DE Help - PLEASE!

[aatkins09]

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

USE SUBSTITUTION u=x^2+y^2

I tried to work it out but got a really ugly answer, please help!

 

[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

 

[uart]

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!

 

[uart]

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

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

Now just substitute in your original expression for dy/dx and it literally just falls into place.

 

[aatkins09]

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

Now just substitute in your original expression for dy/dx and it literally just falls into place.

Thanks so much :)