Solve the initial value problem y′=$\displaystyle \frac{x^2}{2y + 1}$, y(0)=−1 by separation of variables.

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- Jan 25th 2009, 01:50 PMronaldo_07[SOLVED] Differentiation
Solve the initial value problem y′=$\displaystyle \frac{x^2}{2y + 1}$, y(0)=−1 by separation of variables.

- Jan 25th 2009, 01:51 PMJhevon
- Jan 25th 2009, 02:02 PMronaldo_07
2yx+x=$\displaystyle \frac{x^3}{3}$ Then rearange to make y=....... then let y{0)=-1 to find the solution right?

- Jan 25th 2009, 02:12 PMJhevon
- Jan 25th 2009, 02:31 PMronaldo_07
With respect to y, would that be y^2 + y

- Jan 25th 2009, 02:34 PMJhevon
- Jan 25th 2009, 02:38 PMronaldo_07
$\displaystyle y^2+y=\frac{x^3}{3}+c$ Im not sure how to get y on its own at this stage

- Jan 25th 2009, 02:52 PMJhevon
- Jan 25th 2009, 03:10 PMronaldo_07
Im still not able to solve for c using quadratic formulae

- Jan 25th 2009, 03:13 PMJhevon
- Jan 25th 2009, 03:22 PMronaldo_07
c=1? do i need to solve for y to answer this question?

- Jan 25th 2009, 03:26 PMJhevon
- Jan 25th 2009, 03:28 PMronaldo_07
can you help im not sure how to finish this

- Jan 25th 2009, 03:31 PMJhevon
- Jan 25th 2009, 03:40 PMronaldo_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