Solving ODE: y = $\sqrt {x\ln (x)}$ w/ Initial Conditions

[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

 

[HallsofIvy]

Yes, since (x+y^2)/(2xy) does not exist when y=0, the "existence and uniqueness" theorem does not hold here. That means we cannot conclude that a unique solution exists on any open interval around x= 1. But it might be true anyway.

What you have done is find a solution, y= \sqrt{xln(x)}, That is defined only for x> 1 (and y must be positive)- that is not defined on an open interval containing x= 1. But, in fact, what happens here is that there DOES exist a solution on an interval around x= 1: y= 0 for all x is such a second solution. The solution is not unique on the x> 1.

 

[supercali]

But, in fact, what happens here is that there DOES exist a solution on an interval around x= 1: y= 0 for all x is such a second solution. The solution is not unique on the x> 1.

if can please elaborate more i don't think i understand what you mean

 

[HallsofIvy]

What part do you not understand? Surely it is obvious that y= 0 for all x is a solution to that problem.

 

[supercali]

well y=0 is not a solution as i see it if i put it in my original equation it doesn't solve it

what i don't understand is if for these initial conditions the theorem on uniqueness and existence apply or not? and if it doesn't apply what can i say on the solution i found hence is it a solution or not?