Understanding Inverse Laplace Transforms with Initial Conditions

[Knissp]

Homework Statement

y'' + 2y' + 2y = t^2 + 4t
y(0) = 0
y'(0) = -1

Homework Equations

L(y) = Y
L(y') = sY - y(0)
L(y'') = s^2 Y - s y(0) - y'(0)
L(t^n) = \frac {n!}{s^{n+1}}
L(a*b) = AB
a*b = L^{-1}(AB)

The Attempt at a Solution

Take the Laplace transform of both sides.

L(y'' + 2y' + 2y) = L(t^2 + 4t)

By linearity:

L(y'') + 2L(y') + 2L(y) = L(t^2) + 4L(t)

Substitute:

(s^2 Y - s y(0) - y'(0)) + 2(sY - y(0)) + 2Y = \frac{2}{s^3} + \frac{4}{s^2}

Plug in initial conditions:

(s^2 Y + 1) + 2(sY) + 2Y = \frac{2}{s^3} + \frac{4}{s^2}

Solve for Y:

Y (s^2 + 2s + 2) -1 = \frac{2}{s^3} + \frac{4}{s^2}

Y = \frac {\frac{2}{s^3} + \frac{4}{s^2} + 1}{s^2 + 2s + 2}

Now I must take the inverse Laplace transform. This is where I get confused.

Rewrite:


Y = \frac{1}{s^2 + 2s + 2} (\frac{2}{s^3} + \frac{4}{s^2} + 1)

Distribute:

Y = \frac{1}{s^2 + 2s + 2} \frac{2}{s^3} + \frac{1}{s^2 + 2s + 2} \frac{4}{s^2} + \frac{1}{s^2 + 2s + 2}

Completing the square:

Y = \frac{1}{(s+1)^2 + 1} \frac{2}{s^3} + \frac{1}{(s+1)^2 + 1} \frac{4}{s^2} + \frac{1}{(s+1)^2 + 1}

Now to take the inverse Laplace transform of both sides:

y = L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3}) + L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2}) + L^{-1}(\frac{1}{(s+1)^2 + 1})

y = L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3}) + L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2}) + e^{-t}sin(t)

So now I need to do the separate inverse transforms (via convolutions):

L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{2}{s^3})

= e^{-t}sin(t) * t^2

= \int_0^t (e^{-(t-v)}sin(t-v))(v^2)) dv

L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2})[/tex<br /> <br /> = e^{-t}sin(t) * 4t<br /> <br /> = \int_0^t (e^{-(t-v)}sin(t-v))(4v) dv <br /> <br /> I don&#039;t know how to figure those convolution integrals. Any ideas? <br /> Alternatively, is there a better way to approach the problem?

 

[Feldoh]

Perhaps you might have better luck using partial fractions, instead of completing the square... I don't know for sure since I didn't do it, but it's a suggestion you might want to look into.