Question

Just a hint or link to similar problem please!
Solve X'=Ax+F. More below

Answer 1

\[A=\left[\begin{matrix}0 & 4 \\ -1 & 0\end{matrix}\right]\] \[F=\left[\begin{matrix}e^{2t} & 0\\0 &1\end{matrix}\right]\]

Answer 2

x=[x(t);y(t)]

Answer 3

Are you sure about the dimensions of \(F\)?

Answer 4

nope you're right

Answer 5

F=[sint ; 0]

Answer 6

Okay, yeah I was wondering how a 2x2 would work on the RHS...

Anyway, this seems like a fairly standard undetermined-coefficients problem.

Answer 7

I know A has complex eigenvalues

Answer 8

What are you getting for your homogeneous solution?

Answer 9

\[X'= c_1e^{2it}v_1+c_2e^{-2it}v_2\]

Answer 10

v1 and v2 are the repective eigenvectorss

Answer 11

Okay, looks right. Euler's formula will help with the rewriting of the solution.

As for the nonhomogeneous part: have you worked out any undetermined-coefficients approaches before for systems?

Answer 12

Can do the rewrite, but issue is the other part. No I haven't :/

Answer 13

Also upon rewrite I got
\[x=c_1v_1(cos(2t)+isin(2t))+c_2v_2(cos(-2t)+isin(-2t))\]

Answer 14

The idea is the same as for single ODEs. Say the nonhomogeneous function is something like \(ae^t\). Then we might guess a solution of the form \(\vec{a}e^t\).
\[{\bf{x}}'-\begin{pmatrix}0&4\\-1&0\end{pmatrix}{\bf{x}}=\begin{pmatrix}1\\0\end{pmatrix}\sin t\]
For a regular ODE we might consider a trial solution of \(a\sin t+b\cos t\), so here we'll try \(\vec{a}\sin t+\vec{b}\cos t\).

So, if our trial solution is \({\bf x}_p\), then
\[{\bf x}_p=\vec{a}\sin t+\vec{b}\cos t~~\implies~~{{\bf x}_p}'=\vec{a}\cos t-\vec{b}\sin t\]
Substituting into the original system, we have
\[\left(\vec{a}\cos t-\vec{b}\sin t\right)-\begin{pmatrix}0&4\\-1&0\end{pmatrix}\left(\vec{a}\sin t+\vec{b}\cos t\right)=\begin{pmatrix}1\\0\end{pmatrix}\sin t\]
From here we'd simplify and solve for \(\vec{a}\) and \(\vec{b}\).

Answer 15

let me try

Answer 16

we can't do the matrix multiplication though?

Answer 17

Why's that? We have a 2x2 matrix multiplied by a 2x1 vector. We would end up with another 2x1 vector.

Answer 18

ohhh cause it's a vector

Answer 19

duh

Answer 20

k so I get \[[acost~~~-bsint]-[4bcost~~~-asint]=[1;0]sint\] Is that right?

Answer 21

so b= 1/3?

Answer 22

You're half-right... Keep in mind that \(\vec{b}=\begin{pmatrix}b_1\\b_2\end{pmatrix}\). What does it means for \(\vec{b}=\dfrac{1}{3}\)?

Answer 23

b1=1 b2=3?

Answer 24

No, you found one component of \(\vec{b}\), but we're still missing the other. Here's my approach:
\[\begin{align*}
\left(\vec{a}\cos t-\vec{b}\sin t\right)-\underbrace{\begin{pmatrix}0&4\\-1&0\end{pmatrix}}_{A}\left(\vec{a}\sin t+\vec{b}\cos t\right)&=\begin{pmatrix}1\\0\end{pmatrix}\sin t\\\\

\left(\vec{a}-A\vec{b}\right)\cos t-\left(\vec{b}+A\vec{a}\right)\sin t&=\begin{pmatrix}1\\0\end{pmatrix}\sin t
\end{align*}\]
which gives us two equations:
\[\vec{a}-A\vec{b}=\begin{pmatrix}0\\0\end{pmatrix}\quad\text{and}\quad \vec{b}+A\vec{a}=\begin{pmatrix}1\\0\end{pmatrix}\]
where \(\vec{a}\) and \(\vec{b}\) are 2x1 vectors.

Answer 25

Solving for \(\vec{a}\) in the first equation, we can substitute into the second to solve for \(\vec{b}\):
\[\vec{a}=A\vec{b}~~\implies~~\vec{b}+A^2\vec{b}=(I_2+A^2)\vec{b}=\begin{pmatrix}1\\0\end{pmatrix}~~\implies~~\vec{b}=(I_2+A^2)^{-1}\begin{pmatrix}1\\0\end{pmatrix}\]
Computing gives \(\vec{b}=\begin{pmatrix}1/3\\0\end{pmatrix}\).

Answer 26

You can then plug this back into the first equation and solve for \(\vec{a}\), and you'll have your nonhomogeneous solution.

Answer 27

ok, and was my rewrite using euler's right?

Answer 28

Yes, provided that \(v_1\) and \(v_2\) are correct. You can do some simplification to have a homogeneous solution in terms of \(C_1v_1\sin t+C_2v_2\cos t\).

Answer 29

and isn't a just [-1/3; 0]?

Answer 30

Close, you have your components switched around.

Answer 31

gosh darnit

Answer 32

Here's another example worked out if you need some more practice: http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx

Answer 33

Thank you. Yea, I need more practice to say the least

Answer 34

No problem!

Answer 35

I'm so screwed for this test

Answer 36

this is the review sheet, and we have not really learned probably 75% of it

Answer 37

Well if you have any other questions, there are at least a few of us here that can help you out.

It's possible that undetermined coefficients might not have been the intended approach. There might be a different way using elimination techniques.

Answer 38

there is no guide to how to approach it. you are probably right

Answer 39

the v1 and v2 are \[v_1=[-2i;1] ~~and~~v_2=[2i;i]\] right?

Answer 40

Yeah those are correct.

Answer 41

2i,1*****

Answer 42

wait, was the i right?

Answer 43

The first one, the eigenvectors are \(\pm\begin{pmatrix}2i\\1\end{pmatrix}\).

Answer 44

If you're interested in another approach, it's very doable. Possibly more to your liking.

Answer 45

ok thats what I thought. No this is fine. He has a tendency to make the test identical to the review sheet. I will learn the crap out of this

Answer 46

thanks again sithsies

Answer 47

Alright then, good luck. You're welcome!

Answer 48

so doesn't the solution have to be real?

Answer 49

@zepdrix I still have a few questions on this one because it is complex, can you help?

Answer 50

nvm, I figured it out! :D