Use the differential equation given by dy/dx=xy/3, y > 0. find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 4.

dy/dx=xy/3, y > 0  can be rewritten in terms of only one variable (x or y) each:

3 dy
   = x*dx  (first term is all in y; second term all in x).
  y
                                  x^2
Integrating, 3 ln y = + ln C,  Then ln y^3 = (1/2)x^2 + ln C.
                                     2

Combining the log terms:

ln y^3-ln C = (1/2)x^2.

Given:  if x=0, y=4.  Subst. these v alues into the equation in y given above:
                                                               4^3
ln 4^3 - ln C = (1/2)(0)^2 = 0.   Then ln = 0, which tells us that 
                                                                 C

4^3 = C.  Thus, the solution is    ln y^3 = (1/2)x^2 + ln (4/3).

There are other ways in which you could write this same expression.  You could, for example, solve for either y^3 or y alone.

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