(3x-5y)dx + (x+y) dy = 0

[J4e8a16n]

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

[German_EE]

And the point of this math exercise is??

[J4e8a16n]

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

[Mechatrommer]

And the point of this math exercise is??

wasting your time and mine... in one possibility... y=A, x=B...

[onlooker]

A thing having no purpose may still have solutions,
 
(y-3x)² = C(y-x)

[Christe4nM]

It's a differential equation. Write it in the form of dy/dx + y = x. Then solve like a first order linear differential equation.

Edit: tried to rewrite and solve it, but it's a non-linear differential equation. My abilities unfortunately only include linear ones with constant coefficients

[Jope]

Solution by Wolfram Alpha: (3x-5y)dx + (x+y) dy = 0 .

[IanB]

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.

[J4e8a16n]

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
-----
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

[IanB]

Something like this, I believe:

[free_electron]

to calculate my circuits.

i use this :

[free_electron]

Laplace Transformations are for beginners. Real men use a Widlar Transformations.

[J4e8a16n]

Something like this, I believe:

Yes.

[IanB]

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
-----
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...

[J4e8a16n]

Thanks for your support )


JPD