Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 65

Given ,
y' - y = e^xroot3(x) ----------(1)
this equation is already in linear form of first order ,so
this is of the form
y' +Py=Q ------------(2)
then the general solution y=((int (I.F) (Q) dx)+c)/(I.F)
where I.F is the integration function = e^(int p dx)

so on comapring the equations (1) and (2) we get,
P =-1 , Q= e^xroot3(x)

so the I.F = e^(int p dx) = e^(int -1 dx)= e^(-x)
now the general solution is
y=((int (I.F) *Q dx )+c)/(I.F)
=>y= ((int (e^(-x)) (e^xroot3(x)) dx)+c)/(e^(-x))
= (int root3(x) dx + c)/e^(-x)
= ((x^(4/3))/(4/3) +c)/(e^(-x))
=((x^(4/3))/(4/3) +c)*e^(x)
is the general solution.