ODE Canonical form again,
so I have this system $u''+u'=uv^2$ and $v''+v'=-uv^2+v$ So, I am just not sure the protocol for this style of problem, I can do the eqs individually, but I am not sure if that is the answer

Question

ODE Canonical form again, 
so I have this system $u''+u'=uv^2$ and $v''+v'=-uv^2+v$  So, I am just not sure the protocol for this style of problem, I can do the eqs individually, but I am not sure if that is the answer

fibonaccichick666 · Answer

@Kainui

fibonaccichick666 · Answer

Follow up question, will I need to find bounds for u,v,u',v'?

fibonaccichick666 · Answer

@ganeshie8 you're welcome to chime in, if you have any ideas

ganeshie8 · Answer

idk these stuff :/

fibonaccichick666 · Answer

:( it's ok, it doesn't seem like many people do

ganeshie8 · Answer

you want to write it as system of first order equations is it

 $\mathbb{x}' = A\mathbb{x}$

fibonaccichick666 · Answer

well, That's the thing, I just don't know.  canonical form is just if a de is of degree n, $y^{(n)}=f(t,y,y',y'',...,y^{n-1})$

fibonaccichick666 · Answer

if a matrix then similar is ok

fibonaccichick666 · Answer

like could do a substitution and then get everything in terms of that substitution, but I'm just not sure how to approach this

ganeshie8 · Answer

Ohk i think we just need to make them friendly for matrix method

fibonaccichick666 · Answer

ok, I'm game

fibonaccichick666 · Answer

so I guess et's make a mapping let $$x_1=u ~~x_2=u'~~x_3=v~~x_4=v'$$ then we create a matrix

anonymous · Answer

So you can find the solutions for the equations separately?

ganeshie8 · Answer

lets introduce two more variables

$x = u' \implies   x' = u''$
$y = v'\implies  y' = v''$

anonymous · Answer

Find the solutions for each equation, and keep the ones that fit both equations.

ganeshie8 · Answer

substitute and we will get that canonical whatever form, right ?

fibonaccichick666 · Answer

|dw:1420867110410:dw|

fibonaccichick666 · Answer

I don't need a solution here

anonymous · Answer

Wait, is this a PDE?

fibonaccichick666 · Answer

no... but for some reason we are doing this

anonymous · Answer

We are differentiating with respect to what?

fibonaccichick666 · Answer

we are not differentiating at all

anonymous · Answer

What does $'$ mean?

fibonaccichick666 · Answer

prime

ganeshie8 · Answer

lol

fibonaccichick666 · Answer

lol

anonymous · Answer

As in some other variable that is independent of $u$?

kainui · Answer

Haha I'm crying a little

fibonaccichick666 · Answer

well, since we are not solving it there is no differentiation involved technically.  We are just creating an isomorphism

ganeshie8 · Answer

Alright! this looks easy, atleast cooking up the canonical form : 

$$x_1=u ~~x_2=u'~~x_3=v~~x_4=v'$$

$$x_1' = u'~x_2' = u''~x_3'=v'~x_4' = v''$$

fibonaccichick666 · Answer

but anyways, is that matrix alright?

fibonaccichick666 · Answer

|dw:1420867569457:dw|

ganeshie8 · Answer

i haven't reached that far, let me see..

fibonaccichick666 · Answer

kk

anonymous · Answer

Oh, I thought $u' = \frac{du}{dv}$ and $v'=\frac{dv}{du}$, but it makes a bit more sense now.

fibonaccichick666 · Answer

yea, we don't know,  it's probably like dv/dt

ganeshie8 · Answer

$u''+u'=uv^2$ changes to 
$x_2' +x_2 = x_1x_3^2 \implies  x_2' =  -x_2+x_1x_3^2 $

anonymous · Answer

You are assuming that they're all being differentiated with respect to the same variable though, and my assumption makes that unlikely

fibonaccichick666 · Answer

check

ganeshie8 · Answer

|dw:1420867856621:dw|