# 2nd order ode w/complex conjugate roots

1. Jan 13, 2009

### danbone87

x''+x'+2x=0 x(0)=2 x'(0)=0

I've taken the characteristic equation and reduced the roots to

1/2 +- Sqrt(7/4)i of the form

a +- bi (i = sqrt(-1)

Then i put the homogeneous solution into the form of e$$^{}at$$*(B1cos(bt)+B2sin(bt))

for B1 i used the first i.c. and found that B1=2

for B2 i used the i.c. for x' and found B2 to be 1/(sqrt7/4)

now i need to get it into the form of Ae$$^{}at$$sin(bt+phi)

where A = sqrt(B1^2+B2^2) and phi is acos(B1/A)

the problem is that A is not the hypotenuse of of a right triangle. I didn't know if i screwed up my b values along the way or if i should go to the law of cosines, but i saw no mention of such a problem so I figured i may have made an error beforehand. any help is appreciated.

2. Jan 13, 2009

### Thaakisfox

Actually I dont understand what your problem is... :D

What do you mean by "A is not the hypotenuse of a right triangle"??, What right triangle?

3. Jan 13, 2009

### danbone87

I mean that in this case law of cosines would be required to find A

so i can't reduce the homogeneous solution by the method i was given (A = sqrt(B1^2+B2^2) , so i'm wondering if i made a mistake beforehand.

;]

4. Jan 13, 2009

### NoMoreExams

I think you made a tiny arithmetic error, you have x'' + x' + 2x = 0, let's say the char. equation is a^2 + a + 2 = 0, quadratic formula gives us roots

$$\frac{-1 \pm \sqrt{-7}}{2} = -\frac{1}{2} \pm i\sqrt{\frac{7}{4}}$$

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