- #1

manenbu

- 103

- 0

yy''-(y')

^{2}= y

^{3}

After doing some work it gets to this point:

(y')

^{2}=2y

^{3}

Taking the root:

y' = +-(2y

^{3})

^{1/2}

My question is, what do I do with the +-?

Do I solve 2 different integrals? Assume that this is positive?

Does it matter if it has starting conditions (y(x)=a, y'(x)=b)?