \( f(t) \), the Laplace transform since \( f(t) = 1 \ t > 0 \) will be \( \frac{1}{s} \).Thus, the equation becomes: \[ s^2 Y(s) - sy(0) - y'(0) - a^2 Y(s) = \frac{1}{s} \]Given initial conditions \( y(0) = 0 \) and \( y'(0) = 0 \), we simplify to: \[ (s^2 - a^2) Y(s) = \frac{1}{s} \]

- Solve for \( Y(s) \)

Isolate \( Y(s) \): \[ Y(s) = \frac{1}{s(s^2 - a^2)} \]Factor the denominator: \[ Y(s) = \frac{1}{s(s - a)(s + a)} \]

- Apply Partial Fraction Decomposition

Express \( Y(s) \) in terms of partial fractions: \[ \frac{1}{s(s - a)(s + a)} = \frac{A}{s} + \frac{B}{s - a} + \frac{C}{s + a} \]Find constants \( A, B, C \) by solving: \[ 1 = A(s - a)(s + a) + Bs(s + a) + Cs(s - a) \]Solving for constants gives: \[ A = \frac{1}{a^2}, B = \frac{1}{2a}, C = -\frac{1}{2a} \]Thus: \[ Y(s) = \frac{1/a^2}{s} + \frac{1/2a}{s - a} - \frac{1/2a}{s + a} \]

- Inverse Laplace Transform

Now find the Inverse Laplace Transform: \[ y(t) = \mathcal{L}^{-1} \left\{ \frac{1/a^2}{s} + \frac{1/2a}{s - a} - \frac{1/2a}{s + a} \right\} \]Using standard Laplace Transform pairs: \[ \mathcal{L}^{-1} \left( \frac{1}{s} \right) = 1, \mathcal{L}^{-1} \left( \frac{1}{s - a} \right) = e^{at}, \mathcal{L}^{-1} \left( \frac{1}{s + a} \right) = e^{-at} \] gives: \[ y(t) = \frac{1}{a^2} \cdot 1 + \frac{1}{2a} e^{at} - \frac{1}{2a} e^{-at} \]Simplify to: \[ y(t) = \frac{1}{a^2} + \frac{1}{2a} (e^{at} - e^{-at}) \]or more compactly using sinh function: \[ y(t) = \frac{1}{a^2} + \frac{1}{a} \sinh(at) \]

Key concepts

Laplace transform

The Laplace Transform is a powerful tool for solving differential equations. It converts a function of time, denoted as \(f(t)\), into a function of a complex variable, denoted as \(F(s)\). This transformation makes it easier to work with differential equations because it turns derivatives into algebraic terms.
For example, the Laplace Transform of a second derivative \(y''\) is given by \(s^2 Y(s) - sy(0) - y'(0)\). This simplifies the differential equation by converting it from the time domain to the s-domain. In our case, the initial conditions \(y(0) = 0\) and \(y'(0) = 0\) help eliminate the initial terms, simplifying the equation further.
This simplification allows us to focus on solving algebraic equations rather than differential ones, providing a streamlined path to solving the problem at hand. For solving the given differential equation, we apply the Laplace Transform to both sides of the equation to convert it into an algebraic equation in \(Y(s)\).

Inverse Laplace transform

After applying the Laplace Transform, we must revert back to the time domain using the Inverse Laplace Transform. This step involves converting \(Y(s)\) back to \(y(t)\). To achieve this, we use standard Laplace Transform pairs and properties.
For instance, the Inverse Laplace Transform of \(\frac{1}{s} \) is 1, which means it corresponds to a step function in the time domain. Similarly, \(\mathcal{L}^{-1} \left(\frac{1}{s-a}\right) = e^{at}\) and \(\mathcal{L}^{-1} \left(\frac{1}{s+a}\right) = e^{-at}\).
By reversing the Laplace transform solved in the s-domain, we obtain our solution in the time domain. In our example, we use these pairs to find the inverse transform of partial fractions such as \(\frac{1}{a^2}\), \(\frac{1/2a}{s-a}\), and \(\frac{1/2a}{s+a}\), resulting in the final solution for \(y(t)\).

Partial fraction decomposition

Partial Fraction Decomposition is crucial when dealing with the Laplace Transform because it simplifies complex rational expressions. The goal is to express a complex fraction as a sum of simpler fractions, making the Inverse Laplace Transform easier to apply.
In our problem, we decomposed \(\frac{1}{s(s^2 - a^2)}\) into partial fractions: \(\frac{1}{s(s - a)(s + a)} = \frac{A}{s} + \frac{B}{s - a} + \frac{C}{s + a}\).
We determined the constants \(A, B, C\) by solving the equation: \(1 = A(s - a)(s + a) + Bs(s + a) + Cs(s - a)\), which resulted in \(A = \frac{1}{a^2}\), \(B = \frac{1}{2a}\), and \(C = -\frac{1}{2a}\).
This step is vital because it breaks down a single complex term into smaller, manageable terms that can be easily transformed back to the time domain using known Laplace Transform pairs.

Initial conditions

Initial conditions play a critical role in solving differential equations using the Laplace Transform. They provide the necessary information to solve for the constants that arise during the transformation process.
For our differential equation \(y'' - a^2y = f(t)\) with initial conditions \(y(0) = 0\) and \(y'(0) = 0\), these values help simplify the Laplace-transformed equation. When we take the Laplace Transform of \(y''\), the initial conditions eliminate the \(sy(0)\) and \(y'(0)\) terms, reducing it to \(s^2 Y(s)\).
This simplification is crucial because it allows us to solve for \(Y(s)\) without dealing with additional unknowns. The initial conditions thus provide a clear starting point for solving the equation.