$$y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$$
$$y'\sqrt{(x^2+1)}-\frac{2xy}{\sqrt{(x^2+1)}}=4\sqrt{y}$$
$$\frac{y'\sqrt{(x^2+1)}-\frac{2xy}{\sqrt{(x^2+1)}}}{x^2+1}= \frac{4\sqrt{y}}{x^2+1}$$
I tried to solve this but I can't get rid of the 2 which would have enabled me to wrtie $$(\frac{y}{x^2+1})'$$
how do I solve this?
It's Bernouilli's equation $$y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$$ do this rather $$y'y^{-1/2}(x^2+1)-2xy^{1/2}=4\sqrt{(x^2+1)}$$ $$2(y^{1/2})'(x^2+1)-2x(y^{1/2})=4\sqrt{(x^2+1)}$$ Substitute $z=y^{1/2}$ $$z'(x^2+1)-xz=2\sqrt{(x^2+1)}$$ Now you can solve it easily...
Edit $$(\frac{y}{x^2+1})'= \frac{4\sqrt{y}}{x^2+1}$$ You could do this starting from where you were stuck $$(\frac{y}{x^2+1})'= 4\sqrt{\frac{{y}}{x^2+1}}\frac{1}{\sqrt{x^2+1}}$$ $$\frac {d(\frac{y}{x^2+1})}{\sqrt{\frac{{y}}{x^2+1}}}= \frac{4dx}{\sqrt{x^2+1}}$$
And simply integrate... $$\sqrt{\frac{{y}}{x^2+1}}= 2 \int \frac{dx}{\sqrt{x^2+1}}$$
But Bernouilli's method is easier