Question

find partial solution of x'= Ax + b where \(A=\left[\begin{matrix}2&4\\-2&-2\end{matrix}\right]\) and \(b=\left[\begin{matrix}8sin2t\\8cos2t\end{matrix}\right]\)
Please, help

Answer 1

I have characteristic equation for A is r^2 +4 =0 therefore, r = \(\pm2i\) so, eigenvector for r=2i is \(\left[\begin{matrix}1+i\\-1\end{matrix}\right]\) and eigenvector for r = -2i is \(\left[\begin{matrix}1-i\\-1\end{matrix}\right]\)

Answer 2

and .... stuck!! :)

Answer 3

My net is so bad here, if you don't see me, it doesn't mean that I don't care about the help. I 'll be back right after I can. ( In case I am kicked out of the net) Thanks in advance.

Answer 4

@TuringTest

Answer 5

man I'm rusty on DE systems, but my answer to representing 1 + i would be\[\cos(0)+i\sin(\frac\pi2)\]if you can use two different angles. I'd have to review DE's to know if that's allowed, so you tell me if that helps

Answer 6

you need it in terms of t, huh...

Answer 7

so, to plug them in fundamental matrix, what should I do?

Answer 8

that's where I'd have to review DE's, I feel like I'm on shaky ground with this...

Answer 9

I am waiting for you. Take time to review and teach me then, please, please, please. final is upcoming, I must master those stuff

Answer 10

I have a few things to study myself, but it's worth looking over
if you haven't tried this resource I'd recommend it; it's what I'll be using:
http://tutorial.math.lamar.edu/Classes/DE/ComplexEigenvalues.aspx

Answer 11

Thanks for the link. I will.

Answer 12

I know your eigenvectors are right, but I keep getting the wrong answer for the first eigenvector"\[(2-2i)\eta_1+4\eta_2=0\implies\eta_2=\frac12(i-1)\implies\vec\eta=\binom 2{i-1}\]I know it's stupid to ask you to find my mistake when I'm the one supposed to be helping you, but I don't feel confident moving on until I can at least catch up to where you are :P

Answer 13

could somebody just point out my mistake so I can move on? I know it's an algebra error or something dumb....

Answer 14

@TuringTest I am sorry when not being here, I cannot ignore the Asker when I can do something for them.

Answer 15

Oh it's fine, I just figured the message system isn;t working very well...
I'm reading up on my own stuff in the meantime, no rush

Answer 16

I got the same thing turing

\[\left(\begin{matrix}1 \\ (i-1)/2\end{matrix}\right)\]

Answer 17

yeah, but wolfram disagrees :(

Answer 18

In the end you are going to have to use one of the four main methods for solving differential equations either diagonalization, undetermined coefficients, Laplace transforms or variation of parameters.

Answer 19

I was thinking variation of paramaters would be easiest as well, but I was trying a different approach

Answer 20

Also, on math portal i got the answer above whereas on wolfram gave me his answer :P.I'm sure they are just multiples of each other by some factor :P

Answer 21

I choose for x to be i and simplified and got the answer same as on math portal.

Answer 22

well, whatever my mistake is, or isn't, the next step is to get one of the solutions like so\[\vec x_1(t)=e^{2i}\binom{-i-1}1=[\cos(2t)+i\sin(2t)]\binom{-i-1}1\]we multiply this out and factor out the i to get an answer of the form\[\vec x_1=\vec u(t)+i\vec v(t)\]

Answer 23

Yes that is correct, taking the real and imaginary parts of x1 will you a fundamental set of solutions if the wronskian is not 0.

Answer 24

is the wronskian zero? I'm not sure how to check with systems...

Answer 25

hoping that the wronskian isn't zero...
that gives\[\vec x_1(t)=\binom{-\cos(2t)-\sin(2t)}{\cos(2t)}+i\binom{-\cos(2t)-\sin(2t)}{\cos(2t)}\]and, again assuming the wronskian isn't zero, these guys are linearly independent, so the general complimentary solution is\[\vec x_1(t)=c_1\binom{-\cos(2t)-\sin(2t)}{\cos(2t)}+c_2\binom{-\cos(2t)-\sin(2t)}{\cos(2t)}\]if people could checkme on that I'd appreciate it, I don't trust my algebra :P

Answer 26

ohohoh,,, that's what my book says, I just don't get why from those eigenvector they can get it

Answer 27

oh I see a mistake, that should be sine on the second row in the second solution

Answer 28

\[\vec x_1(t)=\binom{-\cos(2t)-\sin(2t)}{\cos(2t)}+i\binom{-\cos(2t)-\sin(2t)}{\sin(2t)}\]which implies\[\vec x(t)=c_1\binom{-\cos(2t)-\sin(2t)}{\cos(2t)}+c_2\binom{-\cos(2t)-\sin(2t)}{\sin(2t)}\]

Answer 29

yeah, @Schrodingers_Cat and I both got the same other answer... But wolfram agrees with yur book, so I'd stick with that for now

Answer 30

This is what I got however I got sin(2t) is positive on the top row of of the real part of the solution because i(-i) =1 right?

Answer 31

I didn't multiply by i, I just factored it out

Answer 32

oh I see

Answer 33

\[\vec x_1(t)=[\cos(2t)+i\sin(2t)]\binom{-i-1}1\\=\binom{-i\cos(2t)+\sin(2t)-\cos(2t)-i\sin(2t)}{\cos(2t)+i\sin(2t)}\\=\binom{-\cos(2t)+\sin(2t)}{\cos(2t)}+i\binom{-\cos(2t)-\sin(2t)}{\sin(2t)}\]did I fix it?

Answer 34

Yep that looks good :)

Now to find the wronskian you have to do this

\[\det \left[\begin{matrix}-\cos(2t)+\sin(2t) & -\cos(2t) -\sin(2t) \\ \cos(2t)& \sin(2t)\end{matrix}\right]\]

Answer 35

doesn't look like what I'd expect it to for a system, but I'll take your word for it :)

Answer 36

well that's 1, so I guess we're good to continue...

Answer 37

Yeah the Wronskian looks like this

W(u,v)(t) = det[u,v]

Answer 38

Yep it is one.

Answer 39

I thought wronskian was

W=det[x1 x2]
[x1' x2']
which wouldn't look like that, but I have a feeling that they wouldn't make this prob;em any harder than it already is by throwing us a cruveball now, we still have the non-homogenious part to do :P

Answer 40

my book says we can always just call the second matrix linearly independent... whatever I trust you and I want to press on XD

Answer 41

Yeah your just trying to find if the vectors u and v in the solution x1 are multiples of eachother.

Answer 42

So, next would be the method to choose to solve the nonhomgenous system

Answer 43

good enough for me\[\vec x_c(t)=c_1\binom{-\cos(2t)+\sin(2t)}{\cos(2t)}+c_2\binom{-\cos(2t)-\sin(2t)}{\sin(2t)}\]is our complimentary is where we got last I think

Answer 44

yep

Answer 45

gotta read again lol

Answer 46

I'm looking at this btw:
http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx

Answer 47

I was thinking probably variation of parameters but the only problem with that is we have to find the inverse of the our fundamental which might isn't to frustrating hopefully :)

Answer 48

We already know the determinant which is nice :)

Answer 49

I agree, but either way we needed the complimentary, so I figured I'd get that out of the way before even thinking about what comes next.
I'm taking a quick break, back in 5. Feel free to forge ahead :)

Answer 50

feel free to discuss and please, don't delete anything. I 'll be back to get. Thanks for discussing

Answer 51

yeah variation of parameters is totally the way to go
we need to get the particular formula with\[\vec x_p(t)=X\int X^{-1}\vec g(t)dt\]where \(X\) is the combination of the two solution vectors \([\vec x_1|\vec x_2]\)
we can get the inverse \(X^{-1}\) pretty easily with the formula\[X=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\implies X^{-1}=\frac1{\det X}\left[\begin{matrix}d&-b\\-c&a\end{matrix}\right]\]

Answer 52

we already found that \(\det X=1\) above, so I am hoping you can actually take it from here. Let me if you get stuck!

Answer 53

Yep that is correct its like this problem was set up so the determinant was 1 :)

Answer 54

lol

Answer 55

at least they threw us one bone :P long problems, these types are....

Answer 56

Yep also Loser66 its important that you learn the other methods as well because if have simple solutions to your characteristic equation diagonaliztion will save you alot of time especially on a test.

Answer 57

I got everything but the step from eigenvector form\(\left[\begin{matrix}-1-i\\1\end{matrix}\right]\) to cos, sin, sin cos,.... If I past that step. I am ok with the rest.

Answer 58

the solution to your characteristic equation is \(e^{rit}\), use Euler's formula...

Answer 59

I know how to take wronskian, apply Cramer's rule to get u' t and then take integral to get u (t) , then plug back to get partial, but the so called Euler's formula

Answer 60

or rather as an eigenvalue \(\large e^{\lambda it}\)
Euler's formula, which you absolutely *must* know in any DE class, is\[\large e^{i\theta}=\cos\theta+i\sin\theta\]

Answer 61

we have \(\lambda = \pm2i \) that mean e^0 (cos 2t + sin 2t) , then?? do what to eigenvectors to get fundamental matrix?

Answer 62

Yes remember loser66 that a general solution to system of differential equations takes the form

\[x = c_{n}e^{\lambda t}\]

The problem is evaluating the exponential if you have imaginary solutions

Using Euler's formula as turing stated above allows us to do this!!!!!!!!!!!!!!!!!!!!

Answer 63

it can be shown that if the roots to the characteristic or eigenvalue is complex, \[\lambda=\mu\pm\theta i\]we can write the solution set as\[\large y_{1,2}(t)=e^{(\mu\pm\theta i)t}=e^{\mu t}(\cos(\theta t)+i\sin(\theta t) )\]from which it can be shown that this can be written\[\large y(t)=c_1e^{\mu}\cos(\theta t)+c_2\sin(\theta t)\]

Answer 64

Oops there should be an eige vector in there after the Cn :)

Answer 65

Yeah, my bad thanks. knew I was bound for some typos

it can be shown that if the roots to the characteristic or eigenvalue is complex, \[\lambda=\mu\pm\theta i\]we can write the solution set as\[\large y_{1,2}(t)=e^{(\mu\pm\theta i)t}=e^{\mu t}(\cos(\theta t)+i\sin(\theta t) )\]from which it can be shown that this can be written\[\large y(t)=c_1e^{\mu t}\cos(\theta t)+c_2e^{\mu t}\sin(\theta t)\]

Answer 66

Oh no I was talking about the one I posted above lol

Answer 67

oh mine has typos too :P
speaking generally above here, not just about systems^

Answer 68

in any event, Loser66 the point is we just got one eigenvector and multiplied it by the complimentary solution to the characteristic... erm is that the right term
we multiplied one of the eigenvectors by \[\large e^{\lambda t}=e^{(\mu\pm\theta) t}=\cos(\theta t)+i\sin(\theta t)\]

Answer 69

oh, I got it. thanks a lot

Answer 70

cool, welcome!

Answer 71

and thanks @Schrodingers_Cat too!

Answer 72

Yep, hope this helps good luck :)

Answer 73

Thanks for your time.