Question

Need help finding the complex solutions for the differential equation y''' + 3y'' +28y' +26y = 0

Answer 1

Sorry, general solution

Answer 2

if i recall right \[y = y_c\] since it's homogeneous

Answer 3

well.....I basically need help factoring to get one regular solution and 2 complex solutions.

Answer 4

and not a clue, sorry

Answer 5

jumping to the good stuff you have
\[\huge m^3 +3m^2 + 28m + 26 = 0\]
right?

Answer 6

yes

Answer 7

i know the real root will be negative...

Answer 8

-1

Answer 9

really?

Answer 10

so that means i factor out a (m+1)?

Answer 11

hmm yep that's right

Answer 12

the actual answer is ce^(-x) + ce^(-x)cos5x + ce^(-x)sin5x

Answer 13

so what exactly is your problem?

Answer 14

so i think that means there are to factorials (wording?) with complex numbers 'i'?

Answer 15

two*

Answer 16

i'm looking at alphawolf and it says that they are: -1-5i and -1+5i. do you know how that is so?

Answer 17

divide m^3 + 3m^2 + 28m + 26 by m+1

Answer 18

m^2 + 2m + 26
______________________________
m+1 | m^3 + 3m^2 + 28m + 26
m^3 + m^2
-------------------
2m^2 + 28m
2m^2 + 2m
----------
26m + 26

Answer 19

so the quotient is m^2 + 2m + 26

Answer 20

it's not factorable so you use quadratic formula

Answer 21

\[\huge m = \frac{-2 \pm \sqrt{2^2 - 4(26)}}{2}\]
\[\huge \implies m=\frac{-2 \pm \sqrt{4 - 104}}{2}\]
\[\huge \implies m=\frac{-2 \pm \sqrt{-100}}{2}\]
\[\huge \implies m= \frac{-2 \pm 10i}{2}\]
\[\huge \implies m=-2 \pm 5i\]

Answer 22

uhhh that should be \[\huge \implies m = -1 \pm 5i\]

Answer 23

does that help?

Answer 24

absolutely! this is where my teacher skipped

Answer 25

haha now you know ^_^

Answer 26

thank you

Answer 27

welcome :D