Laplace transform problem help

Question

I'm stuck with this Laplace transform problem.

Using Laplace transform, solve initial problem: $$ \ y''' - y = 1 \\ y(0) = y'(0) = 1 \\ y''(0) = 0 $$

And from that we would get:

$$ \ s^3Y - s^2 - s - Y = \frac{1}{s} $$

I've got it until the point I get: $$ \ Y = \frac{s^3 + s^2 + 1}{s(s-1)(s^2 + s + 1)} = \frac{s^3 + s^2 + 1}{s(s^3 - 1)} $$ and now I'm not sure what to do next.

I know that I should get something like:

$$ \ Y = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s^2 + s + 1} $$

But I'm not able to get that, and if I got that I'm not sure what to do with fraction that has denominator like this:

$$ \ \frac{C}{s^2 + s + 1} $$

That also makes me curios if I have overlooked on something and that term should actually cancel with something? (As it looks very similar to the numerator).

Any help and explanation of the situation would be of great help and use!

To summarize: I'm stuck at one point in solving this Laplace transform problem and my question would be how to solve it.

Answer 1

Are you familiar with the general partial fraction method? With a quadratic (order 2) polynomial in the denominator, you should write a general linear (order 1) polynomial in the numerator. Then putting things over a common denominator, $$\frac{A}{s} + \frac{B}{s-1} + \frac{Cs + D}{s^2+s+1} = \frac{(A+B+C)s^3+(B-C+D)s^2+(B-D)s-A}{s(s-1)(s^2+s+1)}$$ from which $A=-1$ and $B=C=D=1$. (Alternatively, you could factorize $s^2+s+1$ over the complex numbers.)

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Then the inverse Laplace transform of $$\frac{1}{s^2+s+1} = \frac{1}{\left(s+\frac{1}{2}\right)^2 + \frac{3}{4}}$$ is $e^{-x/2}$ multiplying the inverse Laplace transform of $$\frac{1}{s^2 + \frac{3}{4}}$$ which is itself $2/\sqrt{3} \times \sin \left(\sqrt{3}x/2\right)$.

Then multiplying by $s$ is the same as differentiating by $x$.

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Note that the complex method (factorizing the quadratic with complex roots) is maybe more intuitive and has less arbitrary rule-following. It gives $$ \frac{s+1}{s^2+s+1} = \frac{\frac{1}{2} + \frac{i}{2\sqrt{3}}}{s+\frac{1}{2} + \frac{i\sqrt{3}}{2}} +\frac{\frac{1}{2} - \frac{i}{2\sqrt{3}}}{s+\frac{1}{2} - \frac{i\sqrt{3}}{2}} = {\rm Re} \left[ \frac{1 + \frac{i}{\sqrt{3}}}{s+\frac{1}{2} + \frac{i\sqrt{3}}{2}} \right] $$ whose inverse Laplace transform is $$ {\rm Re} \left[ \left(1 + \frac{i}{\sqrt{3}}\right) e^{-\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)x} \right] = e^{-x/2} \cos\left(\frac{\sqrt{3}}{2}x\right) + \frac{1}{\sqrt{3}}e^{-x/2} \sin\left(\frac{\sqrt{3}}{2}x\right) $$

Answer 2

$$\frac 1 {(s^2+s+1)} =\frac 1{(s+.5)^2 +.75}$$

Taking the inverse lappace transform of this give $$\sqrt 3\sin \left (\frac {\sqrt3}2 (t+.5)\right )$$ If you endup with a linear term in the fraction, you will end up with a cosine intead of sine.

Answer 3

Taking advantage of the fraction... $$\ Y = \frac{s^3 + s^2 + 1}{s(s-1)(s^2 + s + 1)}$$ $$\ Y = \frac{s^3 + s^2 \color {red}{+s-s} + 1}{s(s-1)(s^2 + s + 1)}$$ $$\ Y = \frac{s(s^2 + s+1) -(s- 1)}{s(s-1)(s^2 + s + 1)}$$ $$\ Y = \frac1{(s- 1)}- \frac 1{s(s^2 + s + 1)}$$ $$\ Y = \frac1{(s- 1)}+ \left (\frac {s+1}{(s^2 + s + 1)} -\frac 1s\right )$$ $$\ Y = \frac1{(s- 1)}-\frac 1s+ \left (\frac {s+1}{(s^2 + s + 1)} \right )$$ $$\ Y = \frac1{(s- 1)}-\frac 1s+ \left (\frac {s+1/2}{(s +1/2)^2 +3/4)} \right )+\left (\frac {1/2}{(s +1/2)^2 +3/4)} \right )$$ Then take inverse Laplace Transform gives: $$y(t)=e^t-1+e^{-t/2} \left(\frac {\sqrt 3}3\sin (\frac {\sqrt 3}2 t)+\cos (\frac {\sqrt 3}2 t) \right)$$