Solving Laplace Transform Homework: y''+y = f(t)

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SUMMARY

The discussion focuses on solving the differential equation y'' + y = f(t) using the Laplace Transform method. The initial conditions are y(0) = 0 and y'(0) = 1, with the function f(t) defined as f(t) = 1 for 0 ≤ t < π/2 and f(t) = 0 for t ≥ π/2. The incorrect Laplace Transform expression presented was (s^2 + 1)L{y} = (s - e^(-π/2s))/s + 1, which requires correction for accurate results. Participants emphasized the need to correctly compute the Laplace Transform of f(t) before proceeding to find y(t) through inverse transformation.

PREREQUISITES
  • Understanding of Laplace Transform techniques
  • Familiarity with solving second-order differential equations
  • Knowledge of initial value problems in differential equations
  • Ability to perform inverse Laplace Transform calculations
NEXT STEPS
  • Review the correct Laplace Transform of piecewise functions
  • Practice solving second-order linear differential equations with varying initial conditions
  • Learn about the properties of the Laplace Transform relevant to discontinuous functions
  • Explore examples of inverse Laplace Transform applications in engineering problems
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Students studying differential equations, educators teaching Laplace Transforms, and anyone involved in mathematical modeling or engineering applications requiring differential equation solutions.

myusernameis
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Homework Statement



y''+y = f(t)

y(0) = 0; y'(0)=1

f(t) = 1, 0<=t<pi/2
0, pi/2<=t


The Attempt at a Solution


so far, i have


(s^2+1)*L{y} = [tex]\frac{s-e^(-pi/2s)}{s}[/tex] +1

what is next ?
 
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myusernameis said:
so far, i have (s^2+1)*L{y} = [tex]\frac{s-e^(-pi/2s)}{s}[/tex] +1

what is next ?

That doesn't look quite right; the Laplace Transform of [itex]f(t)[/itex] is not [tex]\frac{s-e^{-\pi s/2}}{s}[/tex] (although it's close). Double check that calculation.

Once you correct your expression, solve for [itex]\mathcal{L}[y(t)][/itex] and then take the inverse Laplace transform to get [itex]y(t)[/itex]
 

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