DE Help - Solve dy/dx=(-x+sqrt(u))/y

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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!
 
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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.
 
Nex Vortex said:
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
 
uart said:
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 don't understand where to go from there. how can I replace dy/dx with it?
 
aatkins09 said:
du/dx would be 2x+2y but I don't 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.
 
Nex Vortex said:
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!
 
Nex Vortex said:
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!
 
aatkins09 said:
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.
 
  • #10
uart said:
\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 :)
 
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