aatkins09
dy/dx=(-x+sqrt(x^2+y^2))/y

USE SUBSTITUTION u=x^2+y^2

Nex Vortex
If you use the correct substitution then several things should cancel. It would be helpful to also post your attempt at the solution in order to see where you are having problems.

aatkins09
If you use the correct substitution then several things should cancel. It would be helpful to also post your attempt at the solution in order to see where you are having problems.

I got
int(y dy)=int(-x+sqrt(u))dx
which turned into
y^2/2=(1/-2+2y)*int(sqrt(u) du)
which equals
-1/2+2y*2/3u^3/2
which then gets down to
y^2/2=(-u^(3/2))/3-3y

and then it gets uglier when I try to get y by itself
I know I am messing up somewhere but cannot pin point it

Instead of starting out by trying to separate and integrate, try finding an expression for du/dx and using it to replace dy/dx

aatkins09
Instead of starting out by trying to separate and integrate, try finding an expression for du/dx and using it to replace dy/dx

du/dx would be 2x+2y but I dont understand where to go from there. how can I replace dy/dx with it?

Nex Vortex
du/dx would be 2x+2y but I dont understand where to go from there. how can I replace dy/dx with it?

Remember that you have to use implicit differentiation.
$$du=2xdx+2ydy$$

Then you can solve for $$dy/dx$$
You should find that when you find this and substitute, several things should cancel.

aatkins09
Remember that you have to use implicit differentiation.
$$du=2xdx+2ydy$$

Then you can solve for $$dy/dx$$
You should find that when you find this and substitute, several things should cancel.

I did it (in my head, not on paper yet) and I'm pretty sure I've got it! thank you so much, I appreciate all of your help!

aatkins09
Remember that you have to use implicit differentiation.
$$du=2xdx+2ydy$$

Then you can solve for $$dy/dx$$
You should find that when you find this and substitute, several things should cancel.

nvm, I still can't get it. it's okay though, thank you, I appreciate all of your help!

$$\frac{du}{dx} = 2x+2y \, \frac{dy}{dx}$$
$$\frac{du}{dx} = 2x+2y \, \frac{dy}{dx}$$