Second Order Homogeneous Eq's, Auxiliary Eq for complex roots - Help

[_Greg_]

Hi

I'v got a maths exam on Tuesday for my 2nd year of chemial engineering.
Been going through a past paper and have been going over 2nd order homogenous DE's

Im at the stage of calculating the roots (wether repeated or 2 distinct roots) I take the easy path like so:

E.g m2 + 4m + 4 = 0

(m + 2)(m + 2) = 0 so m=-2 (twice)

Thats easy but that doesn't work for complex roots.

My lechtrer does it this way which I don't really understand (for complex numbers that is, fine for everything else)

So something like:

m2 - 6m + 10 = 0

He then does:


m = ( 6 ± SQRT[36-40] / 2 ) = 3 ± i


I don't understand this! A general trend i see here is that the 6m = the 6 before the square root, the 36 is the 6 squared and the 40 is always the
last number * 4 (so 10*4 = 40) This holds true for every problem but I don't have a clue how to get that answer?

I don't understand really what ' i ' is but if i got that answer I can work throuh my textbook and work out the rest, I just need to know how to get 3 ± i in the first place.

Thanks in advance

 

[rock.freak667]

You should post this in the correct forum. But I will answer you anyway.

m² - 6m + 10 = 0

your teacher used the quadratic equation formula

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

and then used the fact that i=√-1 for the minus sign in the the square root when it is computed

generally for complex roots λ±μi, your general solution (or homogeneous solution) is given by

e^λx(Asin(μx)+Bcos(μx)) where A & B are constants.

 

[_Greg_]

Fantastic, thanks very much!