91Ó°ÊÓ

A differential equation is a mathematical equation that relates a function with its derivatives. These equations play a crucial role in modeling situations where a quantity changes over time, space, or another variable. For the given exercise, the differential equation is:

\[y'' + 5y' + 4y = 0\]

This is a second-order linear homogeneous differential equation because:

• "Second order" means the highest derivative is a second derivative \(y''\).
• "Linear" signifies that the equation involves no powers or products of \(y\) and its derivatives.
• "Homogeneous" indicates that every term involves \(y\) or its derivatives and there is no external forcing function (right-hand side equals zero).

Such equations often describe systems that return to an equilibrium state, like damped harmonic oscillators. Solving these helps predict future behavior of the modeled system.

Initial conditions specify values of the function or its derivatives at a particular point, usually to identify a unique solution for a differential equation. For our case, the initial conditions are:

• \(y(0) = 1\)
• \(y'(0) = 0\)

These conditions tell us how the system starts. Applying the Laplace transform to the differential equation involves substituting these initial conditions into the transformed equations:

• The Laplace transform of \(y(t)\) includes both \(y(0)\) and \(y'(0)\).
• This transforms the problem from a differential equation in terms of \(t\) into an algebraic equation in terms of \(s\).

These conditions ensure that the solution is specific to the given particular instance of the problem.

Partial fraction decomposition is a technique used to express a rational function as a sum of simpler fractions. This method simplifies finding the inverse Laplace transform of the expression. In our exercise, after factorizing the quadratic denominator:
\[Y(s) = \frac{s + 5}{(s+1)(s+4)}\]

The next steps involve writing this as a sum of fractions:
\[\frac{s + 5}{(s+1)(s+4)} = \frac{A}{s+1} + \frac{B}{s+4}\]

By solving equations derived from setting these equal over a common denominator, one finds values for \(A\) and \(B\). This separation into partial fractions allows us to easily compute the inverse Laplace transform, leading to a simple time-domain solution.

The inverse Laplace transform is a method for transforming a function from the Laplace domain back into the time domain. This is essential for solving differential equations because it translates the solution from a transformed algebraic form back into its original time-based form. In the problem:

\[Y(s) = \frac{4}{3(s+1)} - \frac{1}{3(s+4)}\]

The inverse Laplace transform is applied to each term separately. Knowing standard Laplace transform pairs, like \(\mathcal{L}\{e^{-at}\} = \frac{1}{s+a}\), is crucial. It facilitates finding:
\[y(t) = \mathcal{L}^{-1}\left\{ \frac{4}{3(s+1)} \right\} - \mathcal{L}^{-1}\left\{ \frac{1}{3(s+4)} \right\}\]

This results in:

• \(\frac{4}{3}e^{-t}\) from \(\frac{4}{3(s+1)}\)
• \(\frac{1}{3}e^{-4t}\) from \(-\frac{1}{3(s+4)}\)

Thus, the final solution in terms of time is \(y(t) = \frac{4}{3}e^{-t} - \frac{1}{3}e^{-4t}\). This provides us with the predicted behavior of the equation over time.