\[ x y''(x)+2 y'(x)-x y(x)-e^x=0 \] ✓ Mathematica : cpu = 0.0302767 (sec), leaf count = 37
\[\left \{\left \{y(x)\to \frac {e^{-x} \left (e^{2 x} \left (2 c_2+2 x-1\right )+4 c_1\right )}{4 x}\right \}\right \}\]
✓ Maple : cpu = 0.039 (sec), leaf count = 23
\[ \left \{ y \left ( x \right ) ={\frac {\sinh \left ( x \right ) {\it \_C2}}{x}}+{\frac {\cosh \left ( x \right ) {\it \_C1}}{x}}+{\frac {{{\rm e}^{x}}}{2}} \right \} \]
\begin {equation} xy^{\prime \prime }+2y^{\prime }-xy=e^{x}\tag {1} \end {equation}
First method, much shorter, using transformation. Let \(y_{h}=\frac {u\left ( x\right ) }{x}\), hence (now we are solving only the homogeneous part).
\begin {align*} y^{\prime } & =\frac {u^{\prime }}{x}-\frac {u}{x^{2}}\\ y^{\prime \prime } & =\frac {u^{\prime \prime }}{x}-\frac {u^{\prime }}{x^{2}}-\frac {u^{\prime }}{x^{2}}+2\frac {u}{x^{3}} \end {align*}
And (1) becomes
\begin {align*} x\left ( \frac {u^{\prime \prime }}{x}-\frac {u^{\prime }}{x^{2}}-\frac {u^{\prime }}{x^{2}}+2\frac {u}{x^{3}}\right ) +2\left ( \frac {u^{\prime }}{x}-\frac {u}{x^{2}}\right ) -x\left ( \frac {u}{x}\right ) & =0\\ u^{\prime \prime }-2\frac {u^{\prime }}{x}+2\frac {u}{x^{2}}+\frac {2u^{\prime }}{x}-\frac {2u}{x^{2}}-u & =0\\ u^{\prime \prime }-u & =0 \end {align*}
Hence the roots of the characteristic equation are \(\pm 1\) and the solution is
\[ u=Ae^{x}+Be^{-x}\]
Hence
\[ y_{h}=\frac {1}{x}\left ( Ae^{x}+Be^{-x}\right ) \]
The particular solution is found below, and given in the second method. The transformation method is much simpler.
Second method, much longer, using series method. This is used if a transformation is now known or can not be found. There is singularity at \(x=0\). We need to check if it regular or not. Writing in standard form \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0\) gives (we are looking at the homogeneous part now only)\[ y^{\prime \prime }+\frac {2}{x}y^{\prime }-y=0 \]
Hence \(\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x\frac {2}{x}=2\) which is analytic at \(x=0\). And \(\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}-x^{2}=0\) which is analytic. Hence the singularity is regular (removable). Using Frobenius series, assume that \[ y=\sum _{n=-\infty }^{\infty }c_{n}x^{n+r}\] Where \(c_{n}=0\) for \(n<0\). Hence\begin {align*} y^{\prime } & =\sum \left ( n+r\right ) c_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum \left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2} \end {align*}
Substituting back in the original ODE gives\[ \sum \left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-1}+\sum 2\left ( n+r\right ) c_{n}x^{n+r-1}-\sum c_{n}x^{n+r+1}=0 \] Adjusting so that all have same power \(x^{n+r}\) gives\[ \sum \left ( n+r+1\right ) \left ( n+r\right ) c_{n+1}x^{n+r}+\sum 2\left ( n+r+1\right ) c_{n+1}x^{n+r}-\sum c_{n-1}x^{n+r}=0 \] Hence\begin {align} \left ( n+r+1\right ) \left ( n+r\right ) c_{n+1}+2\left ( n+r+1\right ) c_{n+1}-c_{n-1} & =0\nonumber \\ \left ( n+r+1\right ) \left ( 2+\left ( n+r\right ) \right ) c_{n+1}-c_{n-1} & =0\tag {2} \end {align}
We want equation with \(c_{0}\) in it. Hence let \(n=-1\)\[ \left ( -1+r+1\right ) \left ( 2+\left ( -1+r\right ) \right ) c_{0}-c_{-2}=0 \] But \(c_{n}=0\) for all \(n<0\) hence\[ \left ( -1+r+1\right ) \left ( 2+\left ( -1+r\right ) \right ) c_{0}=0 \] But \(c_{0}\neq 0\), as this is the basis for this method. Therefore, we obtain the indicial equation for \(r\)\begin {align*} \left ( -1+r+1\right ) \left ( 2+\left ( -1+r\right ) \right ) & =0\\ r\left ( r+1\right ) & =0 \end {align*}
Hence \(r=0\) or \(r=-1\) are the roots. Now for each \(r\) we find a solution. Using \(r=0\), we go back the recurrence equation (2)\begin {align*} \left ( n+1\right ) \left ( 2+n\right ) c_{n+1}-c_{n-1} & =0\\ c_{n+1} & =\frac {c_{n-1}}{\left ( n+1\right ) \left ( 2+n\right ) } \end {align*}
For \(n=0\)\[ c_{1}=\frac {c_{-1}}{\left ( n+1\right ) \left ( 2+n\right ) }=0 \] For \(n=1\)\[ c_{2}=\frac {c_{0}}{\left ( 2\right ) \left ( 3\right ) }\] For \(n=2\)\[ c_{3}=\frac {c_{1}}{\left ( n+1\right ) \left ( 2+n\right ) }=0 \] For \(n=3\)\[ c_{4}=\frac {c_{2}}{\left ( 4\right ) \left ( 5\right ) }=\frac {c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 4\right ) \left ( 5\right ) }\] And so on. Hence, for \(r=0\) we have\begin {align} y_{r=0} & =\sum _{n=0}^{\infty }c_{n}x^{n}=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3}+\cdots \nonumber \\ & =c_{0}+\frac {c_{0}}{6}x^{2}+\frac {c_{0}}{120}x^{4}+\cdots \nonumber \\ & =A\left ( 1+\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) \tag {3} \end {align}
Where \(A\) is used as arbitrary constant instead of \(a_{0}\). Now we find the solution for \(r=-1.\) we go back the recurrence equation (2)\begin {align*} \left ( n-1+1\right ) \left ( 2+\left ( n-1\right ) \right ) c_{n+1}-c_{n-1} & =0\\ n\left ( 1+n\right ) c_{n+1}-c_{n-1} & =0\\ c_{n+1} & =\frac {c_{n-1}}{n\left ( 1+n\right ) } \end {align*}
For \(n=0\)\[ c_{1}=\frac {c_{-1}}{n\left ( 1+n\right ) }=0 \] For \(n=1\)\[ c_{2}=\frac {c_{0}}{2}\] For \(n=2\)\[ c_{3}=\frac {c_{1}}{n\left ( 1+n\right ) }=0 \] For \(n=3\)\[ c_{4}=\frac {c_{2}}{\left ( 3\right ) \left ( 4\right ) }=\frac {c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 4\right ) }\] For \(n=4\)\[ c_{5}=\frac {c_{3}}{n\left ( 1+n\right ) }=0 \] For \(n=5\)\[ c_{6}=\frac {c_{4}}{\left ( 5\right ) \left ( 6\right ) }=\frac {c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 4\right ) \left ( 5\right ) \left ( 6\right ) }\] And so on. Hence solution is\begin {align*} y_{r=-1} & =\frac {1}{x}\sum _{n=0}^{\infty }c_{n}x^{n}=\frac {1}{x}\left ( c_{0}+\frac {c_{0}}{2}x^{2}+\frac {c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 4\right ) }x^{4}+\frac {c_{0}}{\left ( 2\right ) \left ( 3\right ) \left ( 4\right ) \left ( 5\right ) \left ( 6\right ) }x^{6}+\cdots \right ) \\ & =\frac {B}{x}\left ( 1+\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\frac {1}{720}x^{6}+\cdots \right ) \end {align*}
Where \(B\) is used as arbitrary constant instead of \(a_{0}\). Therefore, the homogeneous solution found is\begin {align*} y_{h} & =y_{r=0}+y_{r=-1}\\ & =A\left ( 1+\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) +\frac {B}{x}\left ( 1+\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\frac {1}{720}x^{6}+\cdots \right ) \end {align*}
But \begin {equation} e^{x}=1+x+\frac {1}{2}x^{2}+\frac {1}{6}x^{3}+\frac {1}{24}x^{4}+\frac {1}{120}x^{5}+\cdots \tag {3} \end {equation} And \begin {equation} e^{-x}=1-x+\frac {1}{2}x^{2}-\frac {1}{6}x^{3}+\frac {1}{24}x^{4}-\frac {1}{120}x^{5}+\cdots \tag {4} \end {equation} Hence adding (3)+(4) gives\begin {align*} e^{x}+e^{-x} & =2+2\frac {1}{2}x^{2}+2\frac {1}{24}x^{4}+\cdots \\ & =2\left ( 1+\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\frac {1}{720}x^{6}\cdots \right ) \end {align*}
But \(y_{r=-1}=\frac {B}{x}\left ( 1+\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\frac {1}{720}x^{6}+\cdots \right ) \), therefore comparing the result we found above, we see that we can write \(y_{r=-1}\) as\[ y_{r=-1}=\frac {B}{x}\left ( \frac {e^{x}+e^{-x}}{2}\right ) \] Similarly, we obtain \(y_{r=0}\) expression\begin {align} \frac {1}{x}e^{x} & =\frac {1}{x}\left ( 1+x+\frac {1}{2}x^{2}+\frac {1}{6}x^{3}+\frac {1}{24}x^{4}+\frac {1}{120}x^{5}+\cdots \right ) \nonumber \\ & =\frac {1}{x}+1+\frac {1}{2}x+\frac {1}{6}x^{2}+\frac {1}{24}x^{3}+\frac {1}{120}x^{4}+\cdots \tag {3A} \end {align}
And\begin {align} \frac {1}{x}e^{-x} & =\frac {1}{x}\left ( 1-x+\frac {1}{2}x^{2}-\frac {1}{6}x^{3}+\frac {1}{24}x^{4}-\frac {1}{120}x^{5}+\cdots \right ) \nonumber \\ & =\frac {1}{x}-1+\frac {1}{2}x-\frac {1}{6}x^{2}+\frac {1}{24}x^{3}-\frac {1}{120}x^{4}+\cdots \tag {4A} \end {align}
Now (3A)-(4A) gives\begin {align*} \frac {1}{x}e^{x}-\frac {1}{x}e^{-x} & =\left ( \frac {1}{x}+1+\frac {1}{2}x+\frac {1}{6}x^{2}+\frac {1}{24}x^{3}+\frac {1}{120}x^{4}+\cdots \right ) -\left ( \frac {1}{x}-1+\frac {1}{2}x-\frac {1}{6}x^{2}+\frac {1}{24}x^{3}-\frac {1}{120}x^{4}+\cdots \right ) \\ & =2+2\frac {1}{6}x^{2}+2\frac {1}{120}x^{4}+\cdots \\ & =2\left ( 1+\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) \end {align*}
Hence\[ \left ( 1+\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) =\frac {1}{2x}\left ( e^{x}-e^{-x}\right ) \] But \(y_{r=0}=A\left ( 1+\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) \), therefore comparing the result we found above, wee see that we can write \(y_{r=0}\) as\begin {align*} y_{r=0} & =A\left ( \frac {1}{2x}\left ( e^{x}-e^{-x}\right ) \right ) \\ & =\frac {A}{2x}\left ( e^{x}-e^{-x}\right ) \end {align*}
Therefore\begin {align*} y_{h} & =y_{r=0}+y_{r=-1}\\ & =\frac {A}{2x}\left ( e^{x}-e^{-x}\right ) +\frac {B}{2x}\left ( e^{x}+e^{-x}\right ) \\ & =\frac {1}{x}\left ( \frac {A}{2}e^{x}-\frac {A}{2}e^{-x}+\frac {B}{2}e^{x}+\frac {B}{2}e^{-x}\right ) \\ & =\frac {1}{x}\left ( e^{x}\left ( \frac {A}{2}+\frac {B}{2}\right ) +e^{-x}\left ( -\frac {A}{2}+\frac {B}{2}\right ) \right ) \end {align*}
Let \(\frac {A+B}{2}=A_{0},\frac {B-A}{2}=B_{0}\) hence the above becomes\[ y_{h}=\frac {1}{x}\left ( A_{0}e^{x}+B_{0}e^{-x}\right ) \]
We see this is the same solution using the transformation method given above. We now need to find particular solution. Let \(y_{1}=\frac {e^{x}}{x},y_{2}=\frac {e^{-x}}{x}\), hence \(y_{1}^{\prime }=\frac {e^{x}}{x}-\frac {e^{x}}{x^{2}}\) and \(y_{2}^{\prime }=\frac {-e^{-x}}{x^{2}}-\frac {e^{-x}}{x}\), hence the Wronskian is\begin {align*} W & =\begin {vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end {vmatrix} \\ & =\begin {vmatrix} \frac {e^{x}}{x} & \frac {e^{-x}}{x}\\ \frac {e^{x}}{x}-\frac {e^{x}}{x^{2}} & \frac {-e^{-x}}{x^{2}}-\frac {e^{-x}}{x}\end {vmatrix} \\ & =\frac {e^{x}}{x}\left ( \frac {-e^{-x}}{x^{2}}-\frac {e^{-x}}{x}\right ) -\frac {e^{-x}}{x}\left ( \frac {e^{x}}{x}-\frac {e^{x}}{x^{2}}\right ) \\ & =\left ( \frac {-1}{x^{3}}-\frac {1}{x^{2}}\right ) -\left ( \frac {1}{x^{2}}-\frac {1}{x^{3}}\right ) \\ & =-\frac {2}{x^{2}} \end {align*}
Therefore, let \(y_{p}=u_{1}y_{1}+u_{2}y_{2}\) and hence\[ u_{1}=-\int \frac {y_{2}}{aW}e^{x}dx \] Where \(a=x\) since the original ODE is \(xy^{\prime \prime }+2y^{\prime }-xy=e^{x}\), and \(a\) is the coefficient of \(y^{\prime \prime }\) always. Hence the above becomes\[ u_{1}=-\int \frac {\frac {e^{-x}}{x}}{x\left ( -\frac {2}{x^{2}}\right ) }e^{x}dx=\int \frac {e^{-x}}{2}e^{x}dx=\int \frac {1}{2}dx=\frac {x}{2}\] And\[ u_{2}=\int \frac {\frac {e^{x}}{x}}{x\left ( -\frac {2}{x^{2}}\right ) }e^{x}dx=-\int \frac {\frac {e^{x}}{x}}{\frac {2}{x}}e^{x}dx=-\frac {1}{2}\int e^{2x}dx=-\frac {1}{2}\frac {e^{2x}}{2}=\frac {-1}{4}e^{2x}\] Hence \begin {align*} y_{p} & =u_{1}y_{1}+u_{2}y_{2}\\ & =\frac {x}{2}\frac {e^{x}}{x}-\frac {1}{4}e^{2x}\frac {e^{-x}}{x}\\ & =\frac {1}{2}e^{x}-\frac {1}{4x}e^{x} \end {align*}
Therefore \[ y_{p}=e^{x}\left ( \frac {1}{2}-\frac {1}{4x}\right ) \] Hence the general solution is\begin {align*} y & =y_{h}+y_{p}\\ & =\frac {1}{x}\left ( A_{0}e^{x}+B_{0}e^{-x}\right ) +e^{x}\left ( \frac {1}{2}-\frac {1}{4x}\right ) \end {align*}
Verification
restart; ode:=x*diff(diff(y(x),x),x)+2*diff(y(x),x)-x*y(x)=exp(x); y0:=1/x*( _C1* exp(x)+ _C2*exp(-x))+ (1/2-1/(4*x))*exp(x); odetest(y(x)=y0,ode); 0