2.4.15 problem 15
Internal
problem
ID
[8580]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
15
Date
solved
:
Thursday, December 12, 2024 at 09:31:25 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} \left (x -2\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+\left (x +1\right ) y&=0 \end{align*}
Using series expansion around \(x=2\)
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
\[ \tau = x -2 \]
The ode is converted to be in terms of the
new independent variable \(\tau \). This results in
\[
\tau \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+\frac {\frac {d}{d \tau }y \left (\tau \right )}{\tau +2}+\left (\tau +3\right ) y \left (\tau \right ) = 0
\]
With its expansion point and initial
conditions now at \(\tau = 0\). The transformed ODE is now solved. The type of the expansion
point is first determined. This is done on the homogeneous part of the ODE.
\[ \tau \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+\frac {\frac {d}{d \tau }y \left (\tau \right )}{\tau +2}+\left (\tau +3\right ) y \left (\tau \right ) = 0 \]
The following is summary of singularities for the above ode. Writing the ode as
\begin{align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p(\tau ) \frac {d}{d \tau }y \left (\tau \right ) + q(\tau ) y \left (\tau \right ) &=0 \end{align*}
Where
\begin{align*} p(\tau ) &= \frac {1}{\left (\tau +2\right ) \tau }\\ q(\tau ) &= \frac {\tau +3}{\tau }\\ \end{align*}
Table 2.69: Table \(p(\tau ),q(\tau )\) singularites.
| |
| \(p(\tau )=\frac {1}{\left (\tau +2\right ) \tau }\) |
| |
|
singularity | type |
| |
| \(\tau = -2\) | \(\text {``regular''}\) |
| |
| \(\tau = 0\) |
\(\text {``regular''}\) |
| |
| |
| \(q(\tau )=\frac {\tau +3}{\tau }\) |
| |
|
singularity | type |
| |
| \(\tau = 0\) | \(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([-2, 0]\)
Irregular singular points : \([\infty ]\)
Since \(\tau = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ \left (\tau +2\right ) \tau \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )+\frac {d}{d \tau }y \left (\tau \right )+\left (\tau +2\right ) \left (\tau +3\right ) y \left (\tau \right ) = 0 \]
Let the solution be represented as Frobenius power series of the form
\[
y \left (\tau \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +r}
\]
Then
\begin{align*}
\frac {d}{d \tau }y \left (\tau \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} \tau ^{n +r -1} \\
\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} \tau ^{n +r -2} \\
\end{align*}
Substituting the above back into the ode gives
\begin{equation}
\tag{1} \left (\tau +2\right ) \tau \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} \tau ^{n +r -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} \tau ^{n +r -1}\right )+\left (\tau +2\right ) \left (\tau +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +r}\right ) = 0
\end{equation}
Which simplifies to
\begin{equation}
\tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +r} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} \tau ^{n +r -1} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} \tau ^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +r +2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 \tau ^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} \tau ^{n +r}\right ) = 0
\end{equation}
The next step is to
make all powers of \(\tau \) be \(n +r -1\) in each summation term. Going over each summation
term above with power of \(\tau \) in it which is not already \(\tau ^{n +r -1}\) and adjusting the power and
the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +r} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) \tau ^{n +r -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +r +2} &= \moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} \tau ^{n +r -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}5 \tau ^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} \tau ^{n +r -1} \\
\moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} \tau ^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \tau ^{n +r -1} \\
\end{align*}
Substituting all the above in Eq (2A) gives the
following equation where now all powers of \(\tau \) are the same and equal to \(n +r -1\).
\begin{equation}
\tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) \tau ^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} \tau ^{n +r -1} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} \tau ^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} \tau ^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}5 a_{n -2} \tau ^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \tau ^{n +r -1}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 2 a_{n} \tau ^{n +r -1} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} \tau ^{n +r -1} = 0 \]
When \(n = 0\) the above becomes
\[ 2 a_{0} \tau ^{-1+r} r \left (-1+r \right )+r a_{0} \tau ^{-1+r} = 0 \]
Or
\[ \left (2 \tau ^{-1+r} r \left (-1+r \right )+r \,\tau ^{-1+r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ r \,\tau ^{-1+r} \left (2 r -1\right ) = 0 \]
Since the above is true for all \(\tau \) then the
indicial equation becomes
\[ 2 r^{2}-r = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ r \,\tau ^{-1+r} \left (2 r -1\right ) = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not an integer, then we can construct two linearly independent solutions
\begin{align*} y_{1}\left (\tau \right ) &= \tau ^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n}\right )\\ y_{2}\left (\tau \right ) &= \tau ^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} \tau ^{n}\right ) \end{align*}
Or
\begin{align*} y_{1}\left (\tau \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} \tau ^{n +\frac {1}{2}}\\ y_{2}\left (\tau \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} \tau ^{n} \end{align*}
We start by finding \(y_{1}\left (\tau \right )\). 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
\(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives
\[ a_{1} = \frac {-r^{2}+r -6}{2 r^{2}+3 r +1} \]
Substituting \(n = 2\) in Eq. (2B) gives
\[ a_{2} = \frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6} \]
For \(3\le n\) the recursive
equation is
\begin{equation}
\tag{3} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -3}+5 a_{n -2}+6 a_{n -1} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {n^{2} a_{n -1}+2 n r a_{n -1}+r^{2} a_{n -1}-3 n a_{n -1}-3 r a_{n -1}+a_{n -3}+5 a_{n -2}+8 a_{n -1}}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \]
Which for the root \(r = {\frac {1}{2}}\)
becomes
\[ a_{n} = \frac {-4 n^{2} a_{n -1}+8 n a_{n -1}-4 a_{n -3}-20 a_{n -2}-27 a_{n -1}}{8 n^{2}+4 n}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) | \(-{\frac {23}{12}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {127}{160}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{3}={\frac {1621}{40320}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) | \(-{\frac {23}{12}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {127}{160}\) |
| | |
| \(a_{3}\) |
\(\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )}\) |
\(\frac {1621}{40320}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {r^{8}+8 r^{7}+24 r^{6}+11 r^{5}-53 r^{4}-153 r^{3}-75 r^{2}-1458 r -897}{\left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{4}=-{\frac {426599}{5806080}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) | \(-{\frac {23}{12}}\) |
| | |
| \(a_{2}\) | \(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {127}{160}\) |
| | |
| \(a_{3}\) | \(\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )}\) | \(\frac {1621}{40320}\) |
| | |
| \(a_{4}\) |
\(\frac {r^{8}+8 r^{7}+24 r^{6}+11 r^{5}-53 r^{4}-153 r^{3}-75 r^{2}-1458 r -897}{\left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) |
\(-{\frac {426599}{5806080}}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=\frac {-r^{10}-15 r^{9}-92 r^{8}-270 r^{7}-316 r^{6}+276 r^{5}+970 r^{4}+207 r^{3}-4303 r^{2}+2370 r +13486}{\left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )} \]
Which for the root \(r = {\frac {1}{2}}\) becomes
\[ a_{5}={\frac {4670443}{425779200}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) |
\(-{\frac {23}{12}}\) |
| | |
| \(a_{2}\) |
\(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {127}{160}\) |
| | |
| \(a_{3}\) | \(\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )}\) | \(\frac {1621}{40320}\) |
| | |
| \(a_{4}\) |
\(\frac {r^{8}+8 r^{7}+24 r^{6}+11 r^{5}-53 r^{4}-153 r^{3}-75 r^{2}-1458 r -897}{\left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) |
\(-{\frac {426599}{5806080}}\) |
| | |
| \(a_{5}\) |
\(\frac {-r^{10}-15 r^{9}-92 r^{8}-270 r^{7}-316 r^{6}+276 r^{5}+970 r^{4}+207 r^{3}-4303 r^{2}+2370 r +13486}{\left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) |
\(\frac {4670443}{425779200}\) |
| | |
Using the above table, then the solution \(y_{1}\left (\tau \right )\) is
\begin{align*} y_{1}\left (\tau \right )&= \sqrt {\tau } \left (a_{0}+a_{1} \tau +a_{2} \tau ^{2}+a_{3} \tau ^{3}+a_{4} \tau ^{4}+a_{5} \tau ^{5}+a_{6} \tau ^{6}\dots \right ) \\ &= \sqrt {\tau }\, \left (1-\frac {23 \tau }{12}+\frac {127 \tau ^{2}}{160}+\frac {1621 \tau ^{3}}{40320}-\frac {426599 \tau ^{4}}{5806080}+\frac {4670443 \tau ^{5}}{425779200}+O\left (\tau ^{6}\right )\right ) \end{align*}
Now the second solution \(y_{2}\left (\tau \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients.
The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary
and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives
\[ b_{1} = \frac {-r^{2}+r -6}{2 r^{2}+3 r +1} \]
Substituting \(n = 2\) in Eq. (2B) gives
\[ b_{2} = \frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6} \]
For \(3\le n\) the
recursive equation is
\begin{equation}
\tag{3} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) b_{n}+b_{n -3}+5 b_{n -2}+6 b_{n -1} = 0
\end{equation}
Solving for \(b_{n}\) from recursive equation (4) gives
\[ b_{n} = -\frac {n^{2} b_{n -1}+2 n r b_{n -1}+r^{2} b_{n -1}-3 n b_{n -1}-3 r b_{n -1}+b_{n -3}+5 b_{n -2}+8 b_{n -1}}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \]
Which for the root \(r = 0\)
becomes
\[ b_{n} = \frac {-n^{2} b_{n -1}+3 n b_{n -1}-b_{n -3}-5 b_{n -2}-8 b_{n -1}}{n \left (2 n -1\right )}\tag {5} \]
At this point, it is a good idea to keep track of \(b_{n}\) in a table both before
substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) | \(-6\) |
| | |
| \(b_{2}\) | \(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {31}{6}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )} \]
Which for the root \(r = 0\) becomes
\[ b_{3}=-{\frac {37}{45}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) | \(-6\) |
| | |
| \(b_{2}\) | \(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {31}{6}\) |
| | |
| \(b_{3}\) |
\(\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )}\) |
\(-{\frac {37}{45}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {r^{8}+8 r^{7}+24 r^{6}+11 r^{5}-53 r^{4}-153 r^{3}-75 r^{2}-1458 r -897}{\left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )} \]
Which for the root \(r = 0\) becomes
\[ b_{4}=-{\frac {299}{840}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) | \(-6\) |
| | |
| \(b_{2}\) | \(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {31}{6}\) |
| | |
| \(b_{3}\) | \(\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )}\) | \(-{\frac {37}{45}}\) |
| | |
| \(b_{4}\) |
\(\frac {r^{8}+8 r^{7}+24 r^{6}+11 r^{5}-53 r^{4}-153 r^{3}-75 r^{2}-1458 r -897}{\left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) |
\(-{\frac {299}{840}}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=\frac {-r^{10}-15 r^{9}-92 r^{8}-270 r^{7}-316 r^{6}+276 r^{5}+970 r^{4}+207 r^{3}-4303 r^{2}+2370 r +13486}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+19 r +45\right )} \]
Which for the root \(r = 0\) becomes
\[ b_{5}={\frac {6743}{56700}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(\frac {-r^{2}+r -6}{2 r^{2}+3 r +1}\) |
\(-6\) |
| | |
| \(b_{2}\) |
\(\frac {r^{4}+r^{2}-15 r +31}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {31}{6}\) |
| | |
| \(b_{3}\) | \(\frac {-r^{6}-3 r^{5}-3 r^{4}+17 r^{3}+26 r^{2}+182 r -74}{\left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+7 r +6\right )}\) | \(-{\frac {37}{45}}\) |
| | |
| \(b_{4}\) |
\(\frac {r^{8}+8 r^{7}+24 r^{6}+11 r^{5}-53 r^{4}-153 r^{3}-75 r^{2}-1458 r -897}{\left (2 r^{2}+15 r +28\right ) \left (4 r^{4}+20 r^{3}+35 r^{2}+25 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) |
\(-{\frac {299}{840}}\) |
| | |
| \(b_{5}\) |
\(\frac {-r^{10}-15 r^{9}-92 r^{8}-270 r^{7}-316 r^{6}+276 r^{5}+970 r^{4}+207 r^{3}-4303 r^{2}+2370 r +13486}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+3 r +1\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+19 r +45\right )}\) |
\(\frac {6743}{56700}\) |
| | |
Using the above table, then the solution \(y_{2}\left (\tau \right )\) is
\begin{align*} y_{2}\left (\tau \right )&= b_{0}+b_{1} \tau +b_{2} \tau ^{2}+b_{3} \tau ^{3}+b_{4} \tau ^{4}+b_{5} \tau ^{5}+b_{6} \tau ^{6}\dots \\ &= 1-6 \tau +\frac {31 \tau ^{2}}{6}-\frac {37 \tau ^{3}}{45}-\frac {299 \tau ^{4}}{840}+\frac {6743 \tau ^{5}}{56700}+O\left (\tau ^{6}\right ) \end{align*}
Therefore the homogeneous solution is
\begin{align*}
y_h(\tau ) &= c_1 y_{1}\left (\tau \right )+c_2 y_{2}\left (\tau \right ) \\
&= c_1 \sqrt {\tau }\, \left (1-\frac {23 \tau }{12}+\frac {127 \tau ^{2}}{160}+\frac {1621 \tau ^{3}}{40320}-\frac {426599 \tau ^{4}}{5806080}+\frac {4670443 \tau ^{5}}{425779200}+O\left (\tau ^{6}\right )\right ) + c_2 \left (1-6 \tau +\frac {31 \tau ^{2}}{6}-\frac {37 \tau ^{3}}{45}-\frac {299 \tau ^{4}}{840}+\frac {6743 \tau ^{5}}{56700}+O\left (\tau ^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y \left (\tau \right ) &= y_h \\
&= c_1 \sqrt {\tau }\, \left (1-\frac {23 \tau }{12}+\frac {127 \tau ^{2}}{160}+\frac {1621 \tau ^{3}}{40320}-\frac {426599 \tau ^{4}}{5806080}+\frac {4670443 \tau ^{5}}{425779200}+O\left (\tau ^{6}\right )\right )+c_2 \left (1-6 \tau +\frac {31 \tau ^{2}}{6}-\frac {37 \tau ^{3}}{45}-\frac {299 \tau ^{4}}{840}+\frac {6743 \tau ^{5}}{56700}+O\left (\tau ^{6}\right )\right ) \\
\end{align*}
Replacing \(\tau \) in the above
with the original independent variable \(x\)using \(\tau = x -2\) results in
\[
y = c_1 \sqrt {x -2}\, \left (\frac {29}{6}-\frac {23 x}{12}+\frac {127 \left (x -2\right )^{2}}{160}+\frac {1621 \left (x -2\right )^{3}}{40320}-\frac {426599 \left (x -2\right )^{4}}{5806080}+\frac {4670443 \left (x -2\right )^{5}}{425779200}+O\left (\left (x -2\right )^{6}\right )\right )+c_2 \left (13-6 x +\frac {31 \left (x -2\right )^{2}}{6}-\frac {37 \left (x -2\right )^{3}}{45}-\frac {299 \left (x -2\right )^{4}}{840}+\frac {6743 \left (x -2\right )^{5}}{56700}+O\left (\left (x -2\right )^{6}\right )\right )
\]
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x -2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\frac {\frac {d}{d x}y \left (x \right )}{x}+\left (x +1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=-\frac {\left (x +1\right ) y \left (x \right )}{x -2}-\frac {\frac {d}{d x}y \left (x \right )}{\left (x -2\right ) x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+\frac {\frac {d}{d x}y \left (x \right )}{\left (x -2\right ) x}+\frac {\left (x +1\right ) y \left (x \right )}{x -2}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {1}{x \left (x -2\right )}, P_{3}\left (x \right )=\frac {x +1}{x -2}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {1}{2} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x -2\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\frac {d}{d x}y \left (x \right )+x \left (x +1\right ) y \left (x \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y \left (x \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y \left (x \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d x}y \left (x \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-3+2 r \right ) x^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (-1+2 r \right )+a_{0} r \left (-1+r \right )\right ) x^{r}+\left (-a_{2} \left (2+r \right ) \left (1+2 r \right )+a_{1} \left (1+r \right ) r +a_{0}\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k -1+2 r \right )+a_{k} \left (k +r \right ) \left (k +r -1\right )+a_{k -1}+a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (1+r \right ) \left (-1+2 r \right )+a_{0} r \left (-1+r \right )=0, -a_{2} \left (2+r \right ) \left (1+2 r \right )+a_{1} \left (1+r \right ) r +a_{0}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} r \left (-1+r \right )}{2 r^{2}+r -1}, a_{2}=\frac {a_{0} \left (r^{3}-r^{2}+2 r -1\right )}{4 r^{3}+8 r^{2}-r -2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +1+r \right ) \left (k +r -\frac {1}{2}\right ) a_{k +1}+a_{k} \left (k +r \right ) \left (k +r -1\right )+a_{k -1}+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -2 \left (k +3+r \right ) \left (k +\frac {3}{2}+r \right ) a_{k +3}+a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )+a_{k +1}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+2 k r a_{k +2}+r^{2} a_{k +2}+3 k a_{k +2}+3 r a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (k +3+r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+3 k a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (k +3\right ) \left (2 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {k^{2} a_{k +2}+3 k a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (k +3\right ) \left (2 k +3\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +2}+6 k a_{k +2}+a_{k}+a_{k +1}+\frac {35}{4} a_{k +2}}{\left (k +\frac {9}{2}\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +3}=\frac {k^{2} a_{k +2}+6 k a_{k +2}+a_{k}+a_{k +1}+\frac {35}{4} a_{k +2}}{\left (k +\frac {9}{2}\right ) \left (2 k +6\right )}, a_{1}=\frac {3 a_{0}}{20}, a_{2}=\frac {25 a_{0}}{224}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y \left (x \right )=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {3}{2}}\right ), a_{k +3}=\frac {k^{2} a_{k +2}+3 k a_{k +2}+a_{k}+a_{k +1}+2 a_{k +2}}{\left (k +3\right ) \left (2 k +3\right )}, a_{1}=0, a_{2}=\frac {a_{0}}{2}, b_{k +3}=\frac {k^{2} b_{k +2}+6 k b_{k +2}+b_{k}+b_{k +1}+\frac {35}{4} b_{k +2}}{\left (k +\frac {9}{2}\right ) \left (2 k +6\right )}, b_{1}=\frac {3 b_{0}}{20}, b_{2}=\frac {25 b_{0}}{224}\right ] \end {array} \]
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)]
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
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
-> hypergeometric
-> heuristic approach
-> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius
-> Mathieu
-> Equivalence to the rational form of Mathieu ODE under a power @ Moebius
trying a solution in terms of MeijerG functions
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius
<- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 `
Maple dsolve solution
Solving time : 0.030 (sec)
Leaf size : 46
dsolve((x-2)*diff(diff(y(x),x),x)+1/x*diff(y(x),x)+y(x)*(x+1) = 0,y(x),
series,x=2)
\[
y = c_{1} \sqrt {x -2}\, \left (1-\frac {23}{12} \left (x -2\right )+\frac {127}{160} \left (x -2\right )^{2}+\frac {1621}{40320} \left (x -2\right )^{3}-\frac {426599}{5806080} \left (x -2\right )^{4}+\frac {4670443}{425779200} \left (x -2\right )^{5}+\operatorname {O}\left (\left (x -2\right )^{6}\right )\right )+c_{2} \left (1-6 \left (x -2\right )+\frac {31}{6} \left (x -2\right )^{2}-\frac {37}{45} \left (x -2\right )^{3}-\frac {299}{840} \left (x -2\right )^{4}+\frac {6743}{56700} \left (x -2\right )^{5}+\operatorname {O}\left (\left (x -2\right )^{6}\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.011 (sec)
Leaf size : 105
AsymptoticDSolveValue[{(x-2)*D[y[x],{x,2}] + 1/x*D[y[x],x] + (x+1)*y[x] ==0,{}},
y[x],{x,2,5}]
\[
y(x)\to c_1 \left (\frac {4670443 (x-2)^5}{425779200}-\frac {426599 (x-2)^4}{5806080}+\frac {1621 (x-2)^3}{40320}+\frac {127}{160} (x-2)^2-\frac {23 (x-2)}{12}+1\right ) \sqrt {x-2}+c_2 \left (\frac {6743 (x-2)^5}{56700}-\frac {299}{840} (x-2)^4-\frac {37}{45} (x-2)^3+\frac {31}{6} (x-2)^2-6 (x-2)+1\right )
\]