# Existence and uniqueness

1. Jun 24, 2008

### supercali

1. The problem statement, all variables and given/known data
given this ODE with initial conditions y(1)=0
$$$(x + y^2 )dx - 2xydy = 0$$$
2. Relevant equations
solving this ODE gives us
$$$y = \sqrt {x\ln (x)}$$$
as we can see this equation is true only for x>=1
in order to use the theorem on existence and uniqueness we isulate for y'=f(x,y)
$$$y' = \frac{{(x + y^2 )}}{{2xy}}$$$
and we can see that when y=0 the equation is not defined
3. The attempt at a solution
my question is
1) if x>=1 does that mean that the bound for y is y>=0?
2)if it meaas that y>=0 then should i conclude that the theorem on existence and uniqueness does not apply here since the function is not continuous thus we cant say that the solution is unique? what does it mean
thanks for the help

2. Jun 25, 2008

### Jitse Niesen

That's just one solution for the ODE. The general solution has a constant in there.

The equation is not defined when $2xy=0$. So y=0 is not the whole story.

No. It means that the equation has a unique solution for y>0 and y<0. Since you only found the solution for y>0, your solution is not complete.

3. Jun 25, 2008

### supercali

so i dont get it
the general solution was:
$$$\ln (x) - \frac{{y^2 }}{x} = c$$$
for the initial conditions y(1)=0
my solution was
$$$y = \sqrt {x\ln (x)}$$$

so i dont get it what is the final answer for this ODE?
does this slolution apply?
i get it the the ODE has a solution only for y>0 or y<0 but those arent my initial conditions!
so i dont quite understand

4. Jun 25, 2008

### Jitse Niesen

I'm sorry, my fault. I hadn't seen that you were given an initial condition. Yes, the solution
$$$\ln (x) - \frac{{y^2 }}{x} = c$$$
is correct (assuming that x > 0, which is the case here). And indeed, the initial condition implies that c = 0. However, the solution
$$$y = \sqrt {x\ln (x)}$$$
is slightly different (hint: the equation $x^2=1$ has two solutions).

To be more precise: the only thing you know is that the existence and uniqueness theorem does not apply for y = 0. It may be that there is no solution, it may be there is more than one solution, it may even be that there is only one solution.

5. Jun 26, 2008

### supercali

great i understand the explanation one more thing though
since i have found a solution that is:
$$$\ln (x) - \frac{{y^2 }}{x} = c$$$
and we know that the theorem does not apply for the initial conditions what can i say about this solution?
does it solve the initial problem?