Given $y'=\frac{e^{-y^2}}{y(2x+x^2)}$
What is the best method to solve this equation? I thought write it in seperable notation, obtaining $ \frac{X_1(x)}{X_2(x)}+\frac{Y_2(y)}{Y_1(y)}y'=0,$ which has a solution $\int\frac{X_1(x)}{X_2(x)}dx+\int\frac{Y_2(y)}{Y_1(y)}dy=c$ I defined $X_1=1; \quad X_2=2x+x^2; \quad Y_2=-y; \quad Y_1= e^{y^2}.$ My solution is $\frac 1 2 \log(x)-\log(x+2) -\frac 1 2 e^{y^2}=c.$ Is this the correct solution?