Originally Posted by

**Tau** Hello

$\displaystyle xy' + y^2 = 1$, $\displaystyle y(1) = \frac{1}{3}$

My thinking:

Rearrange the equation and you will get this:

$\displaystyle \frac{1}{1 - y^{2}}\delta y = \frac{1}{x} \delta x$

Integrate it...

$\displaystyle \frac{1}{2}({\color{red} - } \ln|1 - y| + \ln|1 + y|) = \ln|x| + C$ Mr F says: The mistake is in this line. I've added the correction (in red).

I think that the absolutes can be ignored, in that case I get:

$\displaystyle y = \pm \sqrt{C - x^2}$

But the correct f(x) is something very different.(Wondering)

Thanks for your help.