# LaPlace Transformations to Solve Ordinary Differential Equations

by Mike86
Tags: differential, equation, la place, ordinary, transformations
 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!
P: 109
 Quote by Mike86 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).
Use the partial fractions technique in

$$\mathcal{L} = \frac{e^{-s}}{(s^2 + 2s + 5)}$$

Then, in order to find x in your new expression, you'll have to apply the Inverse Laplace Transform $$\mathcal{L}^{-1}$$.

By the way, do you remember how to use partial fractions? And do you understand how the Inverse Laplace Transform works?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,536 Write the Laplace transform in the form $$X(s) = \frac{e^{-(s+a)}e^a}{(s+a)^2+b^2}$$ 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.
P: 5

## LaPlace Transformations to Solve Ordinary Differential Equations

 Quote by vela Write the Laplace transform in the form $$X(s) = \frac{e^{-(s+a)}e^a}{(s+a)^2+b^2}$$ 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.
Thanks for the replies! :)

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!
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,536 Sorry, I made a mistake. You want it to look like $$X(s) = \frac{e^{-s}}{b} \left[\frac{b}{(s+a)^2+b^2}\right]$$ (It's just a slight algebraic rewrite; your a and b don't change.) Look at the frequency- and time-shifting properties of the transform.

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