Differential Equations: Picard's Existence Theorem

In summary, the problem is to solve the separable differential equation y*y' = 3 and find two different solutions, given the initial condition y(2) = 0. To solve this, we can write the equation as y*dy = 3*dx and integrate both sides. This gives us y^2 = 6x + C, and using the given initial condition, we can find that C = -12. Therefore, the two solutions are y = \sqrt{6x - 12} and y = -\sqrt{6x - 12}.
  • #1
mattst88
29
0

Homework Statement



[tex] y y\prime = 3 [/tex]
[tex] y(2) = 0[/tex]

Homework Equations



Solve and find two different solutions.

The Attempt at a Solution



[tex] F = \frac{3}{y} [/tex]
[tex] \frac{\partial F}{\partial y} = \frac{-3}{y^2} [/tex]

Where do I go from here?
 
Physics news on Phys.org
  • #2
Hey.
The problem is a so called separable diff. equation, so write it as
[tex]dy=\frac{3}{y}dx,y'(x)=\frac{dy}{dx}[/tex]
and integrate on both sides.
I think you now can proceed by yourself.
 
  • #3
mattst88 said:

Homework Statement



[tex] y y\prime = 3 [/tex]
[tex] y(2) = 0[/tex]

Homework Equations



Solve and find two different solutions.

The Attempt at a Solution



[tex] F = \frac{3}{y} [/tex]
[tex] \frac{\partial F}{\partial y} = \frac{-3}{y^2} [/tex]

Where do I go from here?
Where did you get those? What is F?
It should be easy to write
[tex]y\frac{dy}{dt}= 3[/itex]
so
[tex]ydy= 3dt[/itex]
That's basically what eys_physics said.
 
  • #4
Thanks guys.

That problem was so simple. I can't believe I didn't see how to do it. ;)
 
  • #5
To find one solution:
[tex] \int y dy = \int 3 dx [/tex]
[tex] y^2 = 6x + C [/tex]
[tex] y = \pm \sqrt{6x + C} [/tex]

So given y(2) = 0 I find that C = -12.

So my two solutions are
[tex] y = \sqrt{6x + C} [/tex]
[tex] y = - \sqrt{6x + C} [/tex]

Right?
 
Last edited:

1. What is Picard's Existence Theorem?

Picard's Existence Theorem is a fundamental result in the field of differential equations, which states that given a first-order ordinary differential equation (ODE) with an initial condition, there exists a unique solution on a certain interval. It guarantees the existence of a solution to the ODE, which may not always be explicit or easy to find.

2. How does Picard's Existence Theorem work?

Picard's Existence Theorem relies on the concept of a "Picard iteration", which is a process of approximating the solution to a differential equation using a sequence of functions. The theorem states that if a certain condition, known as the Lipschitz condition, is satisfied by the ODE, then the sequence of functions will converge to the unique solution of the equation.

3. What is the Lipschitz condition?

The Lipschitz condition is a mathematical condition that guarantees the uniqueness of a solution to a differential equation. It requires that the rate of change of the function in the ODE is bounded by a constant multiple of the function itself. In simpler terms, it ensures that the function does not grow or shrink too quickly, allowing for a unique solution to exist.

4. What are the applications of Picard's Existence Theorem?

Picard's Existence Theorem has many applications in mathematics, physics, engineering, and other scientific fields. It is used to prove the existence of solutions to various types of differential equations, which are used to model a wide range of phenomena in the natural world. It is also a key tool in the study of dynamical systems and chaos theory.

5. Are there any limitations to Picard's Existence Theorem?

While Picard's Existence Theorem is a powerful tool in the study of differential equations, it does have its limitations. It only guarantees the existence of a solution on a certain interval, and the solution may not be explicit or easy to find. It also assumes that the Lipschitz condition is satisfied, which may not always be the case for more complex equations. In these cases, other methods may need to be used to find a solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
635
  • Calculus and Beyond Homework Help
Replies
6
Views
794
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
560
  • Calculus and Beyond Homework Help
Replies
2
Views
222
  • Calculus and Beyond Homework Help
Replies
2
Views
403
  • Calculus and Beyond Homework Help
Replies
6
Views
505
  • Calculus and Beyond Homework Help
Replies
7
Views
643
  • Calculus and Beyond Homework Help
Replies
4
Views
518
  • Calculus and Beyond Homework Help
Replies
2
Views
480
Back
Top