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

In summary, the ODE given has a general solution of \[\ln (x) - \frac{{y^2 }}{x} = c\] with the initial condition y(1)=0. However, since the theorem on existence and uniqueness does not apply for this initial condition, there may be more than one solution or no solution at all.
  • #1
supercali
53
0

Homework Statement


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

Homework Equations


solving this ODE gives us
[tex]\[y = \sqrt {x\ln (x)} \][/tex]
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)
[tex]\[y' = \frac{{(x + y^2 )}}{{2xy}}\][/tex]
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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  • #2
supercali said:
solving this ODE gives us
[tex]\[y = \sqrt {x\ln (x)} \][/tex]
That's just one solution for the ODE. The general solution has a constant in there.

[tex]\[y' = \frac{{(x + y^2 )}}{{2xy}}\][/tex]
and we can see that when y=0 the equation is not defined
The equation is not defined when [itex]2xy=0[/itex]. 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.
 
  • #3
so i don't get it
the general solution was:
[tex]\[\ln (x) - \frac{{y^2 }}{x} = c\][/tex]
for the initial conditions y(1)=0
my solution was
[tex]\[y = \sqrt {x\ln (x)} \][/tex]

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
 
  • #4
I'm sorry, my fault. I hadn't seen that you were given an initial condition. Yes, the solution
[tex]\[\ln (x) - \frac{{y^2 }}{x} = c\][/tex]
is correct (assuming that x > 0, which is the case here). And indeed, the initial condition implies that c = 0. However, the solution
[tex]\[y = \sqrt {x\ln (x)} \][/tex]
is slightly different (hint: the equation [itex] x^2=1 [/itex] 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.
 
  • #5
great i understand the explanation one more thing though
since i have found a solution that is:
[tex]\[\ln (x) - \frac{{y^2 }}{x} = c\][/tex]
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?
 

1. What is the concept of existence and uniqueness in mathematics?

Existence and uniqueness is a fundamental concept in mathematics that deals with the question of whether a solution to a problem exists and if it is the only possible solution. In other words, it refers to the existence of a unique solution to a mathematical problem.

2. What is the significance of existence and uniqueness in scientific research?

Existence and uniqueness are crucial in scientific research as they provide a means of verifying the validity of mathematical models and theories. It ensures that the solutions obtained are not only accurate but also unique, making them more reliable for further analysis and experimentation.

3. How is the concept of existence and uniqueness applied in real-world problems?

The concept of existence and uniqueness is applied in various fields such as physics, engineering, economics, and biology to solve real-world problems. For example, it is used to determine the stability of a system, predict the behavior of a physical phenomenon, or find the optimal solution to an economic model.

4. Can a problem have more than one solution and still satisfy the conditions of existence and uniqueness?

No, a problem cannot have more than one solution and still satisfy the conditions of existence and uniqueness. If a problem has more than one solution, it violates the uniqueness condition, indicating that the problem is ill-posed and does not have a unique solution.

5. What are the common techniques used to prove existence and uniqueness of solutions in mathematics?

The common techniques used to prove existence and uniqueness of solutions in mathematics include the method of continuity, the method of contraction, and the method of maximum principle. These methods involve using the properties of the problem and theorems from analysis to establish the existence and uniqueness of solutions.

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