Solve \begin {align*} y^{\prime } & =2y+3e^{2x}\\ y^{\prime } & =\omega \left ( x,y\right ) \end {align*}
From the lookup table, since this is linear ode \(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) \) then \begin {align*} \xi & =0\\ \eta & =e^{\int fdx}\\ & =e^{\int 2dx}\\ & =e^{2x}. \end {align*}
If we were to use the integrating factor method, then \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{e^{2x}}\\ & =e^{-2x} \end {align*}
Then the general solution is \begin {align*} \int \mu \left ( x,y\right ) \left ( dy-\omega dx\right ) & =c_{1}\\ \int e^{-2x}\left ( dy-\left ( 2y+3e^{2x}\right ) dx\right ) & =c_{1}\\ \int e^{-2x}dy-\left ( 2ye^{-2x}+3\right ) dx & =c_{1}\\ \int e^{-2x}dy-2ye^{-2x}dx & =\int 3dx+c_{1}\\ \int d\left ( e^{-2x}y\right ) & =\int 3dx+c_{1} \end {align*}
Hence\begin {align*} e^{-2x}y & =3x+c_{1}\\ y & =e^{2x}\left ( 3x+c_{1}\right ) \end {align*}
But if we were to use the basic Lie symmetry method, then the next step is to determine the canonical coordinates \(R,S\). This is done by using the standard characteristic equation by writing\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end {equation} Since \(\xi =0\) then this is the special case where \(R=x\). And \(S\) is found from \[ dS=\frac {dy}{\eta }=e^{-2x}dy \] Integrating gives\begin {align*} S & =e^{-2x}y+c_{1}\\ & =e^{-2x}y \end {align*}
By choosing \(c_{1}=0\). Now the ODE \(\frac {dS}{dR}=F\left ( R\right ) \) is found from\begin {align*} \frac {dS}{dR} & =\frac {\frac {dS}{dx}+\omega \left ( x,y\right ) \frac {dS}{dy}}{\frac {dR}{dx}+\omega \left ( x,y\right ) \frac {dR}{dy}}\\ & =\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}} \end {align*}
But \(S_{x}=-2e^{-2x}y,R_{x}=1,S_{y}=e^{-2x},R_{y}=0\). Substituting these into the above and simplifying gives\begin {align*} \frac {dS}{dR} & =-2e^{-2x}y+\left ( 2y+3e^{2x}\right ) e^{-2x}\\ & =-2e^{-2x}y+2ye^{-2x}+3\\ & =3 \end {align*}
Which is a quadrature. Solving gives\begin {align*} \int dS & =\int 3dR\\ S & =3R+c_{1} \end {align*}
Converting back to \(x,y\) gives\begin {align*} e^{-2x}y & =3x+c_{1}\\ y & =\left ( 3x+c_{1}\right ) e^{2x} \end {align*}
Of course, this ode is first order linear and can be solved much easier using integrating factor method. But this is just to illustrate the Lie symmetry method.