Differential Equations: Picard's Existence Theorem

mattst88
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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?
 
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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.
 
mattst88 said:

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.
 
Thanks guys.

That problem was so simple. I can't believe I didn't see how to do it. ;)
 
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?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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