Math Problem Statement

Use the Laplace transform to solve the following initial value problem: 4y'' + 5y' + 17y = 2cos(2t), y(0) = 0, y'(0) = 0.

Solution

To solve the given initial value problem using the Laplace transform, we proceed as follows:


Part (a): Take the Laplace Transform and Solve for L{y(t)}\mathcal{L}\{y(t)\}

The differential equation is:

4y+5y+17y=2cos(2t),y(0)=0,y(0)=0.4y'' + 5y' + 17y = 2\cos(2t), \quad y(0) = 0, \quad y'(0) = 0.

Step 1: Take the Laplace transform of both sides of the equation. Using the Laplace transform rules:

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), and L{cos(2t)}=ss2+4.\mathcal{L}\{\cos(2t)\} = \frac{s}{s^2 + 4}.

Substituting these into the equation:

4(s2Y(s)sy(0)y(0))+5(sY(s)y(0))+17Y(s)=2ss2+4.4\left(s^2Y(s) - sy(0) - y'(0)\right) + 5\left(sY(s) - y(0)\right) + 17Y(s) = \frac{2s}{s^2 + 4}.

Since y(0)=0y(0) = 0 and y(0)=0y'(0) = 0, this simplifies to:

4s2Y(s)+5sY(s)+17Y(s)=2ss2+4.4s^2Y(s) + 5sY(s) + 17Y(s) = \frac{2s}{s^2 + 4}.

Step 2: Factor Y(s)Y(s) on the left-hand side:

Y(s)(4s2+5s+17)=2ss2+4.Y(s)\left(4s^2 + 5s + 17\right) = \frac{2s}{s^2 + 4}.

Solve for Y(s)Y(s):

Y(s)=2s(4s2+5s+17)(s2+4).Y(s) = \frac{2s}{(4s^2 + 5s + 17)(s^2 + 4)}.

Thus, the Laplace transform of y(t)y(t) is:

L{y(t)}=2s(4s2+5s+17)(s2+4).\mathcal{L}\{y(t)\} = \frac{2s}{(4s^2 + 5s + 17)(s^2 + 4)}.


Part (b): Express y(t)y(t) in Terms of a Convolution Integral

We are instructed to use the convolution theorem. The convolution theorem states:

L1{F(s)G(s)}=f(t)g(t),\mathcal{L}^{-1}\left\{\frac{F(s)}{G(s)}\right\} = f(t) * g(t),

where * denotes convolution, defined as:

(fg)(t)=0tf(τ)g(tτ)dτ.(f * g)(t) = \int_0^t f(\tau)g(t - \tau) \, d\tau.

Here, the denominator (4s2+5s+17)(s2+4)(4s^2 + 5s + 17)(s^2 + 4) suggests the use of two functions, one from each term. Specifically:

F(s)=2ss2+4,G(s)=14s2+5s+17.F(s) = \frac{2s}{s^2 + 4}, \quad G(s) = \frac{1}{4s^2 + 5s + 17}.

Step 1: Find the inverse Laplace transforms of F(s)F(s) and G(s)G(s).

  • F(s)=2ss2+4F(s) = \frac{2s}{s^2 + 4} corresponds to 2cos(2t)2\cos(2t) in the time domain.

  • G(s)=14s2+5s+17G(s) = \frac{1}{4s^2 + 5s + 17}: This requires more analysis, but its inverse Laplace transform will correspond to a known exponential and sinusoidal function combination.

Step 2: Write y(t)y(t) as a convolution:

y(t)=0tf(τ)g(tτ)dτ,y(t) = \int_0^t f(\tau)g(t - \tau) \, d\tau,

where f(t)=2cos(2t)f(t) = 2\cos(2t) and g(t)g(t) is the inverse Laplace transform of G(s)G(s).


Would you like a detailed computation of g(t)g(t) or the convolution integral? Let me know! Below are some follow-up points for exploration:


  1. Can we perform partial fraction decomposition to simplify Y(s)Y(s)?
  2. How do we determine g(t)g(t) explicitly from G(s)G(s)?
  3. What are the properties of convolution that simplify y(t)y(t)?
  4. How does the Laplace transform help in solving differential equations efficiently?
  5. Can this approach be extended to non-homogeneous equations with other forcing terms?

Tip: The convolution integral method is powerful when direct partial fraction decomposition is cumbersome, especially in systems with repeated roots or higher-order terms.

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Math Problem Analysis

Mathematical Concepts

Laplace Transform
Differential Equations
Convolution Theorem

Formulas

Laplace Transform of y''(t): L{y''(t)} = s^2Y(s) - sy(0) - y'(0)
Laplace Transform of y'(t): L{y'(t)} = sY(s) - y(0)
Laplace Transform of cos(2t): L{cos(2t)} = s/(s^2 + 4)
Convolution Integral: (f * g)(t) = ∫₀ᵗ f(τ)g(t - τ) dτ

Theorems

Convolution Theorem

Suitable Grade Level

Undergraduate