2.4.28 problem 24
Internal
problem
ID
[8593]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
24
Date
solved
:
Thursday, December 12, 2024 at 09:31:43 AM
CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} 2 x^{2} y^{\prime \prime }-x y^{\prime }+\left (-x^{2}+1\right ) y&=\ln \left (x \right ) \end{align*}
Using 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
\[
2 \left (\tau +1\right )^{2} \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-\left (\tau +1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )+\left (-\left (\tau +1\right )^{2}+1\right ) y \left (\tau \right ) = \ln \left (\tau +1\right )
\]
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
\[
y \left (\tau \right ) = \left (1+\frac {1}{6} \tau ^{3}-\frac {5}{48} \tau ^{4}+\frac {37}{480} \tau ^{5}\right ) c_1 +\left (\tau +\frac {1}{4} \tau ^{2}-\frac {1}{24} \tau ^{3}+\frac {19}{192} \tau ^{4}-\frac {119}{1920} \tau ^{5}\right ) c_2 +\frac {\tau ^{3}}{12}-\frac {3 \tau ^{4}}{32}+\frac {89 \tau ^{5}}{960}+O\left (\tau ^{6}\right )
\]
Replacing \(\tau \) in the above with the original
independent variable \(x\)using \(\tau = x -1\) results in
\[
y = \left (1+\frac {\left (x -1\right )^{3}}{6}-\frac {5 \left (x -1\right )^{4}}{48}+\frac {37 \left (x -1\right )^{5}}{480}\right ) y \left (1\right )+\left (x -1+\frac {\left (x -1\right )^{2}}{4}-\frac {\left (x -1\right )^{3}}{24}+\frac {19 \left (x -1\right )^{4}}{192}-\frac {119 \left (x -1\right )^{5}}{1920}\right ) y^{\prime }\left (1\right )+\frac {\left (x -1\right )^{3}}{12}-\frac {3 \left (x -1\right )^{4}}{32}+\frac {89 \left (x -1\right )^{5}}{960}+O\left (\left (x -1\right )^{6}\right )
\]
Maple step by step solution
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`
Maple dsolve solution
Solving time : 0.007 (sec)
Leaf size : 90
dsolve(2*x^2*diff(diff(y(x),x),x)-diff(y(x),x)*x+(-x^2+1)*y(x) = ln(x),y(x),
series,x=1)
\[
y = \left (1+\frac {\left (x -1\right )^{3}}{6}-\frac {5 \left (x -1\right )^{4}}{48}+\frac {37 \left (x -1\right )^{5}}{480}\right ) y \left (1\right )+\left (x -1+\frac {\left (x -1\right )^{2}}{4}-\frac {\left (x -1\right )^{3}}{24}+\frac {19 \left (x -1\right )^{4}}{192}-\frac {119 \left (x -1\right )^{5}}{1920}\right ) y^{\prime }\left (1\right )+\frac {\left (x -1\right )^{3}}{12}-\frac {3 \left (x -1\right )^{4}}{32}+\frac {89 \left (x -1\right )^{5}}{960}+O\left (x^{6}\right )
\]
Mathematica DSolve solution
Solving time : 0.024 (sec)
Leaf size : 105
AsymptoticDSolveValue[{2*x^2*D[y[x],{x,2}]-x*D[y[x],x]+(1-x^2)*y[x]==Log[x],{}},
y[x],{x,1,5}]
\[
y(x)\to \frac {89}{960} (x-1)^5-\frac {3}{32} (x-1)^4+\frac {1}{12} (x-1)^3+c_1 \left (\frac {37}{480} (x-1)^5-\frac {5}{48} (x-1)^4+\frac {1}{6} (x-1)^3+1\right )+c_2 \left (-\frac {119 (x-1)^5}{1920}+\frac {19}{192} (x-1)^4-\frac {1}{24} (x-1)^3+\frac {1}{4} (x-1)^2+x-1\right )
\]