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LaPlace Transformations to Solve Ordinary Differential Equations 
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#1
Oct1810, 06:09 AM

P: 5

1. The problem statement, all variables and given/known data
Consider the initial value problem: x'' + 2x' + 5x = δ(t  1); with: x(0) = 0 and x'(0) = 0. Using Laplace transforms, solve the initial value problem for x(t). 2. Relevant equations L[x''] = (s^2)*L[x]  s*x(0)  x'(0) L[x'] = s*L[x]  x(0) L[δ(t  1)] = e^(s) 3. The attempt at a solution Using the above known Laplace Transformations and the initial conditions I have gotten: x'' + 2x' + 5x = δ(t  1); with: x(0) = 0 and x'(0) = 0. L[x''] + 2L[x'] + 5[x] = L[δ(t  1)] (s^2)*L[x]  s*x(0)  x'(0) + 2 (s*L[x]  x[0]) + 5 (L[x]) = e^(s) (s^2)*L[x] + 2s*L[x] + 5*L[x] = e^(s) (s^2 + 2s + 5)*L[x] = e^(s) L[x] = e^(s) / (s^2 + 2s + 5) From here I am not sure what Laplace Transformation to use to get the answer x. I can't really factorize (s^2 + 2s +5) because I would have to use the quadratic formula and I would get solutions with imaginary parts (from where I have no idea where to go as far as Laplace transformations are concerned). I'm not sure if I made a mistake in the lead up (I cant see where) or there is a way to continue from here with the quadratic formula. Any advice with be immensely appreciated. Thanks! 


#2
Oct1810, 08:20 AM

P: 109

[tex]\mathcal{L} = \frac{e^{s}}{(s^2 + 2s + 5)}[/tex] Then, in order to find x in your new expression, you'll have to apply the Inverse Laplace Transform [tex]\mathcal{L}^{1}[/tex]. By the way, do you remember how to use partial fractions? And do you understand how the Inverse Laplace Transform works? 


#3
Oct1810, 01:38 PM

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PF Gold
P: 11,683

Write the Laplace transform in the form
[tex]X(s) = \frac{e^{(s+a)}e^a}{(s+a)^2+b^2}[/tex] for the appropriate a and b. Then you should be able to use the properties of the Laplace transform and your table to get back to the time domain. 


#4
Oct2010, 05:23 AM

P: 5

LaPlace Transformations to Solve Ordinary Differential Equations
I have obtained values of: a = 1 and b=2. Only problem is I can't make the connection between the properties of the Laplace Transformations and my tables. I've been playing around and looking for an hour or so but I've been stumped! 


#5
Oct2010, 12:39 PM

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PF Gold
P: 11,683

Sorry, I made a mistake. You want it to look like
[tex]X(s) = \frac{e^{s}}{b} \left[\frac{b}{(s+a)^2+b^2}\right][/tex] (It's just a slight algebraic rewrite; your a and b don't change.) Look at the frequency and timeshifting properties of the transform. 


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