Verifying relations

1. Jan 15, 2009

bengaltiger14

1. The problem statement, all variables and given/known data

Verify that the relation x^2 + y^2 = 1 is a solution to the differential equation:
dy/dx = xy/(x^2 - 1)

Can anyone point me in the right direction on how to begin to solve this problem? Do I take integral of the DE and just plug into equation?

Last edited: Jan 15, 2009
2. Jan 15, 2009

rock.freak667

You could take x^2 + y^2 = 1 and find dy/dx and see if it is the same as the DE

3. Jan 15, 2009

bengaltiger14

dy/dx is just 2x+2y. So it does not equal correct?

4. Jan 15, 2009

rock.freak667

2x+2y(dy/dx)=0 is what you get when you differentiate implicitly with repect to x. Now just find dy/dx and do a bit of algebra.

5. Jan 15, 2009

bengaltiger14

I'm sorry but I'm really slow at DE's.

2x+2y(dy/dx)=0

What do you mean find dy/dx. I thought dy/dx was 2x+2y=0.

Are you saying to solve for dy/dx in the first equation?

6. Jan 15, 2009

rock.freak667

Yes, make dy/dx the subject of the equation.

EDIT: is the relation supposed to be y2-x2=1 ?

Last edited: Jan 15, 2009
7. Jan 15, 2009

bengaltiger14

2y(dy/dx) = -2x

divide by 2y:

dy/dx = -2x/2y

8. Jan 15, 2009

rock.freak667

So dy/dx= -x/y right?

Now multiply the numerator and denominator by y.

(Forget my edit in the previous post)

9. Jan 15, 2009

bengaltiger14

dy/dx = -xy

That is the answer to the problem? So the DE is not a solution.

10. Jan 15, 2009

rock.freak667

If dy/dx=-x/y and you multiply by y/y, do you really get dy/dy=-xy or -xy/y2?

11. Jan 15, 2009

bengaltiger14

yeah, your right... dy/dx = -xy/y^2

But the top y would just cancel and the y^2 would become y again so why do that?

12. Jan 15, 2009

rock.freak667

Don't cancel out anything. Keep it the way it is. Now from the relation x2+y2=1, what is y2 equal to?

13. Jan 15, 2009

bengaltiger14

y^2 = 1-x^2

14. Jan 15, 2009

rock.freak667

Now in the equation

$$\frac{dy}{dx}=\frac{-xy}{y^2}$$

Replace y2.

15. Jan 15, 2009

bengaltiger14

dy/dx = -xy/1-x^2

So they do equal. If you multiply by -1, it gives the DE.

16. Jan 15, 2009

rock.freak667

and if it gives the DE, it is a solution of it.

17. Jan 15, 2009

bengaltiger14

Cool..Thank you very much for your help and patience.