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Differential Equations: Picard's Existence Theorem

  • Thread starter mattst88
  • Start date
  • #1
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1. Homework Statement

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

2. Homework Equations

Solve and find two different solutions.

3. 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?
 

Answers and Replies

  • #2
263
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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
HallsofIvy
Science Advisor
Homework Helper
41,738
899
1. Homework Statement

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

2. Homework Equations

Solve and find two different solutions.

3. 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
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Thanks guys.

That problem was so simple. I can't believe I didn't see how to do it. ;)
 
  • #5
29
0
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:

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