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

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Discussion Overview

The discussion revolves around the existence and uniqueness of solutions for the ordinary differential equation (ODE) given by \((x + y^2)dx - 2xydy = 0\) with the initial condition \(y(1)=0\). Participants explore the implications of the initial condition on the solutions derived from the ODE, including the conditions under which the existence and uniqueness theorem applies.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a solution \(y = \sqrt{x\ln(x)}\) and questions whether the existence and uniqueness theorem applies given that the function is not defined at \(y=0\).
  • Another participant clarifies that the general solution includes a constant and that the equation is not defined when \(2xy=0\), suggesting that \(y=0\) is not the only consideration.
  • A later reply indicates that the general solution is \(\ln(x) - \frac{y^2}{x} = c\) and acknowledges the initial condition \(y(1)=0\) implies \(c=0\), but notes that the solution \(y = \sqrt{x\ln(x)}\) is incomplete as it only accounts for \(y>0\).
  • Participants express confusion about how the derived solutions relate to the initial conditions, particularly regarding the implications of the existence and uniqueness theorem not applying at \(y=0\).
  • One participant seeks clarification on whether the derived general solution resolves the initial problem despite the theorem's limitations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the implications of the initial condition for the existence and uniqueness of solutions. There are competing views on whether the derived solutions adequately address the initial conditions and the applicability of the existence and uniqueness theorem.

Contextual Notes

Participants note that the existence and uniqueness theorem does not apply at \(y=0\), leading to uncertainty about the nature of solutions in that region. The discussion highlights the need for further exploration of the implications of the initial conditions on the general solution.

supercali
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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
 
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supercali said:
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.
 
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
 
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).

supercali said:
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.
 
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?
 

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