This one is my own, which supprizingly has a nice solution to it: y' = y^2
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Originally Posted by ThePerfectHacker This one is my own, which supprizingly has a nice solution to it: y' = y^2 This is just a Bernoulli equation: Let z(x) = 1/y(x) Then y = 1/z ==> y' = -z'/z^2 So -z'/z^2 = 1/z^2 z' = -1 z(x) = -x + C Thus y(x) = 1/(C - x) is the general solution. -Dan
Ha ha I forgot about that. I had a different idea.
Originally Posted by ThePerfectHacker Ha ha I forgot about that. I had a different idea. Well then, what's your idea? (I've heard there is more than one way to skin a cat...) -Dan
Originally Posted by topsquark Well then, what's your idea? (I've heard there is more than one way to skin a cat...) -Dan Power series solution. It really is quite elegant here.
Originally Posted by ThePerfectHacker Power series solution. It really is quite elegant here. Ah! I can see that. -Dan
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