Problem 24

2.4.28 Problem 24

✓Maple
✓Mathematica
✗Sympy

Internal problem ID [10191]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 24
Date solved : Thursday, November 27, 2025 at 10:24:44 AM
CAS classiﬁcation : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} 2 x^{2} y^{\prime \prime }-y^{\prime } x +\left (-x^{2}+1\right ) y&=\ln \left (x \right ) \\ \end{align*}

Series expansion around \(x=1\).

The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let

\[ \tau = x -1 \]

The ode is converted to be in terms of the new independent variable \(\tau \). This results in

\[ 2 \left (\tau +1\right )^{2} \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-\left (\frac {d}{d \tau }y \left (\tau \right )\right ) \left (\tau +1\right )+\left (-\left (\tau +1\right )^{2}+1\right ) y \left (\tau \right ) = \ln \left (\tau +1\right ) \]

With its expansion point and initial conditions now at \(\tau = 0\). The transformed ODE is now solved. Solving ode using Taylor series method. This gives review on how the Taylor series method works for solving second order ode.

Let

\[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \]

Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives

\begin{align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\end{align*}

But

\begin{align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end{align}

And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as

\begin{align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6}\end{align}

Therefore (6) can be used from now on along with

\begin{equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7}\end{equation}

To ﬁnd \(y\left ( x\right ) \) series solution around \(x=0\). Hence

\begin{align*} F_0 &= \frac {y \left (\tau \right ) \tau ^{2}+\left (\frac {d}{d \tau }y \left (\tau \right )\right ) \tau +2 y \left (\tau \right ) \tau +\ln \left (\tau +1\right )+\frac {d}{d \tau }y \left (\tau \right )}{2 \left (\tau +1\right )^{2}}\\ F_1 &= \frac {d F_0}{d\tau } \\ &= \frac {\partial F_{0}}{\partial \tau }+ \frac {\partial F_{0}}{\partial y} \frac {d}{d \tau }y \left (\tau \right )+ \frac {\partial F_{0}}{\partial \frac {d}{d \tau }y \left (\tau \right )} F_0 \\ &= \frac {-3 \ln \left (\tau +1\right )+\left (2 \tau ^{3}+6 \tau ^{2}+3 \tau -1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )+2+\left (\tau ^{2}+2 \tau +4\right ) y \left (\tau \right )}{4 \left (\tau +1\right )^{3}}\\ F_2 &= \frac {d F_1}{d\tau } \\ &= \frac {\partial F_{1}}{\partial \tau }+ \frac {\partial F_{1}}{\partial y} \frac {d}{d \tau }y \left (\tau \right )+ \frac {\partial F_{1}}{\partial \frac {d}{d \tau }y \left (\tau \right )} F_1 \\ &= \frac {\left (2 \tau ^{2}+4 \tau +17\right ) \ln \left (\tau +1\right )+\left (4 \tau ^{3}+12 \tau ^{2}+27 \tau +19\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )-18+\left (2 \tau ^{4}+8 \tau ^{3}+5 \tau ^{2}-6 \tau -20\right ) y \left (\tau \right )}{8 \left (\tau +1\right )^{4}}\\ F_3 &= \frac {d F_2}{d\tau } \\ &= \frac {\partial F_{2}}{\partial \tau }+ \frac {\partial F_{2}}{\partial y} \frac {d}{d \tau }y \left (\tau \right )+ \frac {\partial F_{2}}{\partial \frac {d}{d \tau }y \left (\tau \right )} F_2 \\ &= \frac {\left (-4 \tau ^{2}-8 \tau -109\right ) \ln \left (\tau +1\right )+\left (4 \tau ^{5}+20 \tau ^{4}+22 \tau ^{3}-14 \tau ^{2}-139 \tau -119\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )+\left (4 \tau ^{4}+16 \tau ^{3}+63 \tau ^{2}+94 \tau +148\right ) y \left (\tau \right )+4 \tau ^{2}+8 \tau +178}{16 \left (\tau +1\right )^{5}}\\ F_4 &= \frac {d F_3}{d\tau } \\ &= \frac {\partial F_{3}}{\partial \tau }+ \frac {\partial F_{3}}{\partial y} \frac {d}{d \tau }y \left (\tau \right )+ \frac {\partial F_{3}}{\partial \frac {d}{d \tau }y \left (\tau \right )} F_3 \\ &= \frac {\left (4 \tau ^{4}+16 \tau ^{3}+30 \tau ^{2}+28 \tau +955\right ) \ln \left (\tau +1\right )+\left (12 \tau ^{5}+60 \tau ^{4}+252 \tau ^{3}+516 \tau ^{2}+1401 \tau +1089\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )+\left (4 \tau ^{6}+24 \tau ^{5}+30 \tau ^{4}-40 \tau ^{3}-441 \tau ^{2}-738 \tau -1292\right ) y \left (\tau \right )-32 \tau ^{2}-64 \tau -1982}{32 \left (\tau +1\right )^{6}} \end{align*}

And so on. Evaluating all the above at initial conditions \(\tau = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives

\begin{align*} F_0 &= \frac {y^{\prime }\left (0\right )}{2}\\ F_1 &= y \left (0\right )-\frac {y^{\prime }\left (0\right )}{4}+\frac {1}{2}\\ F_2 &= -\frac {5 y \left (0\right )}{2}+\frac {19 y^{\prime }\left (0\right )}{8}-\frac {9}{4}\\ F_3 &= \frac {37 y \left (0\right )}{4}-\frac {119 y^{\prime }\left (0\right )}{16}+\frac {89}{8}\\ F_4 &= -\frac {323 y \left (0\right )}{8}+\frac {1089 y^{\prime }\left (0\right )}{32}-\frac {991}{16} \end{align*}

Substituting all the above in (7) and simplifying gives the solution as

\[ y \left (\tau \right ) = \left (1+\frac {1}{6} \tau ^{3}-\frac {5}{48} \tau ^{4}+\frac {37}{480} \tau ^{5}\right ) y \left (0\right )+\left (\tau +\frac {1}{4} \tau ^{2}-\frac {1}{24} \tau ^{3}+\frac {19}{192} \tau ^{4}-\frac {119}{1920} \tau ^{5}\right ) y^{\prime }\left (0\right )+\frac {\tau ^{3}}{12}-\frac {3 \tau ^{4}}{32}+\frac {89 \tau ^{5}}{960}+O\left (\tau ^{6}\right ) \]

Since the expansion point \(\tau = 0\) is an ordinary point, then this can also be solved using the standard power series method. The ode is normalized to be

\[ \left (2 \tau ^{2}+4 \tau +2\right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+\left (-\tau -1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )+\left (-\tau ^{2}-2 \tau \right ) y \left (\tau \right ) = \ln \left (\tau +1\right ) \]

Let the solution be represented as power series of the form

\[ y \left (\tau \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n} \]

Then

\begin{align*} \frac {d}{d \tau }y \left (\tau \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} \tau ^{n -1}\\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} \tau ^{n -2} \end{align*}

Substituting the above back into the ode gives

\begin{align*} \left (2 \tau ^{2}+4 \tau +2\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} \tau ^{n -2}\right )+\left (-\tau -1\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} \tau ^{n -1}\right )+\left (-\tau ^{2}-2 \tau \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n}\right ) = \ln \left (\tau +1\right )\tag {1} \end{align*}

Expanding \(\ln \left (\tau +1\right )\) as Taylor series around \(\tau =0\) and keeping only the ﬁrst \(6\) terms gives

\begin{align*} \ln \left (\tau +1\right ) &= \tau -\frac {1}{2} \tau ^{2}+\frac {1}{3} \tau ^{3}-\frac {1}{4} \tau ^{4}+\frac {1}{5} \tau ^{5} + \dots \\ &= \tau -\frac {1}{2} \tau ^{2}+\frac {1}{3} \tau ^{3}-\frac {1}{4} \tau ^{4}+\frac {1}{5} \tau ^{5} \end{align*}

Hence the ODE in Eq (1) becomes

\[ \left (2 \tau ^{2}+4 \tau +2\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} \tau ^{n -2}\right )+\left (-\tau -1\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} \tau ^{n -1}\right )+\left (-\tau ^{2}-2 \tau \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n}\right ) = \tau -\frac {1}{2} \tau ^{2}+\frac {1}{3} \tau ^{3}-\frac {1}{4} \tau ^{4}+\frac {1}{5} \tau ^{5} \]

Which simpliﬁes to

\begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 \tau ^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}4 n a_{n} \tau ^{n -1} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 n \left (n -1\right ) a_{n} \tau ^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} \tau ^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} \tau ^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} \tau ^{n +2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \tau ^{1+n} a_{n}\right ) = \tau -\frac {1}{2} \tau ^{2}+\frac {1}{3} \tau ^{3}-\frac {1}{4} \tau ^{4}+\frac {1}{5} \tau ^{5} \end{equation}

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

\begin{align*} \moverset {\infty }{\munderset {n =2}{\sum }}4 n a_{n} \tau ^{n -1} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 \left (1+n \right ) a_{1+n} n \,\tau ^{n} \\ \moverset {\infty }{\munderset {n =2}{\sum }}2 n \left (n -1\right ) a_{n} \tau ^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +2\right ) a_{n +2} \left (1+n \right ) \tau ^{n} \\ \moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} \tau ^{n -1}\right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (1+n \right ) a_{1+n} \tau ^{n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} \tau ^{n +2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \tau ^{n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \tau ^{1+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \tau ^{n}\right ) \\ \end{align*}

Substituting all the above in Eq (2) gives the following equation where now all powers of \(\tau \) are the same and equal to \(n\).

\begin{equation} \tag{3} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 \tau ^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 \left (1+n \right ) a_{1+n} n \,\tau ^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +2\right ) a_{n +2} \left (1+n \right ) \tau ^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-n a_{n} \tau ^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (1+n \right ) a_{1+n} \tau ^{n}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \tau ^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \tau ^{n}\right ) = \tau -\frac {1}{2} \tau ^{2}+\frac {1}{3} \tau ^{3}-\frac {1}{4} \tau ^{4}+\frac {1}{5} \tau ^{5} \end{equation}

\(n=0\) gives

\[ 4 a_{2}-a_{1}=0 \]

\[ a_{2} = \frac {a_{1}}{4} \]

\(n=1\) gives

\begin{align*} \left (6 a_{2}+12 a_{3}-a_{1}-2 a_{0}\right ) \tau &= \tau \\ 6 a_{2}+12 a_{3}-a_{1}-2 a_{0} &= 1 \\ \end{align*}

Which after substituting earlier equations, simpliﬁes to

\[ a_{3} = \frac {a_{0}}{6}-\frac {a_{1}}{24}+\frac {1}{12} \]

For \(2\le n\), the recurrence equation is

\begin{equation} \tag{4} \left (2 n a_{n} \left (n -1\right )+4 \left (1+n \right ) a_{1+n} n +2 \left (n +2\right ) a_{n +2} \left (1+n \right )-n a_{n}-\left (1+n \right ) a_{1+n}-a_{n -2}-2 a_{n -1}\right ) \tau ^{n} = \tau -\frac {1}{2} \tau ^{2}+\frac {1}{3} \tau ^{3}-\frac {1}{4} \tau ^{4}+\frac {1}{5} \tau ^{5} \end{equation}

For \(n = 2\) the recurrence equation gives

\begin{align*} \left (2 a_{2}+21 a_{3}+24 a_{4}-a_{0}-2 a_{1}\right ) \tau ^{2}&=-\frac {\tau ^{2}}{2} \\ 2 a_{2}+21 a_{3}+24 a_{4}-a_{0}-2 a_{1} &= -{\frac {1}{2}} \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{4} = -\frac {3}{32}+\frac {19 a_{1}}{192}-\frac {5 a_{0}}{48} \]

For \(n = 3\) the recurrence equation gives

\begin{align*} \left (9 a_{3}+44 a_{4}+40 a_{5}-a_{1}-2 a_{2}\right ) \tau ^{3}&=\frac {\tau ^{3}}{3} \\ 9 a_{3}+44 a_{4}+40 a_{5}-a_{1}-2 a_{2} &= {\frac {1}{3}} \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{5} = \frac {89}{960}+\frac {37 a_{0}}{480}-\frac {119 a_{1}}{1920} \]

For \(n = 4\) the recurrence equation gives

\begin{align*} \left (20 a_{4}+75 a_{5}+60 a_{6}-a_{2}-2 a_{3}\right ) \tau ^{4}&=-\frac {\tau ^{4}}{4} \\ 20 a_{4}+75 a_{5}+60 a_{6}-a_{2}-2 a_{3} &= -{\frac {1}{4}} \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{6} = -\frac {991}{11520}+\frac {121 a_{1}}{2560}-\frac {323 a_{0}}{5760} \]

For \(n = 5\) the recurrence equation gives

\begin{align*} \left (35 a_{5}+114 a_{6}+84 a_{7}-a_{3}-2 a_{4}\right ) \tau ^{5}&=\frac {\tau ^{5}}{5} \\ 35 a_{5}+114 a_{6}+84 a_{7}-a_{3}-2 a_{4} &= {\frac {1}{5}} \\ \end{align*}

Which after substituting the earlier terms found becomes

\[ a_{7} = \frac {4261}{53760}+\frac {167 a_{0}}{3840}-\frac {11761 a_{1}}{322560} \]

And so on. Therefore the solution is

\begin{align*} y \left (\tau \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n}\\ &= a_{3} \tau ^{3}+a_{2} \tau ^{2}+a_{1} \tau +a_{0} + \dots \end{align*}

Substituting the values for \(a_{n}\) found above, the solution becomes

\[ y \left (\tau \right ) = a_{0}+a_{1} \tau +\frac {a_{1} \tau ^{2}}{4}+\left (\frac {a_{0}}{6}-\frac {a_{1}}{24}+\frac {1}{12}\right ) \tau ^{3}+\left (-\frac {3}{32}+\frac {19 a_{1}}{192}-\frac {5 a_{0}}{48}\right ) \tau ^{4}+\left (\frac {89}{960}+\frac {37 a_{0}}{480}-\frac {119 a_{1}}{1920}\right ) \tau ^{5}+\dots \]

Collecting terms, the solution becomes

\begin{equation} \tag{3} y \left (\tau \right ) = \left (1+\frac {1}{6} \tau ^{3}-\frac {5}{48} \tau ^{4}+\frac {37}{480} \tau ^{5}\right ) a_{0}+\left (\tau +\frac {1}{4} \tau ^{2}-\frac {1}{24} \tau ^{3}+\frac {19}{192} \tau ^{4}-\frac {119}{1920} \tau ^{5}\right ) a_{1}+\frac {\tau ^{3}}{12}-\frac {3 \tau ^{4}}{32}+\frac {89 \tau ^{5}}{960}+O\left (\tau ^{6}\right ) \end{equation}

At \(\tau = 0\) the solution above becomes

Replacing \(\tau \) in the above with the original independent variable \(x\) using \(\tau = x -1\) results in

\[ y = \left (1+\frac {\left (x -1\right )^{3}}{6}-\frac {5 \left (x -1\right )^{4}}{48}+\frac {37 \left (x -1\right )^{5}}{480}\right ) y \left (1\right )+\left (x -1+\frac {\left (x -1\right )^{2}}{4}-\frac {\left (x -1\right )^{3}}{24}+\frac {19 \left (x -1\right )^{4}}{192}-\frac {119 \left (x -1\right )^{5}}{1920}\right ) y^{\prime }\left (1\right )+\frac {\left (x -1\right )^{3}}{12}-\frac {3 \left (x -1\right )^{4}}{32}+\frac {89 \left (x -1\right )^{5}}{960}+O\left (\left (x -1\right )^{6}\right ) \]

Replacing \(\tau \) in the above with the original independent variable \(x\) using \(\tau = x -1\) results in

✓ Maple. Time used: 0.005 (sec). Leaf size: 90

Order:=6; 
ode:=2*x^2*diff(diff(y(x),x),x)-diff(y(x),x)*x+(-x^2+1)*y(x) = ln(x); 
dsolve(ode,y(x),type='series',x=1);

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 
   <- No Liouvillian solutions exists 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      <- Bessel successful 
   <- special function solution successful 
<- solving first the homogeneous part of the ODE successful

✓ Mathematica. Time used: 0.016 (sec). Leaf size: 105

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

\[ y(x)\to \frac {89}{960} (x-1)^5-\frac {3}{32} (x-1)^4+\frac {1}{12} (x-1)^3+c_1 \left (\frac {37}{480} (x-1)^5-\frac {5}{48} (x-1)^4+\frac {1}{6} (x-1)^3+1\right )+c_2 \left (-\frac {119 (x-1)^5}{1920}+\frac {19}{192} (x-1)^4-\frac {1}{24} (x-1)^3+\frac {1}{4} (x-1)^2+x-1\right ) \]

✗ Sympy

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

ValueError : ODE 2*x**2*Derivative(y(x), (x, 2)) - x*Derivative(y(x), x) + (1 - x**2)*y(x) - log(x) does not match hint 2nd_power_series_regular

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