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