Separate the variables:
y' = dy/dx = xy ⇒ 1/y dy = x dx
Integrate both sides:
∫ 1/y dy = ∫ x dx
ln|y| = 1/2 x² + C
Given that y(0) = 1, we have
ln|1| = 1/2 • 0² + C ⇒ C = 0
so that the particular solution is
ln|y| = 1/2 x²
Solving for y gives
y = exp(1/2 x²)
