Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 18

Given,
x^3y' + 2y = e^(1/x^2) and to find the particular solution of differential equation at y(1) = e.
so proceeding further , we get.
x^3 y' + 2y = e^(1/x^2)
=>y' + 2y/(x^3) = e^(1/x^2) /x^3

so , the equation is linear in y
and is of the form
y' +p(x)y=q(x)
so the general solution is given as
y*(I.F)= int q(x) * I.F dx+c
where I.F (integrating factor ) = e^(int p(x) dx)
on comparing we get ,
p(x) = 2/x^3 and q(x) = e^(1/x^2) /x^3
so ,
I.F = e^(int (2/x^3) dx) = e^(2 (x^-3+1 )/ -2) = e^(-(x^-2))
so ,
y (e^(-(x^-2)))= int (e^(1/x^2) /x^3) * (e^(-(x^-2))) dx+c
=>y (e^(-(x^-2)))= int (x^-3) dx+c
=>y (e^(-(x^-2)))= x^((-3+1)/ -2)+c
=> y (e^(-(x^-2)))= x^-2/ -2+c
=> y = (- (x^-2)/2+c)/(e^(-(x^-2)))
= e^((x^-2)) *(c-(x^-2)/2 )
so , now to find the particular soultion at y(1) =e , we have to do as follows
y(x) = e^((x^-2)) *(c-(x^-2)/2 )
=> y(1) = e^((1^-2) ) *(c-(1^-2)/2 )
=> e= (e ) *(c-(1)/2 )
=> 1= c-1/2
=> c= 3/2
so the particular solution is
y= ((e^((x^-2))) ) *(3/2-(x^-2)/2 )
=e^((x^-2)) *((3-(x^-2))/2 )