With reduction of order, we assume a solution of the form [tex]y_2=zy_1=ze^{2x}[/tex], with [tex]z=z(x)[/tex]. Then
[tex]{y_2}'=(z'+2z)e^{2x}[/tex]
[tex]{y_2}''=(z''+4z'+4z)e^{2x}[/tex]
and substituting into the ODE gives
[tex]x(z''+4z'+4z)e^{2x}-(2x+1)(z'+2z)e^{2x}+2ze^{2x}=0[/tex]
[tex]x(z''+4z'+4z)-(2x+1)(z'+2z)+2z=0[/tex]
[tex]xz''+(2x-1)z'=0[/tex]
Let [tex]\xi(x)=z'(x)[/tex], so that [tex]\xi'=z''[/tex]. This gives the linear ODE
[tex]x\xi'+(2x-1)\xi=0[/tex]
This equation is also separable, so you can write
[tex]\dfrac{\xi'}{\xi}=\dfrac{1-2x}x[/tex]
Integrating both sides with respect to [tex]x[/tex] gives
[tex]\ln|\xi|=-2x+\ln x+C_1[/tex]
[tex]\xi=C_1xe^{-2x}[/tex]
Next, solve [tex]z'=\xi[/tex] for [tex]z[/tex] by integrating both sides again with respect to [tex]x[/tex].
[tex]z'=\xi[/tex]
[tex]\implies z=\displaystyle\int C_1xe^{-2x}\,\mathrm dx[/tex]
[tex]\implies z=C_1e^{-2x}(2x+1)+C_2[/tex]
And finally, solve for [tex]y_2[/tex].
[tex]y_2=zy_1=C_1(2x+1)+C_2e^{2x}[/tex]
and note that [tex]y_1[/tex] is already taken into account as part of [tex]y_2[/tex], so this is the general solution to the ODE.
