Differential Equations: Picard's Existence Theorem

[mattst88]

Homework Statement

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

Homework Equations

Solve and find two different solutions.

The Attempt at a Solution

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

Where do I go from here?

 

[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.

 

[HallsofIvy]

Homework Statement

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

Homework Equations

Solve and find two different solutions.

The Attempt at a Solution

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

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]<br /> so <br /> [tex]ydy= 3dt[/itex] <br /> That's basically what eys_physics said.[/tex][/tex]

 

[mattst88]

Thanks guys.

That problem was so simple. I can't believe I didn't see how to do it. ;)

 

[mattst88]

To find one solution:
$$\int y dy = \int 3 dx$$
$$y^2 = 6x + C$$
$$y = \pm \sqrt{6x + C}$$

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

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

Right?