Find the curve whose gradient is given by dy/dx=xy and which passes through the point (0,3)

First "Separate the Variables" by rearranging the equation to get the ys on the LHS and the xs on the RHS:

(1/y) dy=x dx

Now Integrate:

Integral(1/y) dy = Integral(x) dx

ln(y)=x²/2 + constant of integration (c)

Rearrange to get y=:

e^(lny)=e^(x2/2)+c

y=e^{(x^2/2)+c} = e^c * e^{x^2/2} = Ae^{0.5x^2}

This is your GENERAL SOLUTION (GS)

Now plug in the coordinates:

3=Ae^0.50=A1=A

A=3

So:

y=3e^{0.5x^2}

This is the PARTICUAR SOLUTION (PS) and also the answer to original question