# Y= (x^2 + y^2)/xy ODE

asdf1
for this O.D.E. :
y= (x^2 + y^2)/xy
it's unseparable, so what other methods can there be taken?

EvLer
It's a change of variables case (actually excercise #1 in my book on this topic), the way you do this is check to see whether f(x,y) = f(tx,ty), where f(x,y) = dy/dx.
So if you substitute tx and ty for x and y respectively, you'll see that equality holds. Then do following substitution: y = x V(x) into the DE and see what you get with applying dy/dx and some simplifications. It should reduce to separable equation.

Homework Helper
This is just like the post about the other ODE. You can write it as:

$$y'=x/y+y/x$$

Again, with the substitution z=y/x it becomes:

$$z+xz'=1/z+z$$
or
$$z'=1/(xz)$$
which is separable.

Homework Helper
it's homogenous, separable, use v=y/x

asdf1
ok, thanks!

asdf1
here's my work:
y=x/y +x/y
suppose y=vx
=>v=y/x
=> y=v+xv
so v+xv=1/v+v
=> vdv=dx/x
=> v^2=ln[absolute value(x)]+c
=> (y^2)/(x^2)=ln[absolute value(x)]+c
=> y^2=2(x^2)ln[absolute value(x)]+cx^2
now if i take the square root on both sides, there should be a positive and negative sign on the right~
the correct answer should only have the positive sign, but how can you be sure that it should be positive?

Homework Helper
asdf1 said:
here's my work:
y=x/y +x/y
suppose y=vx
=>v=y/x
=> y=v+xv
so v+xv=1/v+v
=> vdv=dx/x
=> v^2=ln[absolute value(x)]+c
Whoopsie, you've forgot a factor 1/2 on the left side.

You can use either sign, both will be valid solutions to the ODE. This becomes clear when you plug y back into the ODE to check if it works out. Then, with the benefit of hindsight, you could foresee this, since if y is one solution, then the other is -y and it's derivative is -y'. If you write the ODE as xyy'=x^2+y^2 you can see that if y is a solution, then -y is too.

Ofcourse, if you're given a boundary value or initial value/condition then there will be only one solution. (otherwise the problem is ill-stated).

asdf1
thank you very much! :)