91Ó°ÊÓ

Differential equations are mathematical equations that relate a function with its derivatives. They are widely used to model phenomena in physics, engineering, biology, and economics. There are several types of differential equations, but this exercise focuses on linear ordinary differential equations (ODEs). A second-order linear ODE, like the one in this exercise, involves the second derivative of the function. The general form for a second-order linear ODE is: \[ a\ y'' + b\ y' + c\ y = f(t) \], where \(a, b, c\) are constants and \(f(t)\) is some given function of \(t\). In the given problem: \[ y'' + 5y' + 6y = 12, \] the equation relates the second derivative (\(y''\)), the first derivative (\(y'\)), and the function (\(y\)) itself. Solving these equations typically involves finding the function \(y(t)\) that satisfies this relationship. To solve differential equations, several techniques can be used, such as separation of variables, integrating factor method, or using transforms like the Laplace Transform.

When solving differential equations, the solution is not always unique. Initial conditions help provide specific solutions that satisfy the given constraints at a particular point. Initial conditions define the values of the function and its derivatives at a starting point, often \(t = 0\). For the given problem, two initial conditions are provided: - \(y(0) = 2\), which means the value of the function at \(t = 0\) is 2. - \(y'(0) = 0\), meaning the first derivative of the function at \(t = 0\) is 0. These initial conditions are crucial because they allow us to determine the specific solution to the differential equation. In the Laplace Transform method, initial conditions are used to transform the differential equation into an algebraic equation. When the Laplace Transforms of the derivatives are taken, terms involving the initial conditions appear, which are then substituted into the equation to solve for the transformed function. Without initial conditions, we would only have the general solution, which does not cater to the specifics of the particular problem we are solving.

The Laplace Transform is a powerful tool that converts differential equations into algebraic equations, which are often easier to solve. Once the differential equation is transformed, initial conditions are applied, and the algebraic equation is solved for the Laplace Transform of the function, denoted typically by \(Y(s)\). However, our goal is to find the original function \(y(t)\), not its transform \(Y(s)\). To achieve this, we use the Inverse Laplace Transform. Performing an Inverse Laplace Transform converts \(Y(s)\) back into \(y(t)\). In the given problem, after solving for \(Y(s)\), it is written as: \[ Y(s) = \frac{\frac{12}{s} + 2s + 10}{s^2 + 5s + 6} \]. This expression is then decomposed into partial fractions for easier inversion. Partial fraction decomposition helps break down complex fractions into simpler terms that correspond to standard Laplace Transform pairs. Each term in the partial fraction can be inverted using known Laplace Transform pairs or tables. For example, terms like \(\frac{1}{s}\) correspond to a step function, and terms like \(\frac{1}{s + a}\) correspond to an exponential function \(e^{-at}\). The Inverse Laplace Transform converts these terms back to their time domain equivalents, yielding the function \(y(t)\). Thus, it completes the solution of the differential equation, satisfying both the transformed algebraic equation and the initial conditions specified at the beginning.