Question

ODE Assistance please

Answer 1

So, I have \[x'=\left[\begin{matrix}3 & 6 \\ -1 & -2\end{matrix}\right]x\].
I don't usually ask for a full walkthrough, but I need someone to take my hand here. (I missed class) I need to find the general solution of the system. I know it has something to do with eigenvalues and vectors

Answer 2

So, what is my game plan?

Answer 3

find eigenvalues and vectors first, may be we can see what to do later

Answer 4

thats what you do in linear algebra the moment you see a matrix :)

Answer 5

Yea, I hated Linear... Didn't learn a thing for all of the proofs...

Answer 6

So, for eigenvaluses I do : \(det(A-\lambda)\) right?

Answer 7

if you prefer you can solve this by converting the system into a single second order equation

Answer 8

solve below for eigenvalues :
\[\large \left|A - \lambda I\right| = 0\]

Answer 9

yep... don' know how to convert into a single second order eq. Give me a sec to try the det thing though

Answer 10

\[\left|\begin{matrix}3-\lambda & 6 \\ -1 & -2-\lambda\end{matrix}\right| = 0\]

Answer 11

I got lambda = 2 or 1?... that doesn't seem right

Answer 12

http://www.wolframalpha.com/input/?i=eigen+values+%7B%7B3%2C6%7D%2C%7B-1%2C-2%7D%7D

Answer 13

...dang, how'd they get that?

Answer 14

must've multiplied wrong

Answer 15

familiar with finding eigen vectors ?

Answer 16

so I have lambda squared minus lambda

Answer 17

yes

Answer 18

h/o I have to get this part first

Answer 19

kk

Answer 20

so \[\lambda^2-\lambda=0\]
I need to solve

Answer 21

that's where I'm at

Answer 22

1 and zero obviously work, but I can't get the math to

Answer 23

are you kidding

Answer 24

\[\lambda^2 - \lambda = \lambda (\lambda -1) = 0 \implies \lambda = 0 \lor \lambda = 1\]

Answer 25

no.... I wish I was

Answer 26

Omy god....

Answer 27

Yea, ok factor it

Answer 28

yes next find the corresponding eigenvectors and we will be done with solving the system

Answer 29

so now, I have to set up eigenvectors with psi times x or something?

Answer 30

\[\left[\begin{matrix}3-\lambda & 6 \\ -1 & -2-\lambda\end{matrix}\right] \begin{pmatrix}a\\b\end{pmatrix} = 0\]

Answer 31

plugin each \(\lambda \) value and find a corresponding solution (a, b)

the solution (a, b) will be the eigenvector

Answer 32

ok thanks. I'm sorry I seem like such an idiot at the moment. can you check for me in a minute?

Answer 33

for finding an eigenvector corresponding to \(\lambda = 0\), we solve :
\[\left[\begin{matrix}3-0 & 6 \\ -1 & -2-0\end{matrix}\right] \begin{pmatrix}a\\b\end{pmatrix} = 0\]

Answer 34

wait, don't tell me

Answer 35

kk :)

Answer 36

ok so I got \[\left(\begin{matrix}3a+6b \\ -a-2b\end{matrix}\right)=0\] SO I tried to solve for a and b but the answer is [0;0]?

Answer 37

that can't be right

Answer 38

did I multiply wrong?

Answer 39

0,0 is a trivial answer which we don't want

Answer 40

we will always get this 0,0 when solving the matrix for eigen vectors

Answer 41

the rows in matrix are actually dependent, notice that the determinant is 0

Answer 42

... but then... how? how do we get the two eigenvectors? is there only one?

Answer 43

\[\left(\begin{matrix}3a+6b \\ -a-2b\end{matrix}\right)=0 \]

a+2b = 0

a/b = -2/1

so an eigen vector is (-2, 1)

Answer 44

Notice that (-2, 1) satisfies the earlier matrix equation :
\[\left[\begin{matrix}3-0 & 6 \\ -1 & -2-0\end{matrix}\right] \begin{pmatrix}a\\b\end{pmatrix} = 0 \]

Answer 45

how did you get the a/b=-2/1?

Answer 46

a + 2b = 0

a = -2b

a/b = -2/1

Answer 47

it could be :

a/b = -4/2 or -10/5

or any other scalar multiple

Answer 48

ok... well I am now a walking study for the cognitive failures and detriment from chronic lack of sleep

Answer 49

lol you're funny

Answer 50

so basically we just need to find eigen values and eigenvectors to write out the solution of given DE system

Answer 51

You would never believe I am in my second to last math class for a bachelor's in math atm.... I'm an idiot right now, but anyways, how do you correlate a/b=-2/1 into an eigenvector? how does that work?

Answer 52

see you forget things if u don't review, it is ok. im here because i need to review these stuff, so in a way you're helping me

Answer 53

meh, well that's good at least, but I didn't understand Linear first time around like 3 years ago, so add that and memory loss and taadaaa one incompetent idiot.

Answer 54

an eigen vector is any solution that satisfies this equation :

\[\left[\begin{matrix}3-0 & 6 \\ -1 & -2-0\end{matrix}\right] \begin{pmatrix}a\\b\end{pmatrix} = 0\]

Answer 55

ohhhh ok, just substitute into the original ok

Answer 56

since a=-2, b=1 satisfies above matrix equation, (-2, 1) is an eigen vector

Answer 57

let me find the other

Answer 58

notice that ANY scalar multiple of (-2, 1) satisfy the above equation

Answer 59

ok or simply use wolfram..

Answer 60

http://www.wolframalpha.com/input/?i=eigen+vectors+ {{3%2C6}%2C{-1%2C-2}}

Answer 61

no, cause I need to be able to do it for my tests

Answer 62

ohkk

Answer 63

I got (-3;1)

Answer 64

thats right, lets cookup the solution

Answer 65

ok cool

Answer 66

so we have below :

1) \(\lambda = 0\), \( (a, b) = (-2, 1)\)
2) \(\lambda = 1\), \( (a, b) = (-3, 1)\)

Answer 67

a solution to given system is :
\[\large x = \begin{pmatrix}a\\b\end{pmatrix}e^{\lambda t}\]

Answer 68

since you have two independent solutions, take the linear combination for the general solution

Answer 69

wait wait don't tell me

Answer 70

I'm getting \[y=\left(\begin{matrix}-2 \\ 1\end{matrix}\right)+e^t \left(\begin{matrix}-3 \\ 1\end{matrix}\right)\]

Answer 71

oh wait +C!

Answer 72

that looks good !

Answer 73

general solution :
\[\large x = c_1\begin{pmatrix}-2\\1\end{pmatrix}e^{0t} + c_2\begin{pmatrix}-3\\1\end{pmatrix}e^{1t}\]

Answer 74

awesome! Thank you so sososososossoossosososososososososososososososososososososos much

Answer 75

np :D

Answer 76

I am so unbelievably grateful for you putting up with my stupidity

Answer 77

And now, I can do my homework!

Answer 78

lol no it was nice to review these with u :) btw if u have time i can try explaining why that general solution works... but im sure you need to finish other problems and rest a bit

Answer 79

actually this setup works only for "homogeneous linear systems with constant coefficients"

Answer 80

That, I know. It is because every set of eigenvectors is LI and therefore the addition of two LI solutions gives you the general solution to the homogeneous eq

Answer 81

same idea as having a complex gradient when doing second orders

Answer 82

sort of

Answer 83

exactly! but how on earth we have concluded that below is one solution to the given system ?
\[x = \begin{pmatrix}a\\b\end{pmatrix}e^{\lambda t}\]

Answer 84

(a, b) is eigen vector corresponding to the eigen value lambda

Answer 85

you may review ur textbook later when you're free, its really a nice proof :)

Answer 86

because it seemed reasonable since we used it in variation of parameters

Answer 87

luckily, we don't have to know proofs for this class

Answer 88

Oh yeah its more or less same as variation of parameters

Answer 89

**the intuition

Answer 90

But anyways, thank you again. So much. I can replicate this without help now I hope you have fun reviewing!!

Answer 91

:)

Answer 92

:)

Answer 93

are u in my class by any chance :O

Answer 94

1-first assume solution as
\(x=CVe^\lambda \)
2-find eigen value and vectors :)

Answer 95

hehe since rsadhavica already find it
i'll graph it for u :)
http://prntscr.com/53vbbw

Answer 96

V is eigen vector ?

Answer 97

\[x'=\left[\begin{matrix}3 & 6 \\ -1 & -2\end{matrix}\right]x\]

becomes :

\[(CVe^{\lambda})'=\left[\begin{matrix}3 & 6 \\ -1 & -2\end{matrix}\right]CVe^{\lambda}\]

Answer 98

CV is a constant multiplied by a vector