Math Problem Statement
Solution
Let's solve this problem step by step.
Problem Analysis:
We are tasked with solving the second-order differential equation using the Laplace transform. The equation given is:
with initial conditions:
The Laplace transform method involves:
- Taking the Laplace transform of both sides of the equation.
- Solving for .
- Writing the solution in terms of a convolution integral.
(a) Taking the Laplace Transform:
Using Laplace transform properties:
- ,
- ,
- ,
we apply the Laplace transform to both sides of the differential equation:
Substituting the initial conditions and :
[ 5[s^2Y(s)] + 2[sY(s)] + 16Y(s) = \frac{3s}{s^2 + 4}. ]
Simplify the equation:
Solve for :
(b) Convolution Integral:
We will write in terms of a convolution integral. The solution involves:
Instead of partial fraction decomposition, we'll directly express as a product of transforms. Observe:
The Laplace transform corresponds to some function , and corresponds to . Thus, by the convolution theorem:
where is the inverse Laplace transform of .
Final Answer:
(a) The Laplace transform is:
(b) The solution in terms of a convolution integral is:
where .
Do you need further details on the inverse Laplace transform or simplifying the convolution integral? Here are related questions to expand on:
- How is the convolution theorem applied in Laplace transforms?
- What is the process to compute explicitly using partial fractions?
- Can the result be numerically solved or simulated?
- What physical systems are modeled by such second-order differential equations?
- How does this method compare to other solution techniques like undetermined coefficients?
Tip: The Laplace transform is powerful for initial value problems, especially when non-homogeneous terms (e.g., ) are present. Always keep the convolution theorem in mind for compact solutions.
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Convolution Integral
Differential Equations
Formulas
Laplace transform of derivatives: \( \mathcal{L} \{ y''(t) \} = s^2Y(s) - sy(0) - y'(0) \), \( \mathcal{L} \{ y'(t) \} = sY(s) - y(0) \)
Convolution theorem: \( \mathcal{L}^{-1} \{ F(s)G(s) \} = \int_0^t f(\tau)g(t-\tau) d\tau \)
Theorems
Laplace Transform
Convolution Theorem
Suitable Grade Level
Undergraduate Mathematics or Engineering
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