Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 22

Given y'+(2x-1)y=0
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'+(2x-1)y=0--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = (2x-1) and q(x)=0
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 (2x-1) dx) *(0)) dx +c)/e^(int (2x-1) dx)
first we shall solve
e^(int (2x-1) dx)=e^(x^2 -x)
so
proceeding further, we get
y(x) =((int e^(int (2x-1) dx) *(0)) dx +c)/e^(int (2x-1) dx)
= ((int e^(x^2 -x) *(0)) dx +c)/(e^(x^2 -x) )
=0+c/e^(x^2 -x) = e^(x-x^2+c )
y(x) =e^(x-x^2+c)
to find the particular differential equation we have
y(1)=2
=> y(1)=e^(1-1^2+c)
=> e^c =2
=> c = ln(2)
y(x) = e^(x-x^2+ln(2))
y(x) = e^ln2e^(x-x^2)
So,
y(x) = 2e^(x-x^2)