2.4.39 Problem 36

Internal problem ID [8928]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 36
Date solved : Sunday, March 30, 2025 at 01:55:33 PM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

$$\begin{array}{rl}x(x-1)y^{\prime \prime }+3x{y}^{\prime }+y& =0\end{array}$$

Using series expansion around $x=0$

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

$$(x^2-x)y^{\prime \prime }+3x{y}^{\prime }+y=0$$

The following is summary of singularities for the above ode. Writing the ode as

$$\begin{array}{rl}{y}^{\prime \prime }+p(x)y^{\prime }+q(x)y& =0\end{array}$$

Where

$$\begin{array}{rl}p(x)& =\frac{3}{x-1}\\ q(x)& =\frac{1}{x(x-1)}\end{array}$$

Table 2.92: Table $p(x),q(x)$ singularites.

$p(x)=\frac{3}{x-1}$
singularity	type
$x=1$	$\text{“regular”}$

$q(x)=\frac{1}{x(x-1)}$
singularity	type
$x=0$	$\text{“regular”}$
$x=1$	$\text{“regular”}$

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : $[1,0,\mathrm{\infty }]$

Irregular singular points : $[]$

Since $x=0$ is regular singular point, then Frobenius power series is used. The ode is normalized to be

$$x(x-1)y^{\prime \prime }+3x{y}^{\prime }+y=0$$

Let the solution be represented as Frobenius power series of the form

$$y=\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}$$

Then

$$\begin{array}{rl}{y}^{\prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1}\\ y^{\prime \prime }& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

Substituting the above back into the ode gives

$$\begin{array}{}\text{(1)}& x(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)(n+r-1)a_n{x}^{n+r-2})+3x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(n+r)a_n{x}^{n+r-1})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{n+r-1}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}3x^{n+r}{a}_n(n+r))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r})=0\end{array}$$

The next step is to make all powers of $x$ be $n+r-1$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^{n+r-1}$ and adjusting the power and the corresponding index gives

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n+r}{a}_n(n+r)(n+r-1)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}(n+r-1)(n+r-2)x^{n+r-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}3x^{n+r}{a}_n(n+r)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}3a_{n-1}(n+r-1)x^{n+r-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r}& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}{x}^{n+r-1}\end{array}$$

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n+r-1$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}(n+r-1)(n+r-2)x^{n+r-1})+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-x^{n+r-1}{a}_n(n+r)(n+r-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}3a_{n-1}(n+r-1)x^{n+r-1})+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{a}_{n-1}{x}^{n+r-1})=0\end{array}$$

The indicial equation is obtained from $n=0$. From Eq (2B) this gives

$$-x^{n+r-1}{a}_n(n+r)(n+r-1)=0$$

When $n=0$ the above becomes

$$-x^{-1+r}{a}_{0}r(-1+r)=0$$

Or

$$-x^{-1+r}{a}_{0}r(-1+r)=0$$

Since $a_0\ne 0$ then the above simplifies to

$$-x^{-1+r}r(-1+r)=0$$

Since the above is true for all $x$ then the indicial equation becomes

$$-r(-1+r)=0$$

Solving for $r$ gives the roots of the indicial equation as

$$\begin{array}{rl}{r}_1& =1\\ r_2& =0\end{array}$$

Since $a_0\ne 0$ then the indicial equation becomes

$$-x^{-1+r}r(-1+r)=0$$

Solving for $r$ gives the roots of the indicial equation as $[1,0]$.

Since $r_1-r_2=1$ is an integer, then we can construct two linearly independent solutions

$$\begin{array}{rl}{y}_1\left(x\right)& =x^{r_1}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+x^{r_2}(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^n)\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Or

$$\begin{array}{rl}{y}_1\left(x\right)& =\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+1}\\ y_2\left(x\right)& =C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)\end{array}$$

Where $C$ above can be zero. We start by finding $y_1$. Eq (2B) derived above is now used to find all $a_n$ coefficients. The case $n=0$ is skipped since it was used to find the roots of the indicial equation. $a_0$ is arbitrary and taken as $a_0=1$. For $1\le n$ the recursive equation is

$$\begin{array}{}\text{(3)}& a_{n-1}(n+r-1)(n+r-2)-a_n(n+r)(n+r-1)+3a_{n-1}(n+r-1)+a_{n-1}=0\end{array}$$

Solving for $a_n$ from recursive equation (4) gives

$$\begin{array}{}\text{(4)}& a_n=\frac{(n+r)a_{n-1}}{n+r-1}\end{array}$$

Which for the root $r=1$ becomes

$$\begin{array}{}\text{(5)}& a_n=\frac{(n+1)a_{n-1}}{n}\end{array}$$

At this point, it is a good idea to keep track of $a_n$ in a table both before substituting $r=1$ and after as more terms are found using the above recursive equation.

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$

For $n=1$, using the above recursive equation gives

$$a_1=\frac{1+r}{r}$$

Which for the root $r=1$ becomes

$$a_1=2$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1+r}{r}$	$2$

For $n=2$, using the above recursive equation gives

$$a_2=\frac{2+r}{r}$$

Which for the root $r=1$ becomes

$$a_2=3$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1+r}{r}$	$2$
$a_2$	$\frac{2+r}{r}$	$3$

For $n=3$, using the above recursive equation gives

$$a_3=\frac{3+r}{r}$$

Which for the root $r=1$ becomes

$$a_3=4$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1+r}{r}$	$2$
$a_2$	$\frac{2+r}{r}$	$3$
$a_3$	$\frac{3+r}{r}$	$4$

For $n=4$, using the above recursive equation gives

$$a_4=\frac{4+r}{r}$$

Which for the root $r=1$ becomes

$$a_4=5$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1+r}{r}$	$2$
$a_2$	$\frac{2+r}{r}$	$3$
$a_3$	$\frac{3+r}{r}$	$4$
$a_4$	$\frac{4+r}{r}$	$5$

For $n=5$, using the above recursive equation gives

$$a_5=\frac{5+r}{r}$$

Which for the root $r=1$ becomes

$$a_5=6$$

And the table now becomes

$n$	$a_{n,r}$	$a_n$
$a_0$	$1$	$1$
$a_1$	$\frac{1+r}{r}$	$2$
$a_2$	$\frac{2+r}{r}$	$3$
$a_3$	$\frac{3+r}{r}$	$4$
$a_4$	$\frac{4+r}{r}$	$5$
$a_5$	$\frac{5+r}{r}$	$6$

Using the above table, then the solution $y_1\left(x\right)$ is

$$\begin{array}{rl}{y}_1\left(x\right)& =x(a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6\dots )\\ & =x(1+2x+3x^2+4x^3+5x^4+6x^5+O\left(x^6\right))\end{array}$$

Now the second solution $y_2\left(x\right)$ is found. Let

$$r_1-r_2=N$$

Where $N$ is positive integer which is the difference between the two roots. $r_1$ is taken as the larger root. Hence for this problem we have $N=1$. Now we need to determine if $C$ is zero or not. This is done by finding $\underset{r\to r_2}{lim}{a}_1\left(r\right)$. If this limit exists, then $C=0$, else we need to keep the log term and $C\ne 0$. The above table shows that

$$\begin{array}{rl}{a}_N& =a_1\\ & =\frac{1+r}{r}\end{array}$$

Therefore

$$\begin{array}{rl}\underset{r\to r_2}{lim}\frac{1+r}{r}& =\underset{r\to 0}{lim}\frac{1+r}{r}\\ & =\mathit{\text{undefined}}\end{array}$$

Since the limit does not exist then the log term is needed. Therefore the second solution has the form

$$y_2\left(x\right)=C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})$$

Therefore

$$\begin{array}{rl}\frac{d}{dx}{y}_2\left(x\right)& =C{y}_1^{\prime }\left(x\right)\ln \left(x\right)+\frac{C{y}_1\left(x\right)}{x}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{b_n{x}^{n+r_2}(n+r_2)}{x})\\ & =C{y}_1^{\prime }\left(x\right)\ln \left(x\right)+\frac{C{y}_1\left(x\right)}{x}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-1+n+r_2}{b}_n(n+r_2))\\ \frac{d^2}{d{x}^2}{y}_2\left(x\right)& =C{y}_1^{\prime \prime }\left(x\right)\ln \left(x\right)+\frac{2C{y}_1^{\prime }\left(x\right)}{x}-\frac{C{y}_1\left(x\right)}{x^2}+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2})\\ & =C{y}_1^{\prime \prime }\left(x\right)\ln \left(x\right)+\frac{2C{y}_1^{\prime }\left(x\right)}{x}-\frac{C{y}_1\left(x\right)}{x^2}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-2+n+r_2}{b}_n(n+r_2)(-1+n+r_2))\end{array}$$

Substituting these back into the given ode $x(x-1)y^{\prime \prime }+3x{y}^{\prime }+y=0$ gives

$$x(x-1)(C{y}_1^{\prime \prime }\left(x\right)\ln \left(x\right)+\frac{2C{y}_1^{\prime }\left(x\right)}{x}-\frac{C{y}_1\left(x\right)}{x^2}+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2}))+3x(C{y}_1^{\prime }\left(x\right)\ln \left(x\right)+\frac{C{y}_1\left(x\right)}{x}+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{b_n{x}^{n+r_2}(n+r_2)}{x}))+C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})=0$$

Which can be written as

$$\begin{array}{}\text{(7)}& ((x(x-1)y_1^{\prime \prime }\left(x\right)+3y_1^{\prime }\left(x\right)x+y_1\left(x\right))\ln \left(x\right)+x(x-1)(\frac{2y_1^{\prime }\left(x\right)}{x}-\frac{y_1\left(x\right)}{x^2})+3y_1\left(x\right))C+x(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2}))+3x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{b_n{x}^{n+r_2}(n+r_2)}{x})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})=0\end{array}$$

But since $y_1\left(x\right)$ is a solution to the ode, then

$$x(x-1)y_1^{\prime \prime }\left(x\right)+3y_1^{\prime }\left(x\right)x+y_1\left(x\right)=0$$

Eq (7) simplifes to

$$\begin{array}{}\text{(8)}& (x(x-1)(\frac{2y_1^{\prime }\left(x\right)}{x}-\frac{y_1\left(x\right)}{x^2})+3y_1\left(x\right))C+x(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}(\frac{b_n{x}^{n+r_2}{(n+r_2)}^2}{x^2}-\frac{b_n{x}^{n+r_2}(n+r_2)}{x^2}))+3x(\stackrel{\mathrm{\infty }}{\sum _{n=0}}\frac{b_n{x}^{n+r_2}(n+r_2)}{x})+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})=0\end{array}$$

Substituting $y_1=\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r_1}$ into the above gives

$$\begin{array}{}\text{(9)}& \frac{(2x(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-1+n+r_1}{a}_n(n+r_1))+(2x+1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+r_1}))C}{x}+\frac{x^2(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-2+n+r_2}{b}_n(n+r_2)(-1+n+r_2))+3x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-1+n+r_2}{b}_n(n+r_2))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})x}{x}=0\end{array}$$

Since $r_1=1$ and $r_2=0$ then the above becomes

$$\begin{array}{}\text{(10)}& \frac{(2x(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^n{a}_n(n+1))+(2x+1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n+1}))C}{x}+\frac{x^2(x-1)(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{-2+n}{b}_{n}n(n-1))+3x^2(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^{n-1}{b}_{n}n)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)x}{x}=0\end{array}$$

Which simplifies to

$$\begin{array}{}\text{(2A)}& (\stackrel{\mathrm{\infty }}{\sum _{n=0}}2C{x}^{n+1}{a}_n(n+1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-2x^n{a}_n(n+1)C)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}2C{x}^{n+1}{a}_n)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n}C)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^n{b}_{n}n(n-1))+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-n{x}^{n-1}{b}_n(n-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}3x^n{b}_{n}n)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n)=0\end{array}$$

The next step is to make all powers of $x$ be $n-1$ in each summation term. Going over each summation term above with power of $x$ in it which is not already $x^{n-1}$ and adjusting the power and the corresponding index gives

$$\begin{array}{rl}\stackrel{\mathrm{\infty }}{\sum _{n=0}}2C{x}^{n+1}{a}_n(n+1)& =\stackrel{\mathrm{\infty }}{\sum _{n=2}}2C{a}_{-2+n}(n-1)x^{n-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}(-2x^n{a}_n(n+1)C)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-2C{a}_{n-1}n{x}^{n-1})\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}2C{x}^{n+1}{a}_n& =\stackrel{\mathrm{\infty }}{\sum _{n=2}}2C{a}_{-2+n}{x}^{n-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}{a}_n{x}^{n}C& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}C{a}_{n-1}{x}^{n-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}{x}^n{b}_{n}n(n-1)& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}(n-1)b_{n-1}(-2+n)x^{n-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}3x^n{b}_{n}n& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}3(n-1)b_{n-1}{x}^{n-1}\\ \stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^n& =\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_{n-1}{x}^{n-1}\end{array}$$

Substituting all the above in Eq (2A) gives the following equation where now all powers of $x$ are the same and equal to $n-1$.

$$\begin{array}{}\text{(2B)}& (\stackrel{\mathrm{\infty }}{\sum _{n=2}}2C{a}_{-2+n}(n-1)x^{n-1})+\stackrel{\mathrm{\infty }}{\sum _{n=1}}(-2C{a}_{n-1}n{x}^{n-1})+(\stackrel{\mathrm{\infty }}{\sum _{n=2}}2C{a}_{-2+n}{x}^{n-1})+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}C{a}_{n-1}{x}^{n-1})+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}(n-1)b_{n-1}(-2+n)x^{n-1})+\stackrel{\mathrm{\infty }}{\sum _{n=0}}(-n{x}^{n-1}{b}_n(n-1))+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}3(n-1)b_{n-1}{x}^{n-1})+(\stackrel{\mathrm{\infty }}{\sum _{n=1}}{b}_{n-1}{x}^{n-1})=0\end{array}$$

For $n=0$ in Eq. (2B), we choose arbitray value for $b_0$ as $b_0=1$. For $n=N$, where $N=1$ which is the difference between the two roots, we are free to choose $b_1=0$. Hence for $n=1$, Eq (2B) gives

$$-C+1=0$$

Which is solved for $C$. Solving for $C$ gives

$$C=1$$

For $n=2$, Eq (2B) gives

$$(4a_0-3a_1)C+4b_1-2b_2=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$-2-2b_2=0$$

Solving the above for $b_2$ gives

$$b_2=-1$$

For $n=3$, Eq (2B) gives

$$(6a_1-5a_2)C+9b_2-6b_3=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$-12-6b_3=0$$

Solving the above for $b_3$ gives

$$b_3=-2$$

For $n=4$, Eq (2B) gives

$$(8a_2-7a_3)C+16b_3-12b_4=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$-36-12b_4=0$$

Solving the above for $b_4$ gives

$$b_4=-3$$

For $n=5$, Eq (2B) gives

$$(10a_3-9a_4)C+25b_4-20b_5=0$$

Which when replacing the above values found already for $b_n$ and the values found earlier for $a_n$ and for $C$, gives

$$-80-20b_5=0$$

Solving the above for $b_5$ gives

$$b_5=-4$$

Now that we found all $b_n$ and $C$, we can calculate the second solution from

$$y_2\left(x\right)=C{y}_1\left(x\right)\ln \left(x\right)+(\stackrel{\mathrm{\infty }}{\sum _{n=0}}{b}_n{x}^{n+r_2})$$

Using the above value found for $C=1$ and all $b_n$, then the second solution becomes

$$y_2\left(x\right)=1\left(x(1+2x+3x^2+4x^3+5x^4+6x^5+O\left(x^6\right))\right)\ln \left(x\right)+1-x^2-2x^3-3x^4-4x^5+O\left(x^6\right)$$

Therefore the homogeneous solution is

$$\begin{array}{rl}{y}_h(x)& =c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)\\ & =c_{1}x(1+2x+3x^2+4x^3+5x^4+6x^5+O\left(x^6\right))+c_2(1\left(x(1+2x+3x^2+4x^3+5x^4+6x^5+O\left(x^6\right))\right)\ln \left(x\right)+1-x^2-2x^3-3x^4-4x^5+O\left(x^6\right))\end{array}$$

Hence the final solution is

$$\begin{array}{rl}y& =y_h\\ & =c_{1}x(1+2x+3x^2+4x^3+5x^4+6x^5+O\left(x^6\right))+c_2(x(1+2x+3x^2+4x^3+5x^4+6x^5+O\left(x^6\right))\ln \left(x\right)+1-x^2-2x^3-3x^4-4x^5+O\left(x^6\right))\end{array}$$

✓ Maple. Time used: 0.023 (sec). Leaf size: 60

Order:=6; 
ode:=x*(x-1)*diff(diff(y(x),x),x)+3*diff(y(x),x)*x+y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 

$$y=c_{1}x(1+2x+3x^2+4x^3+5x^4+6x^5+\mathrm{O}\left(x^6\right))+(x+2x^2+3x^3+4x^4+5x^5+\mathrm{O}\left(x^6\right))\ln \left(x\right)c_2+(1+3x+5x^2+7x^3+9x^4+11x^5+\mathrm{O}\left(x^6\right))c_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
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)] 
<- linear_1 successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x(x-1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x\left(\frac{d}{dx}y\left(x\right)\right)+y\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=-\frac{y\left(x\right)}{x(x-1)}-\frac{3\left(\frac{d}{dx}y\left(x\right)\right)}{x-1}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{3\left(\frac{d}{dx}y\left(x\right)\right)}{x-1}+\frac{y\left(x\right)}{x(x-1)}=0\\ ◻& & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(x\right)=\frac{3}{x-1},P_3\left(x\right)=\frac{1}{x(x-1)}]\\ & \circ & x\cdot P_2\left(x\right)\text{is analytic at}x=0\\ & & (x\cdot P_2\left(x\right))|{}_{x=0}=0\\ & \circ & x^2\cdot P_3\left(x\right)\text{is analytic at}x=0\\ & & (x^2\cdot P_3\left(x\right))|{}_{x=0}=0\\ & \circ & x=0\text{is a regular singular point}\\ & & \text{Check to see if}{x}_0\text{is a regular singular point}\\ & & x_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & x(x-1)\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+3x\left(\frac{d}{dx}y\left(x\right)\right)+y\left(x\right)=0\\ \bullet & & \text{Assume series solution for}y\left(x\right)\\ & & y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}x\cdot \left(\frac{d}{dx}y\left(x\right)\right)\text{to series expansion}\\ & & x\cdot \left(\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)x^{k+r}\\ & \circ & \text{Convert}{x}^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)\text{to series expansion for}m=1..2\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)x^{k+r-2+m}\\ & \circ & \text{Shift index using}k->k+2-m\\ & & x^m\cdot \left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-2+m}}{a}_{k+2-m}(k+2-m+r)(k+1-m+r)x^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & -a_{0}r(-1+r)x^{-1+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}(-a_{k+1}(k+r+1)(k+r)+a_k{(k+r+1)}^2)x^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & -r(-1+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,1\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & (k+r+1)(-a_{k+1}(k+r)+a_k(k+r+1))=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+1}=\frac{a_k(k+r+1)}{k+r}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+1}=\frac{a_k(k+1)}{k}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k,a_{k+1}=\frac{a_k(k+1)}{k}]\\ \bullet & & \text{Recursion relation for}r=1\\ & & a_{k+1}=\frac{a_k(k+2)}{k+1}\\ \bullet & & \text{Solution for}r=1\\ & & [y\left(x\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^{k+1},a_{k+1}=\frac{a_k(k+2)}{k+1}]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(x\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{x}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{x}^{k+1}),a_{k+1}=\frac{a_k(k+1)}{k},b_{k+1}=\frac{b_k(k+2)}{k+1}]\end{array}$$

✓ Mathematica. Time used: 0.044 (sec). Leaf size: 63

ode=x*(x-1)*D[y[x],{x,2}] +3*x*D[y[x],x]+y[x] == 0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 

$$y(x)\to c_1(x^4+x^3+x^2+(4x^3+3x^2+2x+1)x\log (x)+x+1)+c_2(5x^5+4x^4+3x^3+2x^2+x)$$

✓ Sympy. Time used: 0.901 (sec). Leaf size: 37

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*(x - 1)*Derivative(y(x), (x, 2)) + 3*x*Derivative(y(x), x) + y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 

$$y\left(x\right)=C_{2}x(\frac{27x^4}{40}-\frac{9x^3}{8}+\frac{3x^2}{2}-\frac{3x}{2}+1)+C_1+O\left(x^6\right)$$