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* ex^2/2 = Ae0.5x^2

This is your GENERAL SOLUTION (GS)

Now plug in the coordinates:

3=Ae0.50=A1=A

A=3

So:

y=3e0.5x^2

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

CC
Answered by Christian C. Maths tutor

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