f'(x) = 3f(x)
lets say y = f(x) for funsies
y' = 3y
(dy/dx) = 3y
dy/(3y) = dx
Integrate with respect to both sides...
right side = X+ C
Left side, , it equals (1/3)*(lny) (technically absolute value of y, but whatevs)
so
ln(y) = 3x + 3c ... which we can also say 3x + C, since C is an arbitrary constant
ln(y) = 3x+C
y = e^(3x+C)
y = (e^(3x))(e^C)
e^C is a constant, so we can say that's C
so...
y = Ce^(3x)