# Laplace Transforms (Partial Fractions - complex roots)

1. Dec 9, 2008

### Nubcakes

Just when I thought I had Laplace transforms figured out with partial fractions, I once again fall flat on my face.

The following question from my book labeled " Inverse Laplace Transforms - Partial Fractions (Complex Roots)

NOTE: Since I can't draw S(s) and T(s) very well, I have replaced them with X(s) and Y(s) respectively; f(s)=f(x),{s=x} and f(t)=f(y),{t=y}

So I first thought this should be easy... Until I noticed normal factoring wouldn't work. I attempted to use the quadratic equation...

Once I got the factors I figured it should be smooth sailing. I should simply be able to solve for A and B. From there it's simply reversing the Laplace transforms...

This is where I hit a dead end road block. I can't find ANYTHING in my book of how to deal with imaginary numbers in formula(equations) you wish to transform(or un-transform). I certainly didn't notice anything that could cancel the imaginary number (i) out.

This is the answer my book gives, but as with the last example, the book proved to be wrong. I wouldn't put it past them for a second error.

Any help with this would be greatly appreciated!

Thank you for your time!~

2. Dec 9, 2008

### rock.freak667

Try completing the square of the denominator, then rewrite the numerator as (4x-4)+4

3. Dec 9, 2008

### Nubcakes

I don't know why, but for some reason your idea gave me a idea to wrap my attempt up with an answer. By treating the complex number as a mere constant I came up with this;

I can't really do anything with it, but I am pretty sure its a legit answer...

Any input on that?

4. Dec 9, 2008

### rock.freak667

Not sure how beneficial imaginary numbers are in laplace transforms, but if to solve an ordinary differential equation, it would be better to have real numbers.

5. Dec 9, 2008

### gabbagabbahey

hmmm...there are a couple of small errors....

(1)This first one doesn't have any outcome on your final result, but the 'b' in your quadratic equation is -2 not +2, so your roots are $x_{\pm}=+1 \pm i$....Luckily your next line is correct, since $x^2-2x+2=(x-x_{+})(x-x_{-})=(x-(1+i))(x-(1-i))$. it seems as though you made two errors which canceled eachother out.

(2)Your solution for A and B is a little off; go over that part again. This error will propogate through to your final result.

6. Dec 9, 2008

### gabbagabbahey

There is nothing wrong with expressing the solution in terms of complex nubers, although one can also express it in terms of sines and cosines.

7. Dec 9, 2008

### Nubcakes

Hmm... Ah, I see my errors now! Despite the fact they are small, they would no doubt cost me points on a test...

Anyway, thanks a lot abbagabbahey on the verification regarding my final answer represented with complex numbers. Makes me feel a a little more comfortable doing these problems

8. Dec 9, 2008

### gabbagabbahey

Well your final answer is a little off; you should end up with $$(2i+2)e^{y(1-i)}-(2i+2)e^{y(1+i)}$$ which is equivalent to the answer in your text (you can check this using Euler's formula)

9. Dec 1, 2009

### mxpxer7

How did you end up with 2i+2 instead of 2/i+2 like he had?

10. Dec 2, 2009

### gabbagabbahey

$$\frac{2}{-i}=2i$$

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