assuming a typo, i think you have
#y' = y + 2xe^(2x)#
that's not separable so we write it in this form
#y' - y = 2xe^(2x)#
OR
#y' + (-1)* y = 2xe^(2x)#
and we use an integrating factor #eta(x)# to make the equation exact on the LHS
#eta(x) = e^(int \ (-1) dx) = e^(-x)#, note the pattern
Now multiply every term by #eta#
#e^(-x)y' - e^(-x) y = 2xe^(x)#
and magically the LHS is now very useful
#d/dx (y e^(-x)) = 2xe^(x)#
so
#y e^(-x) = int \ 2xe^(x) \ dx#
time for a spot of IBP
#= int \2x d/dx( e^(x)) \ dx#
#= 2x e^x - int \ d/dx(2x) \ e^(x) \ dx#
#= 2x e^x - int \ 2 \ e^(x) \ dx#
#implies y e^(-x) = 2x e^x - 2 e^(x) +C#
# y e^(-x) = 2 e^x (x -1) +C#
# y = 2e^(2x) (x -1) +C e^x#
The theory is as follows
For an equation in form
#y' + f(x) y = g(x)#
using our mysterious integrating factor #eta (x)#, we have
#eta(x) y' +color(red)(eta(x) f(x) y) =eta(x) g(x)#
we then think wouldn't it have handy if the stuff in red was equal to #eta'(x)#
because it would then be a case of
#eta(x) y' +color(red)(eta(x) f(x)) y =eta(x) g(x)#
#implies eta(x) y' +color(red)(eta'(x)) y =eta(x) g(x)#
and by the product rule
#implies d/dx (eta(x) y ) =eta(x) g(x)#
and so
#eta(x) y = int \ eta(x) g(x) \ dx#
so we want
#eta'(x) = eta(x) f(x)#
we can separate the variables here
#(eta'(x))/ (eta(x)) = f(x)#
and integrate
#int \ (eta'(x))/ (eta(x)) \ dx =int f(x) \ dx#
#ln( eta(x) ) =int f(x) \ dx#
#eta(x) =e^ (int f(x) \ dx)#
