\[ y^{\prime }=\frac {-6x+y-3}{2x-y-1}\] Comparing to \(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) shows that \(a_{1}=-6,b_{1}=1,a_{2}=2,b_{2}=-1\). Hence \(\frac {a_{1}}{b_{1}}=-6,\frac {a_{2}}{b_{2}}=-2\). This shows the lines are not parallel. Let\begin {align*} X & =x-x_{0}\\ Y & =y-y_{0} \end {align*}
The constant \(x_{0},y_{0}\) are found by solving \begin {align*} a_{1}x_{0}+b_{1}y_{0}+c_{1} & =0\\ a_{2}x_{0}+b_{2}y_{0}+c_{2} & =0 \end {align*}
Or\begin {align*} -6x_{0}+y_{0}-3 & =0\\ 2x_{0}-y_{0}-1 & =0 \end {align*}
Solving for \(x_{0},y_{0}\) gives \begin {align*} x_{0} & =-1\\ y_{0} & =-3 \end {align*}
Hence\begin {align*} X & =x+1\\ Y & =y+3 \end {align*}
Using this transformation in \(y^{\prime }=\frac {-6x+y-3}{2x-y-1}\) results in the ode\[ \frac {dY}{dX}=\frac {6X-Y}{-2X+Y}\] This is a homogeneous ode\[ \frac {dY}{dX}=\frac {6-\frac {Y}{X}}{-2+\frac {Y}{Y}}\] Let \(u=\frac {Y}{X}\). Now it is solved as was shown in the above sections. At the end, \(Y\) is replaced by \(y-y_{0}\) to obtain the solution in \(y\left ( x\right ) \).