Solving a System of Differential Equations: dx/dt = y^2 - x^2, dy/dt = -2xy

[atomqwerty]

How can I solve the differential equations system

dx/dt = y^2 - x^2

dy/dt = -2xy

?

Sorry about not using LaTex, I know it looks better.

Thanks!

 

[AlephZero]

x' = y^2 - x^2
y' = -2xy

Differentiate the first equation

x''= 2yy' - 2x

Use the second equation to eliminate y'
Then use the first equation again to eliminate y

 

[atomqwerty]

So I obtain

x'' +4x'x +4x^3 + 2x == 0

and now?

thanks

 

[JJacquelin]

Solving in polar coordinates is easier. The main steps only are shown in the attached document.

 

[atomqwerty]

perfect! I got it! Thank you very much :D

 

[AlephZero]

So I obtain

x'' +4x'x +4x^3 + 2x == 0

and now?

thanks

The standard method to solve that type of equation is

Let x' = p
Then x'' = dp/dt = dp/dx dx/dt = p dp/dx
So you get

p dp/dx + 4px + 4x^3 + 2x = 0

Integrating with respect to x gives

p^2/2 + I can't see how to integrate the 4px term + x^4 + x^2 = C

But I like the polar coordinates method better.

Now we know the answer, my method seems to be heading in right direction, but that's no use unless we can see how to finish it.

 

[Delta2]

In Alephzero's method i think there is a mistake, the 2x term should be 2xx' but this doesn't seem to make things easier.

 

[JJacquelin]

In Alephzero's method i think there is a mistake, the 2x term should be 2xx' but this doesn't seem to make things easier.

I agree :
x'' + 6 x x' +4 x^3 = 0
2 y y'' -3 (y')^2 +4 y^4 = 0
OK. far to be easier, but possible, even without knowing the solution.

 

[JJacquelin]

Remark : the obvious particular solution [ x=1/(t+c) ; y=0 ] is included in the set of solutions in the particular case of b=c*a and a=> infinity.