I have to solve the following problem:
$y'=\frac{y \cos(x)}{(1+2y^2)}$
with the initial condition $y(0)=1$. I came up with the following equation:
$y^2(x)+\log(y(x))=\sin(x)+c_1$.
It is the first time I get such a solution and now I am not able to solve for $y(x)$. I used WolframAlpha Step by Step Solver but I really do not understand how the get $y(x)=...$. Can I solve the IVP without explicitly solve for $y(x)$? Thank you.
Your solution is correct. The next step can be to determine the constant $c_1$ by initial condition, that gives: $$ 1^2+\log (1)=\sin(0)+c_1 \Rightarrow c_1=1 $$
if you want find and explicit form for $y(x)$ this can be done only using the Lambert W function. Let $t=y^2$ and write the exponential of your equation as: $$ \sqrt{t}e^t=e^{\sin x +1} $$ squaring: $$ te^{2t}=e^{2(\sin x +1)} \iff 2te^{2t}=2e^{2(\sin x +1)} $$ so, By definition of the W function, $$ 2t=W\left(2e^{2(\sin x +1)} \right) \iff t=\frac{1}{2}W\left(2e^{2(\sin x +1)} \right) $$ and substituting for $y$: $$ y=\sqrt{\frac{W\left(2e^{2(\sin x +1)} \right)}{2}} $$