Existence and Uniqueness of Solutions for ODE with Initial Conditions y(1)=0

[supercali]

Homework Statement

given this ODE with initial conditions y(1)=0
\[<br /> (x + y^2 )dx - 2xydy = 0<br /> \]

Homework 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)
\[y&#039; = \frac{{(x + y^2 )}}{{2xy}}\]
and we can see that when y=0 the equation is not defined

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 can't say that the solution is unique? what does it mean
thanks for the help

 

[Jitse Niesen]

solving this ODE gives us
\[y = \sqrt {x\ln (x)} \]

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

\[y&#039; = \frac{{(x + y^2 )}}{{2xy}}\]
and we can see that when y=0 the equation is not defined

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

1) if x>=1 does that mean that the bound for y is y>=0?

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.

 

[supercali]

so i don't 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 don't 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 don't quite understand

 

[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).

i get it the the ODE has a solution only for y>0 or y<0 but those arent my initial conditions!
so i don't quite understand

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.

 

[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?