Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 24

Given 2xy' - y = x^3 - x
=>y' -(1/2x)y = (x^2-1)/2
when the first order linear ordinary differential equation has the form of
y'+p(x)y=q(x)
then the general solution is ,
y(x) = (int (e^(int p(x) dx) * q(x) dx + c )) / e^(int p(x) dx)
so,
y' -y/(2x) = (x^2-1)/2--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = -1/(2x) and q(x)=(x^2-1)/2
so on solving with the above general solution we get:
y(x) = (int (e^(int p(x) dx) * q(x) dx + c )) / e^(int p(x) dx)
=((int e^(int -1/(2x) dx) *((x^2-1)/2)) dx +c)/ e^(int(-1/2x) dx)
first we shall solve
e^(int -1/(2x) dx)=e^(-ln(2x)/2)=1/sqrt(2x)
so
proceeding further, we get
y(x) =(int 1/sqrt(2x) *((x^2-1)/2) dx +c)/(1/sqrt(2x) )
y(x) =(1/(2*sqrt(2))(int 1/sqrt(x) *((x^2-1)) dx +c)/(1/sqrt(2x)) )
=(1/(2*sqrt(2))(int1/sqrt(x) *((x^2-1)) dx +c)/(1/sqrt(2x) ))
=(1/(2*sqrt(2))(int1/sqrt(x) *(x^2)dx-int1/sqrt(x) *1 dx +c)/(1/sqrt(2x) ))
=((1/(2*sqrt(2))(2x^(5/2)/5-2sqrt(x)) +c)/(1/sqrt(2x) ))
now to find the particular solution of differential equation we have y(4)=2
so we can find the value of c
y(x)=((1/(2*sqrt(2))(2x^(5/2)/5-2sqrt(x)) +c)/(1/sqrt(2x) ))
=>((1/(2*sqrt(2))(2x^(5/2)/5-2sqrt(x)) +c)*(sqrt(2x) ))
=>((1/(2)(2x^(5/2)/5-2sqrt(x)) +c)*(sqrt(x) ))
=>((x^(5/2)/5-sqrt(x) +C)*(sqrt(x) ))
=>((x^(3)/5-(x) +C*sqrt(x))))
=> y(4) =4^(3)/5-(4) +C*sqrt(4)
=> 2 = 64/5-(4) +C*sqrt(4)
=> 2 = 64/5 -4+2c
=> 6- 64/5 =2c
=> -34/5 =2c
=> c= -17/5
y(x)=x^(3)/5-(x) -17/5sqrt(x)