Differential Equations

Separating variables: $\frac{dy}{y} = 2x dx \implies \ln y = x^2 + C \implies y(x) = C e^{x^2}$. With $y(0) = 3$, $C = 3$. At $x = 1$, $y(1) = 3 e^1$. Hence $k = 1$.