Find the solution to the initial value problem in explicit form.

$\displaystyle y' = xy^3(1 + x^2)^{-\frac{1}{2}}$ ; $\displaystyle y(0) = 1$

$\displaystyle \int \frac{1}{y^3}dy = \int \frac{x}{(1+x^2)^{\frac{1}{2}}}dx$

$\displaystyle -\frac{1}{2y^2} = (1 + x^2)^{\frac{1}{2}} + C$

$\displaystyle -2y^2 = \frac{1}{1 + x^2)^{\frac{1}{2}}} + C$

$\displaystyle -y = \left(\frac{1}{2(1 + x^2)^{\frac{1}{2}}}\right)^{\frac{1}{2}} + C$

$\displaystyle y = - \frac{1}{\left(2(1 + x^2)^{\frac{1}{2}}\right)^{\frac{1}{2}}} \right) + C$

I think I screwed up somewhere because when I go to solve for C I get $\displaystyle C = 1 + \frac{1}{2^{\frac{1}{2}}}$.

The Book's answer in explicit form is $\displaystyle y = \left[3 - 2(1 + x^2)^{\frac{1}{2}}\right]^{-\frac{1}{2}}$