1. Originally Posted by ronaldo_07
when I got c=1, I got it because I let -1=Left side $\displaystyle (y^2$+y). So I guess I should have done$\displaystyle (-1)^2$-(-1)=$\displaystyle \frac{(0)}{3}^3$+C. But this would give C to be 0
yes...and that's what C is

by the way, it is + (-1)

and if you learn to use LaTeX properly, you will realize you won't have to be chopping things up like that and putting only one term between each pair of [tex] tags

2. and if you learn to use LaTeX properly, you will realize you won't have to be chopping things up like that and putting only one term between each pair of [tex] tags[/quote]

yea, sometimes error comes up when I try to put it all in 1 go

3. Originally Posted by ronaldo_07
yea, sometimes error comes up when I try to put it all in 1 go
that's because you don't know how to use it properly , hence my last comment

4. do I leave answer as y=$\displaystyle \frac{-1 \pm \sqrt{1+4x^3}}{2}$

5. Originally Posted by ronaldo_07
do I leave answer as y=$\displaystyle \frac{-1 \pm \sqrt{1+4x^3}}{2}$
well, it should be $\displaystyle \frac {4x^3}3$ instead of $\displaystyle 4x^3$, but yes, that's the right form

6. Originally Posted by Jhevon
well, it should be $\displaystyle \frac {4x^3}3$ instead of $\displaystyle 4x^3$, but yes, that's the right form

Thats what I got it was a typo, thanks for the help

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