I want to solve the following IVP and also want to find interval of validity of the solution.

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

Solution:

First separate and then integrate both sides.$y^{-3}dy=x(1+x^2)^{0.5}dx$

$\int y^{-3}dy=\int x(1+x^2)^{0.5} dx$

$\frac{-1}{2y^2}=\frac{\sqrt{(1+x^2)^3}}{3} +c$

Applying initial condition to get the value ofc.

$\frac{-1}{2}=\frac{\sqrt{1}}{3}+c. c=\frac{-5}{6}$

The implicit solution is then

$\frac{-1}{2y^2}=\frac{\sqrt{(1+x^2)^3}}{3}-\frac56$

Now let's solve for y(x).

$\frac{1}{y^2}=\frac{-5}{3}-\frac23 \sqrt{(1+x^2)^3}$

$y^2=\frac{3}{5-2\sqrt{(1+x^2)^3}}$

$y(x)=\pm\frac{\sqrt{3}}{\sqrt{5-2\sqrt{(1+x^2)^3}}}$

Reapplying the initial condition shows us that the "-" is the correct sign. The explicit solution is then,$y(x)=\frac{-\sqrt{3}}{\sqrt{5-2\sqrt{(1+x^2)^3}}}$

I am not giving full computation for the interval of validity.

$x^2=(\frac{25}{4})^{0.3333333}-1$

Hence -0.917614<x<0.917614 is the interval of validity of the solution.

But answer provided to me do not match with my calculated answer. Would any one find here what is wrong with my calculations?