Find an equation of the curve that passes through the point (0,1) and whose slope at (x,y) is xy.

If the slope at the point [tex](x,y)[/tex] is [tex]xy[/tex], then we have

[tex]\displaystyle\frac{dy}{dx} = xy\ \Rightarrow\ \frac{dy}{y} = x\, dx\ [y \ne 0] \Rightarrow \int \frac{dy}{y} = \int x \, dx \ \Rightarrow \\ \\ \textstyle \ln|y| = \frac{1}{2}x^2 + C.\ y(0) = 1\ \Rightarrow\ \ln 1 = 0 + C\ \Rightarrow\ C = 0. \\ \\ \text{Thus } |y| = e^{x^2 / 2}\ \Rightarrow\ y = \pm e^{x^2 / 2},\ \text{so } y = e^{x^2/2} \text{ since } y(0) = 1 \ \textgreater \ 0. \\ \\ \text{Note that $y=0$ is not a solution because it does not satisfy} \\ \text{the initial condition $y(0) = 1$}.[/tex]