ODE with Laplace Transform: Solving for Y(s) and Partial Fraction Expansion

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SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) y''(t) + 4y(t) = 1 + u(t − 2) using Laplace transforms. The initial transformation yields (s² + 4)Y(s) = (e^(-2s) + 1)/s, leading to Y(s) = (e^(-2s) + 1)/(s(s² + 4)). The partial fraction expansion is discussed, revealing that the complete solution includes terms for both the initial response and the delayed response due to the unit step function u(t − 2). The final solution is confirmed as (1/4)(1 − cos(2t)) + (1/4)(1 − cos(2(t − 2)))u(t − 2).

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with Laplace transforms and their properties
  • Knowledge of partial fraction decomposition techniques
  • Basic concepts of unit step functions in control theory
NEXT STEPS
  • Study the application of Laplace transforms in solving linear ODEs
  • Learn about the properties and applications of the unit step function u(t)
  • Explore advanced techniques in partial fraction expansion for complex functions
  • Review the theory behind the inverse Laplace transform and its applications
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are solving differential equations, particularly those utilizing Laplace transforms and partial fraction expansions.

bosox09
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Homework Statement



The solution to the ODE y''(t) + 4y(t) = 1 + u(t − 2), y(0) = 0, y'(0) = 0 is given by...

The Attempt at a Solution



OK well I figured this one is good to solve with Laplace transforms. So I took the Laplace of both sides to obtain (s2 + 4)Y(s) = [e-2s/s] + 1/s, which equals (e-2s + 1)/s. Isolating Y(s) gave me (e-2s + 1)/s(s2 + 4). I used partial fraction expansion to obtain (1/4) - (1/4)cos2t, but this is apparently only half of the whole answer, given as (1/4)(1 − cos 2t) + (1/4)(1 − cos 2(t − 2))u(t − 2). What am I missing?
 
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How are you expanding in partial fractions? What expansion did you get? In particular shat happened to the [itex]e^{-2s}[/itex]? It's very easy to see that [tex]y= \frac{1}{4}(1- cos(2t))[/tex] satisfies y"+ 4y= 1, y(0)= 0, y'(0)= 0, not the equation you have.
 

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