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April 17, 2014

April 17, 2014

Posted by **david** on Tuesday, December 12, 2006 at 10:39pm.

(you already posted this, but i still don't get it)

dy/dx = 2y^2

Integrating...y=2/3 y^3 + C

put 1,-1 into the equation, and solve for C

c=-1/3

so the equation is y=(2y^3)/3-1/3

then i have to find what y equals when x=2. how do i do that. there's no x in the equation.

Now, if the problem is dx/dy = 2y^2

then x=2/3 y^3 + C and you solve.

Recheck which it is.

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