Question 1032274
The differential equation {{{dy/dx=(xy)/3}}} becomes {{{dy/y = (x/3)dx}}}

==> {{{lny = x^2/6 + c}}} for some undetermined constant c.

{{{y[0] = f(0) = 4}}}  ==> lny = 0 + c = c ==>{{{lny = x^2/6 + ln4}}}, or 

{{{ln(y/4) = x^2/6}}}.  (Notice the restriction to the y-values y > 0.)

In exponential form, the solution is {{{y = 4e^(x^2/6)}}}, or {{{y = 4exp(x^2/6)}}}.