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):

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

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

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

[tex]L^{-1}(\frac{1}{(s+1)^2 + 1} \frac{4}{s^2})[/tex

$$= e^{-t}sin(t) * 4t$$

$$= \int_0^t (e^{-(t-v)}sin(t-v))(4v) dv$$

I don't know how to figure those convolution integrals. Any ideas?
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.