# Differential Equations: Picard's Existence Theorem

1. Mar 30, 2008

### mattst88

1. The problem statement, all variables and given/known data

$$y y\prime = 3$$
$$y(2) = 0$$

2. Relevant equations

Solve and find two different solutions.

3. The attempt at a solution

$$F = \frac{3}{y}$$
$$\frac{\partial F}{\partial y} = \frac{-3}{y^2}$$

Where do I go from here?

2. Mar 30, 2008

### eys_physics

Hey.
The problem is a so called separable diff. equation, so write it as
$$dy=\frac{3}{y}dx,y'(x)=\frac{dy}{dx}$$
and integrate on both sides.
I think you now can proceed by yourself.

3. Mar 30, 2008

### HallsofIvy

Staff Emeritus
Where did you get those? What is F?
It should be easy to write
$$y\frac{dy}{dt}= 3[/itex] so [tex]ydy= 3dt[/itex] That's basically what eys_physics said. 4. Mar 30, 2008 ### mattst88 Thanks guys. That problem was so simple. I can't believe I didn't see how to do it. ;) 5. Mar 30, 2008 ### mattst88 To find one solution: [tex] \int y dy = \int 3 dx$$
$$y^2 = 6x + C$$
$$y = \pm \sqrt{6x + C}$$

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

So my two solutions are
$$y = \sqrt{6x + C}$$
$$y = - \sqrt{6x + C}$$

Right?

Last edited: Mar 30, 2008