# Confused by this ODE

1. Aug 27, 2005

### Benny

I've got a test coming up in a few weeks so I've been going through some exercises that I've done and one that I've come across has got me really worried.

$$\frac{{dy}}{{dx}} = \frac{{x^2 + y^2 }}{{2xy}},y\left( 3 \right) = 2$$

The answer is $$y = \sqrt {x^2 - \frac{{5x}}{3}} ,x \ge \frac{5}{3}$$.

The thing which I am unsure about is the restriction on x. If x = 5/3 then y = 0 but then what happens to the original DE? Looking at the original DE, I see that the denominator on the RHS has xy. Then for the solution, the x values I can have are either x > 1 and or x < 1, ie x cannot be equal to zero. Similarly in the solution, I can't have y = 0. The IC y(3) = 2 implies x > 0 doesn't it? Ok that's fine in terms of the answer that's been given. But what about the values of y? As I said before when x = 5/3 > 0, y = 0 but how can that be given the way that the DE has been written?

Edit: When I have an IVP for example y' = f(x), y(x_0) = y_0, does the solution curve pass through points in which the x-coordinate is to the left of x_0(ie. does the solution curve pass through (x,y) where x < x_0)? I'm wondering if it is correct to assume that the initial conditions are there simply to give you one specific piece of info in the solution curve and does NOT entail any other restrictions on the solution curve.

Last edited: Aug 27, 2005
2. Aug 27, 2005

### saltydog

The general solution is:

$$y(x)=\pm\sqrt{x^2+Kx}$$

of which your particular solution is as given.

$x=5/3$ simply is where the slope is vertical. You know it's a parabola on its side right? Just use the piece that fits the initial conditions. If you had an IVP with a negative value for y, then use the negative part. Don't know where you're getting the x>1 and x<1. For your IVP, the domain is $x\ge 5/3$.

The initial condition only specifies "find a solution which passes through the point given". That's it. Says nothing about what x can be. In the particular case above, x cannot be less than that because of the radical.

Try and look at an IVP as a question: Find a function which has the slope given by the ODE and passes through the point given. May only be a part of some complex looping curve in the x-y plane, may be a nice function, maybe no function works which meets the criteria specified, or maybe more than one or even an infinite number. For your IVP, the top half of a sideways parabola meets the criteria and that parabola does not exist for x<5/3.

3. Aug 27, 2005

### Benny

Ok then thanks for your help.