How To Apply Laplace Transform Properties? Stepbystep Solutions

Understanding and applying Laplace transform properties is crucial for solving differential equations and analyzing control systems in engineering and physics. The Laplace transform is a powerful tool that converts differential equations into algebraic equations, making them easier to solve. Here, we will explore how to apply Laplace transform properties step by step, enabling you to tackle complex problems with confidence.

Introduction to Laplace Transform

Before diving into the properties, let’s briefly recall the definition of the Laplace transform. For a function (f(t)) defined for (t \geq 0), the Laplace transform (F(s)) is given by:

[F(s) = \mathcal{L}{f(t)} = \int_{0}^{\infty} e^{-st}f(t)dt]

where (s) is a complex number.

Properties of Laplace Transform

The Laplace transform has several properties that make it useful for solving problems:

1. Linearity Property: The Laplace transform is a linear operator, meaning that for functions (f(t)) and (g(t)) and constants (a) and (b), the following holds: [\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)}]

2. Shift in Time Property: This property relates to the transformation of (f(t - a)), where (a \geq 0). The formula is: [\mathcal{L}{f(t - a)} = e^{-as}F(s)]

3. Scaling Property: For a constant (a), the Laplace transform of (f(at)) is given by: [\mathcal{L}{f(at)} = \frac{1}{a}F\left(\frac{s}{a}\right)]

4. Differentiation and Integration Properties:

   • Differentiation in the Time Domain: [\mathcal{L}{f’(t)} = sF(s) - f(0)]
   • Integration in the Time Domain: [\mathcal{L}\left{\int_{0}^{t}f(\tau)d\tau\right} = \frac{F(s)}{s}]

5. Convolution Property: The Laplace transform of the convolution of two functions (f(t)) and (g(t)) is given by: [\mathcal{L}{f(t) * g(t)} = F(s)G(s)]

Step-by-Step Application of Laplace Transform Properties

Let’s apply these properties to solve a differential equation step by step.

Example: Solve the differential equation (y”(t) + 4y(t) = \sin(2t)) with initial conditions (y(0) = 1) and (y’(0) = 0).

1. Apply the Laplace transform to both sides of the differential equation: [\mathcal{L}{y”(t)} + 4\mathcal{L}{y(t)} = \mathcal{L}{\sin(2t)}]

2. Use the differentiation property: [s^2Y(s) - sy(0) - y’(0) + 4Y(s) = \frac{2}{s^2 + 4}]

3. Substitute the initial conditions: [s^2Y(s) - s(1) - 0 + 4Y(s) = \frac{2}{s^2 + 4}]

4. Simplify and solve for (Y(s)): [(s^2 + 4)Y(s) - s = \frac{2}{s^2 + 4}] [Y(s)(s^2 + 4) = s + \frac{2}{s^2 + 4}] [Y(s) = \frac{s}{s^2 + 4} + \frac{2}{(s^2 + 4)^2}]

5. Find the inverse Laplace transform:

   • The first term (\frac{s}{s^2 + 4}) corresponds to (y_1(t) = \cos(2t)).
   • For the second term, use partial fractions or recognize it as the Laplace transform of a function that results in (\frac{1}{2}t\sin(2t)).

6. Combine the results: [y(t) = \cos(2t) + \frac{1}{2}t\sin(2t)]

Conclusion

The Laplace transform and its properties are powerful tools for solving differential equations and analyzing systems. By applying these properties step by step, you can break down complex problems into manageable parts and derive meaningful solutions. Remember to practice applying these properties to a variety of problems to become proficient in their use.