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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.

[tex]

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

[/tex]

The answer is [tex]y = \sqrt {x^2 - \frac{{5x}}{3}} ,x \ge \frac{5}{3}[/tex].

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

[tex]

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

[/tex]

The answer is [tex]y = \sqrt {x^2 - \frac{{5x}}{3}} ,x \ge \frac{5}{3}[/tex].

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.
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