Math Problem Statement

Use the Laplace transform to solve the following initial value problem: 5y'' + 2y' + 16y = 3cos(2t), y(0) = 0, y'(0) = 0. (a) Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation and then solve for \( \mathcal{L} \{ y(t) \} \). Do not perform partial fraction decomposition since we will write the solution in terms of a convolution integral. (b) Express the solution \( y(t) \) in terms of a convolution integral.

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:

5y+2y+16y=3cos(2t),5y'' + 2y' + 16y = 3\cos(2t),

with initial conditions:

y(0)=0,y(0)=0.y(0) = 0, \quad y'(0) = 0.

The Laplace transform method involves:

  1. Taking the Laplace transform of both sides of the equation.
  2. Solving for L{y(t)}\mathcal{L}\{y(t)\}.
  3. Writing the solution y(t)y(t) in terms of a convolution integral.

(a) Taking the Laplace Transform:

Using Laplace transform properties:

  • L{y(t)}=s2Y(s)sy(0)y(0)\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0),
  • L{y(t)}=sY(s)y(0)\mathcal{L}\{y'(t)\} = sY(s) - y(0),
  • L{y(t)}=Y(s)\mathcal{L}\{y(t)\} = Y(s),

we apply the Laplace transform to both sides of the differential equation:

5L{y(t)}+2L{y(t)}+16L{y(t)}=L{3cos(2t)}.5\mathcal{L}\{y''(t)\} + 2\mathcal{L}\{y'(t)\} + 16\mathcal{L}\{y(t)\} = \mathcal{L}\{3\cos(2t)\}.

Substituting the initial conditions y(0)=0y(0) = 0 and y(0)=0y'(0) = 0:

[ 5[s^2Y(s)] + 2[sY(s)] + 16Y(s) = \frac{3s}{s^2 + 4}. ]

Simplify the equation:

(5s2+2s+16)Y(s)=3ss2+4.(5s^2 + 2s + 16)Y(s) = \frac{3s}{s^2 + 4}.

Solve for Y(s)Y(s):

Y(s)=3s(5s2+2s+16)(s2+4).Y(s) = \frac{3s}{(5s^2 + 2s + 16)(s^2 + 4)}.


(b) Convolution Integral:

We will write y(t)y(t) in terms of a convolution integral. The solution involves:

y(t)=L1{Y(s)}.y(t) = \mathcal{L}^{-1}\{Y(s)\}.

Instead of partial fraction decomposition, we'll directly express Y(s)Y(s) as a product of transforms. Observe:

Y(s)=3s5s2+2s+161s2+4.Y(s) = \frac{3s}{5s^2 + 2s + 16} \cdot \frac{1}{s^2 + 4}.

The Laplace transform 3s5s2+2s+16\frac{3s}{5s^2 + 2s + 16} corresponds to some function h1(t)h_1(t), and 1s2+4\frac{1}{s^2 + 4} corresponds to sin(2t)\sin(2t). Thus, by the convolution theorem:

y(t)=0th1(τ)sin(2(tτ))dτ,y(t) = \int_0^t h_1(\tau)\sin(2(t - \tau)) \, d\tau,

where h1(t)h_1(t) is the inverse Laplace transform of 3s5s2+2s+16\frac{3s}{5s^2 + 2s + 16}.


Final Answer:

(a) The Laplace transform is:

L{y(t)}(s)=3s(5s2+2s+16)(s2+4).\mathcal{L}\{y(t)\}(s) = \frac{3s}{(5s^2 + 2s + 16)(s^2 + 4)}.

(b) The solution in terms of a convolution integral is:

y(t)=0th1(τ)sin(2(tτ))dτ,y(t) = \int_0^t h_1(\tau)\sin(2(t - \tau)) \, d\tau,

where h1(t)=L1(3s5s2+2s+16)h_1(t) = \mathcal{L}^{-1}\left(\frac{3s}{5s^2 + 2s + 16}\right).


Do you need further details on the inverse Laplace transform or simplifying the convolution integral? Here are related questions to expand on:

  1. How is the convolution theorem applied in Laplace transforms?
  2. What is the process to compute h1(t)h_1(t) explicitly using partial fractions?
  3. Can the result y(t)y(t) be numerically solved or simulated?
  4. What physical systems are modeled by such second-order differential equations?
  5. 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., cos(2t)\cos(2t)) 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