TeeEm
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Considering an ODE which reduces to say

(xy' - y -x)(y'-y)=0

with solutions

from the first factor y1 = xlnx+Ax
from the second factor y2 = Bex

What is the form of the general solution?

y1 U y2?

(brain completely stalled)

Thanks.
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DFranklin
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As I understand it, you can also have solutions that "switch" between one curve and the other. (i.e. pick an x-value x0, choose A, B s.t. y1(x0) = y2(x0) and y1'(x0) = y2'(x0)). Since the evolution of the ODE is completely determined by x, y and y', a curve that = y1 for x<=x0 and y2 for x>=x0 will still obey the ODE.

This can happen at multiple points (even an infiinite number of points) as well...

(edit: I'm not sure it actually can happen at multiple points for this particular ODE, because you don't have enough degrees of freedom. But you can easily imagine a ODE where the solutions are periodic and match up periodically, and you could then switch between them at each period).
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TeeEm
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(Original post by DFranklin)
As I understand it, you can also have solutions that "switch" between one curve and the other. (i.e. pick an x-value x0, choose A, B s.t. y1(x0) = y2(x0) and y1'(x0) = y2'(x0)). Since the evolution of the ODE is completely determined by x, y and y', a curve that = y1 for x<=x0 and y2 for x>=x0 will still obey the ODE.

This can happen at multiple points (even an infiinite number of points) as well...

(edit: I'm not sure it actually can happen at multiple points for this particular ODE, because you don't have enough degrees of freedom. But you can easily imagine a ODE where the solutions are periodic and match up periodically, and you could then switch between them at each period).
what I was actually asking is the general solution

Y = y1 + y2

Y = y1 x y2

Y = y1 or y2


I concluded it must be Y = y1 or y2

which now poses another question

Is Y = y1 + y2 the same as Y = y1 or y2?
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DFranklin
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(Original post by TeeEm)
what I was actually asking is the general solution

Y = y1 + y2

Y = y1 x y2

Y = y1 or y2
Well, what I've pointed out is that there's an (infinite) family of solutions that don't fall into any of the above categories. If you want to ignore that, I'm really not sure what you're trying to do here.
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TeeEm
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(Original post by DFranklin)
Well, what I've pointed out is that there's an (infinite) family of solutions that don't fall into any of the above categories. If you want to ignore that, I'm really not sure what you're trying to do here.


maybe I am not expressing myself too well ...
maybe I do not quite understand your explanation ...
not to worry...

Thanks for your time.
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ztibor
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(Original post by TeeEm)
Considering an ODE which reduces to say

(xy' - y -x)(y'-y)=0

with solutions

from the first factor y1 = xlnx+Ax
from the second factor y2 = Bex

What is the form of the general solution?

y1 U y2?

(brain completely stalled)

Thanks.
When you have the solution from the proper factor in the following form

\displaystyle \phi_1(x,y,C)=0 and \displaystyle \phi_2(x,y,C)=0

then the general solution of the original equation is

\displaystyle   \phi_1(x,y,C)\cdot  \phi_2(x,y,C)=0

This is true for equation with more than two factors in y'
e.g
let dy/dx=p

When the equation is n-th order in y'=p and is rational and integer

p^n+f_{n-1}(x,y)p^{n-1}+ ....+ f_1(x,y)p+f_0(x,y)=0

and the equation has 'n' different roots for p, then the factorized form of the
equation is:

(p-F_1)(p-F_2)\cdot ... \cdot (p-F_n)=0

This multiplication is zero when any from the factors is zero so
we get 'n' piece of 1st order equations

\frac{dy}{dx}=F_1(x,y)
\frac{dy}{dx}=F_2(x,y)
.....
\frac{dy}{dx}=F_n(x,y)

When the solutions for these equations are
\phi_1(x,y,C)=0
\phi_2(x,y,C)=0
....
\phi_n(x,y,C)=0

then the general solution for the original equation is

\displaystyle \Pi_{i=1}^n \phi_i(x,y,C) =0
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TeeEm
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(Original post by ztibor)
When you have the solution from the proper factor in the following form

\displaystyle \phi_1(x,y,C)=0 and \displaystyle \phi_2(x,y,C)=0

then the general solution of the original equation is

\displaystyle   \phi_1(x,y,C)\cdot  \phi_2(x,y,C)=0

This is true for equation with more than two factors in y'
e.g
let dy/dx=p

When the equation is n-th order in y'=p and is rational and integer

p^n+f_{n-1}(x,y)p^{n-1}+ ....+ f_1(x,y)p+f_0(x,y)=0

and the equation has 'n' different roots for p, then the factorized form of the
equation is:

(p-F_1)(p-F_2)\cdot ... \cdot (p-F_n)=0

This multiplication is zero when any from the factors is zero so
we get 'n' piece of 1st order equations

\frac{dy}{dx}=F_1(x,y)
\frac{dy}{dx}=F_2(x,y)
.....
\frac{dy}{dx}=F_n(x,y)

When the solutions for these equations are
\phi_1(x,y,C)=0
\phi_2(x,y,C)=0
....
\phi_n(x,y,C)=0

then the general solution for the original equation is

\displaystyle \Pi_{i=1}^n \phi_i(x,y,C) =0
Thank you for your reply and your detailed explanation (will rep you as soon as it lets me)

I did manage to work the primitive as (y-xlnx+Ax)(y+Bex)=0

which is essentially what you are saying.


PS Do you know how to add a comment on the thread as "solved"
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ztibor
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(Original post by TeeEm)
Thank you for your reply and your detailed explanation (will rep you as soon as it lets me)

I did manage to work the primitive as (y-xlnx+Ax)(y+Bex)=0

which is essentially what you are saying.


PS Do you know how to add a comment on the thread as "solved"
Sorry, I do not know
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