\[ x y''(x)-(x+1) y'(x)+y(x)=0 \] ✓ Mathematica : cpu = 0.0158896 (sec), leaf count = 20
\[\left \{\left \{y(x)\to c_1 e^x+c_2 (-x-1)\right \}\right \}\] ✓ Maple : cpu = 0.021 (sec), leaf count = 13
\[y \relax (x ) = c_{2} {\mathrm e}^{x}+c_{1} x +c_{1}\]
Hand solution
\begin {equation} xy^{\prime \prime }-\left (x+1\right ) y^{\prime }+y=0\tag {1} \end {equation}
Taking Laplace transform of each term and using property of \(\mathcal {L}\left (xf\relax (x) \right ) =-\frac {d}{ds}F\relax (s) \) where \(F\relax (s) =\mathcal {L}f\relax (x) \), then\[\mathcal {L}\left (xy^{\prime \prime }\right ) =-\frac {d}{ds}\left ( \mathcal {L}y^{\prime \prime }\right ) \]
Let \(\mathcal {L}y\relax (x) =Y\relax (s) \equiv Y\). Now \(\mathcal {L}y^{\prime \prime }=s^{2}Y-sy\relax (0) -y^{\prime }\relax (0) \). Assuming \(y\relax (0) =A,y^{\prime }\relax (0) =B\) then\begin {align*} \mathcal {L}\left (xy^{\prime \prime }\right ) & =-\frac {d}{ds}\left (s^{2}Y-As-B\right ) \\ & =-\left (2sY+s^{2}Y^{\prime }-A\right ) \end {align*}
And \begin {align*} \mathcal {L}\left (\left (x+1\right ) y^{\prime }\right ) & =\mathcal {L}\left (xy^{\prime }+y^{\prime }\right ) \\ & =-\frac {d}{ds}\left ( \mathcal {L}y^{\prime }\right ) +\mathcal {L}y^{\prime }\\ & =-\frac {d}{ds}\left (sY-y\relax (0) \right ) +\left (sY-y\left ( 0\right ) \right ) \\ & =-\frac {d}{ds}\left (sY-A\right ) +\left (sY-y\relax (0) \right ) \\ & =-\left (Y+sY^{\prime }\right ) +\left (sY-A\right ) \\ & =-Y-sY^{\prime }+sY-A \end {align*}
Hence Laplace transform of the ODE becomes\begin {align*} -\left (2sY+s^{2}Y^{\prime }-A\right ) -\left (-Y-sY^{\prime }+sY-A\right ) +Y & =0\\ -2sY-s^{2}Y^{\prime }+A+Y+sY^{\prime }-sY+A+Y & =0\\ Y^{\prime }\left (s-s^{2}\right ) +Y\left (-2s+1-s+1\right ) & =-2A\\ Y^{\prime }\left (s^{2}-s\right ) +Y\left (3s-2\right ) & =2A\\ Y^{\prime }+\frac {\left (3s-2\right ) }{s\left (s-1\right ) }Y & =\frac {2A}{s\left (s-1\right ) } \end {align*}
The integrating factor is \(\mu =e^{\int \frac {\left (3s-2\right ) }{s\left ( s-1\right ) }}=e^{\ln \left (s-1\right ) +2\ln \relax (s) }=\left ( s-1\right ) s^{2}\), hence\begin {align*} d\left (\left (s-1\right ) s^{2}Y\right ) & =\left (s-1\right ) s^{2}\frac {2A}{s\left (s-1\right ) }\\ \left (s-1\right ) s^{2}Y & =2A\int sds+c_{1}\\ \left (s-1\right ) s^{2}Y & =2A\frac {s^{2}}{2}+c_{1}\\ Y & =\frac {As^{2}+c_{1}}{\left (s-1\right ) s^{2}} \end {align*}
Inverse Laplace transform gives\begin {align*} y\relax (x) & =-c_{1}+\left (A+c_{1}\right ) e^{x}-c_{1}x\\ & =-c_{1}\left (1+x\right ) +\left (A+c_{1}\right ) e^{x} \end {align*}
Let \(-c_{1}=A_{0},A+c_{1}+B_{0}\), hence\[ y\relax (x) =A_{0}\left (1+x\right ) +B_{0}e^{x}\]
Verification