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
y\frac{dy}{dt}= 3[/itex]<br /> so <br /> ydy= 3dt[/itex] &lt;br /&gt; That&amp;#039;s basically what eys_physics said.

 

[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?