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Hi,
(3x-5y)dx + (x+y) dy = 0
dy/dx = (5y-3x)/ (x+y)
(5y-3x)/ (x+y) = 5A/ (x+y) - 3B/ (x+y)
5y-3x = 5A - 3B
5y-5A = 3x - 3B
5(y-A) = 3(x-B)
(y-A) * 5/3= x-B
............... https://www.eevblog.com/forum/Smileys/default/huh.gif
Have a good day,
JPDaviau -
And the point of this math exercise is??
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Is to find what type of solution exist for this equation.
This is from Shepley L Ross 4th edition
Introduction to ordinary differential equations.
It is not exact.
I could not found an integrating factor
I have tried v=y/x method
Partial fraction does not seem to work
It does not seem Bernouilli or homogenous.
I am aiming to understand Laplace Transform.
I do it for my pleasure and the pleasure to calculate my circuits.
Maybe it is crasy but I am doing it for tree years now. I started with trigonometry. It is good for the mind.
Well, that's it.
JPDaviau
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And the point of this math exercise is??
wasting your time and mine... in one possibility... y=A, x=B... -
A thing having no purpose may still have solutions,
(y-3x)2 = C(y-x) -
I have tried v=y/x method
According to Wolfram Alpha that method leads to a solution (i.e. let y(x) = x v(x) and proceed from there). However, the solution does not come out as anything simple. -
Laplace Transformations are for beginners. Real men use a Widlar Transformations.
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dy/dx = (5y-3x / x + y) / x = > -3 + 5y/x / 1+ y/x
y/x = v y = vx ==>d y /dx = d vx/dx ==>v + v' x ==> v + x*dv/dx
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v + x* dv/dx = -3 + 5y/x / 1+ y/x ==>
x dv/dx = -v^2+4v -3/ 1+v
dx/x = 1+v / -v^2 +4v -3 dv
from here complete the square?
I am working at it)
JPD
It looks like you got it right up that point. You might notice that ( - v^2 + 4v - 3) can be factorized...