if the graph of y=f(x) contains the point (0,2), dy/dx=-x/ye^(x^2) and f(x)>0 for all x then f(x) = ?
$\displaystyle y'=-\frac{x}{ye^{x^2}} \implies yy' = -\frac{x}{e^{x^2}}$
Integrate both sides. (Use u-substitution with $\displaystyle u=x^2$ and $\displaystyle du=2x\,dx$ on the righthand integral.)
$\displaystyle \int yy'\,dx = -\int\frac{x}{e^{x^2}}\,dx \implies \frac{1}{2}y^2 = \frac{1}{2}e^{-x^2}+c \implies y = \sqrt{e^{-x^2}+C}$
Plug in $\displaystyle (0,2)$: $\displaystyle 2=\sqrt{1+C} \implies C=3$
So $\displaystyle y=f(x)=\sqrt{e^{-x^2}+3}$