LaPlace Transformations to Solve Ordinary Differential Equations

[Mike86]

Homework Statement

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).

Homework 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)

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 can't see where) or there is a way to continue from here with the quadratic formula. Any advice with be immensely appreciated. Thanks!

 

[Je m'appelle]

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?

 

[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.

 

[Mike86]

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!

 

[vela]

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.