91Ó°ÊÓ

The Laplace transform is a powerful mathematical tool used to transform a function of time, often denoted as \( f(t) \), into a function of a complex variable \( s \), denoted as \( F(s) \). This is extremely useful in solving linear differential equations. Here's the basic formula for the Laplace transform:

\[ \mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t) e^{-st} \, dt \]
In simpler terms, it transforms the time-domain equation into the s-domain (frequency domain), making it easier to manage their algebraic properties.

To put it into practice, consider the differential equation given in the problem. We begin by taking the Laplace transform of both sides:
\[ y'' - 4y = 3e^{-t} \]
The benefit of doing this is that differentiation in the time domain becomes algebraic manipulation in the s-domain. Knowing how to apply the Laplace transform to various functions is crucial in this step.

Differential equations involve functions and their derivatives, and solving them often requires finding a function that satisfies the given equation. In our case, we were given a second-order linear differential equation:

\[ y'' - 4y = 3e^{-t} \]
This equation involves the second derivative \( y'' \) and the original function \( y \). Linear differential equations can be particularly challenging, but the Laplace transform method simplifies the process significantly. By converting the differential equation into an algebraic equation, it becomes much easier to isolate and solve for \( Y(s) \).
Understanding how to manipulate and solve these transformed equations is key to mastering differential equations.

Initial conditions provide specific values for the function and its derivatives at a particular point, usually at \( t = 0 \). In this problem, the initial conditions are:

\[ y(0) = 1 \]
\[ y'(0) = -3 \]
These conditions are vital as they allow us to solve for the constants when applying the Laplace transform. When we took the Laplace transform of the original equation, these initial conditions were substituted into the transformed equation:

\[ s^2Y(s) - s(1) - (-3) - 4Y(s) = \frac{3}{s+1} \]
Using these conditions correctly ensures that we accurately represent the original differential equation in the s-domain, leading us to the correct solution.

Once we solve the algebraic equation for \( Y(s) \), our ultimate goal is to convert it back to the time domain using the inverse Laplace transform:

\[ y(t) = \mathcal{L}^{-1}\{Y(s)\} \]
This step involves using known inverse transforms and partial fraction decomposition to break down complex rational functions into simpler parts. For the given problem, we applied the inverse Laplace transform to:

\[ Y(s) = \frac{-1}{s-2} + \frac{1}{s+2} + \frac{2}{s+1} \]
Using knowledge of standard inverse transforms:
\[ y(t) = -e^{2t} + e^{-2t} + 2e^{-t} \]
This gives us the solution in the original time domain.

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that can be inverted using known Laplace pairs. In this problem, after isolating \( Y(s) \), we had:

\[ Y(s) = \frac{3}{(s+1)(s-2)(s+2)} + \frac{s-3}{s^2-4} \]
The next step was to decompose this into simpler fractions:
\[ Y(s) = \frac{A}{s-2} + \frac{B}{s+2} + \frac{C}{s+1} \]
We solve for constants A, B, and C by matching coefficients from both sides of the equation. We found:
\[ A = -1, B = 1, C = 2 \]
This yields the partial fractions:
\[ Y(s) = \frac{-1}{s-2} + \frac{1}{s+2} + \frac{2}{s+1} \]
By breaking it down, each term can now be easily inverted to the time domain.