Laplace transform IVP 2nd order

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SUMMARY

The discussion centers on solving the initial value problem (IVP) for the second-order differential equation y'' + 6y' + 10y = 0 with initial conditions y(0) = 2 and y'(0) = 1 using the Laplace transform. The solution process involves transforming the equation into the Laplace domain, isolating Y(s), and performing inverse transformations. The final solution presented is y(t) = 2e^-3t cos(t) + 13e^-3t sin(t), although there is a discrepancy regarding the coefficient of the sine term, which should be 7 according to external references. The Laplace transform techniques and the application of inverse transforms are critical to resolving this problem.

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  • Understanding of Laplace transforms and their properties
  • Familiarity with solving second-order linear differential equations
  • Knowledge of initial value problems (IVP) in differential equations
  • Ability to perform inverse Laplace transforms
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  • Study the properties of the Laplace transform, specifically L^{-1}(s/(s+b)^2 + a^2)
  • Learn how to complete the square in the context of Laplace transforms
  • Review the derivation of the Laplace transform for e^{-bt} cos(at)
  • Practice solving more complex initial value problems using Laplace transforms
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Students and educators in mathematics, particularly those focusing on differential equations, as well as engineers and physicists applying Laplace transforms in their work.

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Homework Statement



y''+6'+10y=0
y(0)=2
y'(0)=1

Homework Equations





The Attempt at a Solution



Laplace everything and I get
s^2*Y(s)-2s-1+6s*Y(s)-12+10Y(s)=0

isolate Y(s)
Y(s)=(2s+13)/(s^2+6s+10)

split into 2 terms, bottom can be rearranged by completing the square

2s/[(s+3)^2+1^1] + 13/[(s+3)^2+1^1]

inverse laplace

y(t)=2e^-3t*cost + 13e^-3t*sint

both wolfram and the answer key say the last term should be 7 instead of 13, but i don't see where I made a mistake
 
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[tex]L^{-1}\frac s{(s+b)^2+a^2}=e^{-bx}(\cos ax-\frac ba\sin ax)[/tex]
 
The Laplace transform of ##\cos t## is ##\frac{s}{s^2+1}##, so when you replace s by s+3, you get ##\frac{s+3}{(s+3)^2+1}##. Compare that to what you said was the Laplace transform of ##e^{-3t}\cos t##.
 

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