2.2.8 Problem 8

Internal problem ID [8812]
Book : Own collection of miscellaneous problems
Section : section 2.0
Problem number : 8
Date solved : Sunday, March 30, 2025 at 01:40:37 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solved as second order ode using Kovacic algorithm

Time used: 0.580 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }-x{y}^{\prime }-xy-x& =0\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }-x{y}^{\prime }-xy& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =-xC& =-x\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{x^2+4x-2}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =x^2+4x-2\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =(\frac{1}{4}{x}^2+x-\frac{1}{2})z(x)\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.37: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-2\\ & =-2\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $-2$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Attempting to find a solution using case $n=1$.

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $x^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{x}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{x}^i\end{array}$$

Let $a$ be the coefficient of $x^v=x^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx \frac{x}{2}+1-\frac{3}{2x}+\frac{3}{x^2}-\frac{33}{4x^3}+\frac{51}{2x^4}-\frac{339}{4x^5}+\frac{591}{2x^6}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=\frac{1}{2}$$

From Eq. (9) the sum up to $v=1$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{x}^i\\ \text{(10)}& & =\frac{x}{2}+1\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $x^{v-1}=x^0=1$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=\frac{1}{4}{x}^2+x+1$$

This shows that the coefficient of $1$ in the above is $1$. Now we need to find the coefficient of $1$ in $r$. How this is done depends on if $v=0$ or not. Since $v=1$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2+4x-2}{4}\\ & =Q+\frac{R}{4}\\ & =(\frac{1}{4}{x}^2+x-\frac{1}{2})+\left(0\right)\\ & =\frac{1}{4}{x}^2+x-\frac{1}{2}\end{array}$$

We see that the coefficient of the term $\frac{1}{x}$ in the quotient is $-\frac{1}{2}$. Now $b$ can be found.

$$\begin{array}{rl}b& =(-\frac{1}{2})-\left(1\right)\\ & =-\frac{3}{2}\end{array}$$

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\frac{x}{2}+1\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{-\frac{3}{2}}{\frac{1}{2}}-1)& & =-2\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{-\frac{3}{2}}{\frac{1}{2}}-1)& & =1\end{array}$$

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=\frac{1}{4}{x}^2+x-\frac{1}{2}$$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$-2$	$\frac{x}{2}+1$	$-2$	$1$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=1$, and since there are no poles then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}\\ & =1\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{x-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =0+(-)(\frac{x}{2}+1)\\ & =-1-\frac{x}{2}\\ & =-1-\frac{x}{2}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(x)$ of degree $d=1$ to solve the ode. The polynomial $p(x)$ needs to satisfy the equation

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(x)& =x+a_0\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-1-\frac{x}{2})\left(1\right)+((-\frac{1}{2})+{(-1-\frac{x}{2})}^2-(\frac{1}{4}{x}^2+x-\frac{1}{2}))& =0\\ -2+a_0=0\end{array}$$

Solving for the coefficients $a_i$ in the above using method of undetermined coefficients gives

$$\{a_0=2\}$$

Substituting these coefficients in $p(x)$ in eq. (2A) results in

$$\begin{array}{rl}p(x)& =2+x\end{array}$$

Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(x)& =p{e}^{\int \omega dx}\\ & =(2+x){\mathrm{e}}^{\int (-1-\frac{x}{2})dx}\\ & =(2+x){\mathrm{e}}^{-x-\frac{1}{4}{x}^2}\\ & =(2+x){\mathrm{e}}^{-\frac{x(4+x)}{4}}\end{array}$$

The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{-x}{1}dx}\\ & =z_1{e}^{\frac{x^2}{4}}\\ & =z_1\left({\mathrm{e}}^{\frac{x^2}{4}}\right)\end{array}$$

Which simplifies to

$$y_1=(2+x){\mathrm{e}}^{-x}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-x}{1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{\frac{x^2}{2}}}{{\left(y_1\right)}^2}dx\\ & =y_1(-\frac{{\mathrm{e}}^{-2+\frac{{(2+x)}^2}{2}}}{2+x}-\frac{i\sqrt{\pi }{\mathrm{e}}^{-2}\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)}{2})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left((2+x){\mathrm{e}}^{-x}\right)+c_2\left((2+x){\mathrm{e}}^{-x}(-\frac{{\mathrm{e}}^{-2+\frac{{(2+x)}^2}{2}}}{2+x}-\frac{i\sqrt{\pi }{\mathrm{e}}^{-2}\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)}{2})\right)\end{array}$$

This is second order nonhomogeneous ODE. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the nonhomogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. $y_h$ is the solution to

$$y^{\prime \prime }-x{y}^{\prime }-xy=0$$

The homogeneous solution is found using the Kovacic algorithm which results in

$$y_h=c_1(2+x){\mathrm{e}}^{-x}+c_2(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}})$$

The particular solution $y_p$ can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on $x$ as well. Let

$$\begin{array}{}\text{(1)}& y_p(x)=u_1{y}_1+u_2{y}_2\end{array}$$

Where $u_1,u_2$ to be determined, and $y_1,y_2$ are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as

$$\begin{array}{rl}{y}_1& =(2+x){\mathrm{e}}^{-x}\\ y_2& =-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}}\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(x)}{aW(x)}\text{(3)}& u_2& =\int \frac{y_{1}f(x)}{aW(x)}\end{array}$$

Where $W(x)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}(2+x){\mathrm{e}}^{-x}& -\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}}\\ \frac{d}{dx}\left((2+x){\mathrm{e}}^{-x}\right)& \frac{d}{dx}(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}})\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}(2+x){\mathrm{e}}^{-x}& -\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}}\\ {\mathrm{e}}^{-x}-(2+x){\mathrm{e}}^{-x}& \frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}+{\mathrm{e}}^{-2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}(2+x)-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}}{2}-(1+x){\mathrm{e}}^{\frac{x(2+x)}{2}}\end{array}\right|$$

Therefore

$$W=\left((2+x){\mathrm{e}}^{-x}\right)(\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}+{\mathrm{e}}^{-2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}(2+x)-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}}{2}-(1+x){\mathrm{e}}^{\frac{x(2+x)}{2}})-(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}})({\mathrm{e}}^{-x}-(2+x){\mathrm{e}}^{-x})$$

Which simplifies to

$$W={\mathrm{e}}^{-x}{\mathrm{e}}^{-2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}{x}^2+4{\mathrm{e}}^{-x}{\mathrm{e}}^{-2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}x-{\mathrm{e}}^{-x}{\mathrm{e}}^{\frac{x(2+x)}{2}}{x}^2+4{\mathrm{e}}^{-x}{\mathrm{e}}^{-2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}-4{\mathrm{e}}^{-x}{\mathrm{e}}^{\frac{x(2+x)}{2}}x-3{\mathrm{e}}^{-x}{\mathrm{e}}^{\frac{x(2+x)}{2}}$$

Which simplifies to

$$W={\mathrm{e}}^{\frac{x^2}{2}}$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}})x}{{\mathrm{e}}^{\frac{x^2}{2}}}dx$$

Which simplifies to

$$u_1=-\int -\frac{x(i{\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)+2{\mathrm{e}}^x)}{2}dx$$

Hence

$$u_1=-\frac{i\sqrt{\pi }\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right){\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}}{2}+{\mathrm{e}}^{x}x-{\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}x-\frac{i\sqrt{\pi }\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right){\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}x}{2}-{\mathrm{e}}^x$$

And Eq. (3) becomes

$$u_2=\int \frac{(2+x){\mathrm{e}}^{-x}x}{{\mathrm{e}}^{\frac{x^2}{2}}}dx$$

Which simplifies to

$$u_2=\int x(2+x){\mathrm{e}}^{-\frac{x(2+x)}{2}}dx$$

Hence

$$u_2=-(1+x){\mathrm{e}}^{-\frac{x(2+x)}{2}}$$

Therefore the particular solution, from equation (1) is

$$y_p(x)=(-\frac{i\sqrt{\pi }\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right){\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}}{2}+{\mathrm{e}}^{x}x-{\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}{\mathrm{e}}^{\frac{{(2+x)}^2}{2}}x-\frac{i\sqrt{\pi }\sqrt{2}\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right){\mathrm{e}}^{-2-\frac{1}{2}{x}^2-x}x}{2}-{\mathrm{e}}^x)(2+x){\mathrm{e}}^{-x}-(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}})(1+x){\mathrm{e}}^{-\frac{x(2+x)}{2}}$$

Which simplifies to

$$y_p(x)=-1$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1(2+x){\mathrm{e}}^{-x}+c_2(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}}))+(-1)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =-1+c_1(2+x){\mathrm{e}}^{-x}+c_2(-\frac{i{\mathrm{e}}^{-2-x}\sqrt{\pi }\mathrm{erf}\left(\frac{i\sqrt{2}(2+x)}{2}\right)\sqrt{2}(2+x)}{2}-{\mathrm{e}}^{\frac{x(2+x)}{2}})\end{array}$$

✓ Maple. Time used: 0.007 (sec). Leaf size: 52

ode:=diff(diff(y(x),x),x)-diff(y(x),x)*x-x*y(x)-x = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y={\mathrm{e}}^{-2-x}\pi \mathrm{erf}\left(\frac{i\sqrt{2}(x+2)}{2}\right)c_1(x+2)-i{\mathrm{e}}^{\frac{x(x+2)}{2}}\sqrt{\pi }\sqrt{2}{c}_1-1+{\mathrm{e}}^{-x}(x+2)c_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   checking if the LODE is of Euler type 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   checking if the LODE is missing y 
   -> Trying a Liouvillian solution using Kovacics algorithm 
      A Liouvillian solution exists 
      Reducible group (found an exponential solution) 
      Group is reducible, not completely reducible 
   <- Kovacics algorithm successful 
<- solving first the homogeneous part of the ODE successful
 

✓ Mathematica. Time used: 0.643 (sec). Leaf size: 216

ode=D[y[x],{x,2}]-x*D[y[x],x]-x*y[x]-x==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \frac{1}{2}{e}^{-\frac{1}{2}(x+2{)}^2}(2\sqrt{2}{e}^{\frac{x^2}{2}+x+2}(x+2){\int }_1^x(\frac{e^{K[1]}K[1]}{\sqrt{2}}-\frac{1}{2}{e}^{-\frac{1}{2}K[1{]}^2-K[1]-2}\sqrt{\pi }\text{erfi}\left(\frac{\sqrt{(K[1]+2{)}^2}}{\sqrt{2}}\right)K[1]\sqrt{(K[1]+2{)}^2})dK[1]-\sqrt{2\pi }\sqrt{(x+2{)}^2}(c_2{e}^{\frac{x^2}{2}+x+2}+x+1)\text{erfi}\left(\frac{\sqrt{(x+2{)}^2}}{\sqrt{2}}\right)+2e^{\frac{x^2}{2}+x+2}(e^x(x+1)+\sqrt{2}{c}_1(x+2)+c_2{e}^{\frac{1}{2}(x+2{)}^2}))$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x*y(x) - x*Derivative(y(x), x) - x + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE y(x) + Derivative(y(x), x) + 1 - Derivative(y(x), (x, 2))/x cannot be solved by the factorable group method