2.4.42 problem 39
Internal
problem
ID
[8357]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
39
Date
solved
:
Sunday, November 10, 2024 at 03:39:18 AM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
Solve
\begin{align*} 2 x^{2} y^{\prime \prime }+x y^{\prime }+\left (x -5\right ) y&=0 \end{align*}
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.
\[ 2 x^{2} y^{\prime \prime }+x y^{\prime }+\left (x -5\right ) y = 0 \]
The following is summary of singularities for the above ode. Writing the ode as
\begin{align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end{align*}
Where
\begin{align*} p(x) &= \frac {1}{2 x}\\ q(x) &= \frac {x -5}{2 x^{2}}\\ \end{align*}
Table 2.89: Table \(p(x),q(x)\) singularites.
| |
| \(p(x)=\frac {1}{2 x}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
| |
| \(q(x)=\frac {x -5}{2 x^{2}}\) |
| |
|
singularity | type |
| |
| \(x = 0\) | \(\text {``regular''}\) |
| |
Combining everything together gives the following summary of singularities for the ode
as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to
be
\[ 2 x^{2} y^{\prime \prime }+x y^{\prime }+\left (x -5\right ) y = 0 \]
Let the solution be represented as Frobenius power series of the form
\[
y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}
\]
Then
\begin{align*}
y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\
y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\
\end{align*}
Substituting the above back into the ode gives
\begin{equation}
\tag{1} 2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x -5\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0
\end{equation}
Which simplifies to
\begin{equation}
\tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) = 0
\end{equation}
The next step is to
make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation
term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and
the corresponding index gives
\begin{align*}
\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r} \\
\end{align*}
Substituting all the above in Eq (2A) gives the
following equation where now all powers of \(x\) are the same and equal to \(n +r\).
\begin{equation}
\tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) = 0
\end{equation}
The indicial
equation is obtained from \(n = 0\). From Eq (2B) this gives
\[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-5 a_{n} x^{n +r} = 0 \]
When \(n = 0\) the above becomes
\[ 2 x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -5 a_{0} x^{r} = 0 \]
Or
\[ \left (2 x^{r} r \left (-1+r \right )+x^{r} r -5 x^{r}\right ) a_{0} = 0 \]
Since \(a_{0}\neq 0\) then the above simplifies to
\[ \left (2 r^{2}-r -5\right ) x^{r} = 0 \]
Since the above is true for all \(x\) then the
indicial equation becomes
\[ 2 r^{2}-r -5 = 0 \]
Solving for \(r\) gives the roots of the indicial equation as
\begin{align*} r_1 &= \frac {1}{4}+\frac {\sqrt {41}}{4}\\ r_2 &= \frac {1}{4}-\frac {\sqrt {41}}{4} \end{align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes
\[ \left (2 r^{2}-r -5\right ) x^{r} = 0 \]
Solving for \(r\) gives the roots of the indicial equation
as \(\left [\frac {1}{4}+\frac {\sqrt {41}}{4}, \frac {1}{4}-\frac {\sqrt {41}}{4}\right ]\).
Since \(r_1 - r_2 = \frac {\sqrt {41}}{2}\) is not an integer, then we can construct two linearly independent solutions
\begin{align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end{align*}
Or
\begin{align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{4}+\frac {\sqrt {41}}{4}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{4}-\frac {\sqrt {41}}{4}} \end{align*}
We start by finding \(y_{1}\left (x \right )\). 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
\(a_{0} = 1\). For \(1\le n\) the recursive equation is
\begin{equation}
\tag{3} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -1}-5 a_{n} = 0
\end{equation}
Solving for \(a_{n}\) from recursive equation (4) gives
\[ a_{n} = -\frac {a_{n -1}}{2 n^{2}+4 n r +2 r^{2}-n -r -5}\tag {4} \]
Which for the
root \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) becomes
\[ a_{n} = -\frac {a_{n -1}}{n \left (\sqrt {41}+2 n \right )}\tag {5} \]
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before
substituting \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) 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
\[ a_{1}=-\frac {1}{2 r^{2}+3 r -4} \]
Which for the root \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) becomes
\[ a_{1}=\frac {1}{-2-\sqrt {41}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2-\sqrt {41}}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ a_{2}=\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4} \]
Which for the root \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) becomes
\[ a_{2}=\frac {1}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) | \(1\) |
| | |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2-\sqrt {41}}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ a_{3}=-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40} \]
Which for the root \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) becomes
\[ a_{3}=-\frac {1}{3240+510 \sqrt {41}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2-\sqrt {41}}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}\) |
| | |
| \(a_{3}\) |
\(-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40}\) |
\(-\frac {1}{3240+510 \sqrt {41}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ a_{4}=\frac {1}{16 r^{8}+288 r^{7}+2024 r^{6}+6912 r^{5}+11129 r^{4}+4662 r^{3}-7549 r^{2}-7362 r -920} \]
Which for the root \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) becomes
\[ a_{4}=\frac {1}{187320+29280 \sqrt {41}} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2-\sqrt {41}}\) |
| | |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}\) |
| | |
| \(a_{3}\) | \(-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40}\) | \(-\frac {1}{3240+510 \sqrt {41}}\) |
| | |
| \(a_{4}\) |
\(\frac {1}{16 r^{8}+288 r^{7}+2024 r^{6}+6912 r^{5}+11129 r^{4}+4662 r^{3}-7549 r^{2}-7362 r -920}\) |
\(\frac {1}{187320+29280 \sqrt {41}}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ a_{5}=-\frac {1}{32 r^{10}+880 r^{9}+10160 r^{8}+63800 r^{7}+234546 r^{6}+497255 r^{5}+518640 r^{4}+28325 r^{3}-443678 r^{2}-311960 r -36800} \]
Which for the root \(r = \frac {1}{4}+\frac {\sqrt {41}}{4}\) becomes
\[ a_{5}=-\frac {1}{600 \left (1561+244 \sqrt {41}\right ) \left (10+\sqrt {41}\right )} \]
And the table
now becomes
| | |
| \(n\) |
\(a_{n ,r}\) |
\(a_{n}\) |
| | |
| \(a_{0}\) |
\(1\) |
\(1\) |
| | |
| \(a_{1}\) |
\(-\frac {1}{2 r^{2}+3 r -4}\) |
\(\frac {1}{-2-\sqrt {41}}\) |
| | |
| \(a_{2}\) |
\(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}\) |
| | |
| \(a_{3}\) | \(-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40}\) | \(-\frac {1}{3240+510 \sqrt {41}}\) |
| | |
| \(a_{4}\) |
\(\frac {1}{16 r^{8}+288 r^{7}+2024 r^{6}+6912 r^{5}+11129 r^{4}+4662 r^{3}-7549 r^{2}-7362 r -920}\) |
\(\frac {1}{187320+29280 \sqrt {41}}\) |
| | |
| \(a_{5}\) |
\(-\frac {1}{32 r^{10}+880 r^{9}+10160 r^{8}+63800 r^{7}+234546 r^{6}+497255 r^{5}+518640 r^{4}+28325 r^{3}-443678 r^{2}-311960 r -36800}\) |
\(-\frac {1}{600 \left (1561+244 \sqrt {41}\right ) \left (10+\sqrt {41}\right )}\) |
| | |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
\begin{align*} y_{1}\left (x \right )&= x^{\frac {1}{4}+\frac {\sqrt {41}}{4}} \left (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 \right ) \\ &= x^{\frac {1}{4}+\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2-\sqrt {41}}+\frac {x^{2}}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}-\frac {x^{3}}{3240+510 \sqrt {41}}+\frac {x^{4}}{187320+29280 \sqrt {41}}-\frac {x^{5}}{600 \left (1561+244 \sqrt {41}\right ) \left (10+\sqrt {41}\right )}+O\left (x^{6}\right )\right ) \end{align*}
Now the second solution \(y_{2}\left (x \right )\) 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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is
\begin{equation}
\tag{3} 2 b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n} \left (n +r \right )+b_{n -1}-5 b_{n} = 0
\end{equation}
Solving for \(b_{n}\) from recursive equation (4) gives
\[ b_{n} = -\frac {b_{n -1}}{2 n^{2}+4 n r +2 r^{2}-n -r -5}\tag {4} \]
Which for the root \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) becomes
\[ b_{n} = \frac {b_{n -1}}{n \left (\sqrt {41}-2 n \right )}\tag {5} \]
At this point, it is a good idea to keep track of \(b_{n}\) in a table both
before substituting \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) and after as more terms are found using the above recursive equation.
| | |
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| | |
| \(b_{0}\) | \(1\) | \(1\) |
| | |
For \(n = 1\), using the above recursive equation gives
\[ b_{1}=-\frac {1}{2 r^{2}+3 r -4} \]
Which for the root \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) becomes
\[ b_{1}=\frac {1}{-2+\sqrt {41}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2+\sqrt {41}}\) |
| | |
For \(n = 2\), using the above recursive equation gives
\[ b_{2}=\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4} \]
Which for the root \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) becomes
\[ b_{2}=\frac {1}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) | \(1\) |
| | |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2+\sqrt {41}}\) |
| | |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}\) |
| | |
For \(n = 3\), using the above recursive equation gives
\[ b_{3}=-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40} \]
Which for the root \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) becomes
\[ b_{3}=\frac {1}{-3240+510 \sqrt {41}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2+\sqrt {41}}\) |
| | |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}\) |
| | |
| \(b_{3}\) |
\(-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40}\) |
\(\frac {1}{-3240+510 \sqrt {41}}\) |
| | |
For \(n = 4\), using the above recursive equation gives
\[ b_{4}=\frac {1}{16 r^{8}+288 r^{7}+2024 r^{6}+6912 r^{5}+11129 r^{4}+4662 r^{3}-7549 r^{2}-7362 r -920} \]
Which for the root \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) becomes
\[ b_{4}=\frac {1}{187320-29280 \sqrt {41}} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(-\frac {1}{2 r^{2}+3 r -4}\) | \(\frac {1}{-2+\sqrt {41}}\) |
| | |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}\) |
| | |
| \(b_{3}\) | \(-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40}\) | \(\frac {1}{-3240+510 \sqrt {41}}\) |
| | |
| \(b_{4}\) |
\(\frac {1}{16 r^{8}+288 r^{7}+2024 r^{6}+6912 r^{5}+11129 r^{4}+4662 r^{3}-7549 r^{2}-7362 r -920}\) |
\(\frac {1}{187320-29280 \sqrt {41}}\) |
| | |
For \(n = 5\), using the above recursive equation gives
\[ b_{5}=-\frac {1}{32 r^{10}+880 r^{9}+10160 r^{8}+63800 r^{7}+234546 r^{6}+497255 r^{5}+518640 r^{4}+28325 r^{3}-443678 r^{2}-311960 r -36800} \]
Which for the root \(r = \frac {1}{4}-\frac {\sqrt {41}}{4}\) becomes
\[ b_{5}=-\frac {1}{600 \left (-1561+244 \sqrt {41}\right ) \left (-10+\sqrt {41}\right )} \]
And the table
now becomes
| | |
| \(n\) |
\(b_{n ,r}\) |
\(b_{n}\) |
| | |
| \(b_{0}\) |
\(1\) |
\(1\) |
| | |
| \(b_{1}\) |
\(-\frac {1}{2 r^{2}+3 r -4}\) |
\(\frac {1}{-2+\sqrt {41}}\) |
| | |
| \(b_{2}\) |
\(\frac {1}{4 r^{4}+20 r^{3}+15 r^{2}-25 r -4}\) | \(\frac {1}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}\) |
| | |
| \(b_{3}\) | \(-\frac {1}{8 r^{6}+84 r^{5}+290 r^{4}+315 r^{3}-133 r^{2}-294 r -40}\) | \(\frac {1}{-3240+510 \sqrt {41}}\) |
| | |
| \(b_{4}\) |
\(\frac {1}{16 r^{8}+288 r^{7}+2024 r^{6}+6912 r^{5}+11129 r^{4}+4662 r^{3}-7549 r^{2}-7362 r -920}\) |
\(\frac {1}{187320-29280 \sqrt {41}}\) |
| | |
| \(b_{5}\) |
\(-\frac {1}{32 r^{10}+880 r^{9}+10160 r^{8}+63800 r^{7}+234546 r^{6}+497255 r^{5}+518640 r^{4}+28325 r^{3}-443678 r^{2}-311960 r -36800}\) |
\(-\frac {1}{600 \left (-1561+244 \sqrt {41}\right ) \left (-10+\sqrt {41}\right )}\) |
| | |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
\begin{align*} y_{2}\left (x \right )&= x^{\frac {1}{4}+\frac {\sqrt {41}}{4}} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= x^{\frac {1}{4}-\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2+\sqrt {41}}+\frac {x^{2}}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}+\frac {x^{3}}{-3240+510 \sqrt {41}}+\frac {x^{4}}{187320-29280 \sqrt {41}}-\frac {x^{5}}{600 \left (-1561+244 \sqrt {41}\right ) \left (-10+\sqrt {41}\right )}+O\left (x^{6}\right )\right ) \end{align*}
Therefore the homogeneous solution is
\begin{align*}
y_h(x) &= c_1 y_{1}\left (x \right )+c_2 y_{2}\left (x \right ) \\
&= c_1 \,x^{\frac {1}{4}+\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2-\sqrt {41}}+\frac {x^{2}}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}-\frac {x^{3}}{3240+510 \sqrt {41}}+\frac {x^{4}}{187320+29280 \sqrt {41}}-\frac {x^{5}}{600 \left (1561+244 \sqrt {41}\right ) \left (10+\sqrt {41}\right )}+O\left (x^{6}\right )\right ) + c_2 \,x^{\frac {1}{4}-\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2+\sqrt {41}}+\frac {x^{2}}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}+\frac {x^{3}}{-3240+510 \sqrt {41}}+\frac {x^{4}}{187320-29280 \sqrt {41}}-\frac {x^{5}}{600 \left (-1561+244 \sqrt {41}\right ) \left (-10+\sqrt {41}\right )}+O\left (x^{6}\right )\right ) \\
\end{align*}
Hence the final solution is
\begin{align*}
y &= y_h \\
&= c_1 \,x^{\frac {1}{4}+\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2-\sqrt {41}}+\frac {x^{2}}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}-\frac {x^{3}}{3240+510 \sqrt {41}}+\frac {x^{4}}{187320+29280 \sqrt {41}}-\frac {x^{5}}{600 \left (1561+244 \sqrt {41}\right ) \left (10+\sqrt {41}\right )}+O\left (x^{6}\right )\right )+c_2 \,x^{\frac {1}{4}-\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2+\sqrt {41}}+\frac {x^{2}}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}+\frac {x^{3}}{-3240+510 \sqrt {41}}+\frac {x^{4}}{187320-29280 \sqrt {41}}-\frac {x^{5}}{600 \left (-1561+244 \sqrt {41}\right ) \left (-10+\sqrt {41}\right )}+O\left (x^{6}\right )\right ) \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+\left (x -5\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 -5\right ) y \left (x \right )}{2 x^{2}}-\frac {\frac {d}{d x}y \left (x \right )}{2 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 )}{2 x}+\frac {\left (x -5\right ) y \left (x \right )}{2 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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}{2 x}, P_{3}\left (x \right )=\frac {x -5}{2 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}}}=-\frac {5}{2} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+x \left (\frac {d}{d x}y \left (x \right )\right )+\left (x -5\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 =0..1 \\ {} & {} & 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} x \cdot \left (\frac {d}{d x}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y \left (x \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (2 r^{2}-r -5\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k^{2}+4 k r +2 r^{2}-k -r -5\right )+a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r^{2}-r -5=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{4}-\frac {\sqrt {41}}{4}, \frac {1}{4}+\frac {\sqrt {41}}{4}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (2 k^{2}+\left (4 r -1\right ) k +2 r^{2}-r -5\right ) a_{k}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (2 \left (k +1\right )^{2}+\left (4 r -1\right ) \left (k +1\right )+2 r^{2}-r -5\right ) a_{k +1}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 k^{2}+4 k r +2 r^{2}+3 k +3 r -4} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4}-\frac {\sqrt {41}}{4} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 k^{2}+4 k \left (\frac {1}{4}-\frac {\sqrt {41}}{4}\right )+2 \left (\frac {1}{4}-\frac {\sqrt {41}}{4}\right )^{2}+3 k -\frac {13}{4}-\frac {3 \sqrt {41}}{4}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{4}-\frac {\sqrt {41}}{4} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{4}-\frac {\sqrt {41}}{4}}, a_{k +1}=-\frac {a_{k}}{2 k^{2}+4 k \left (\frac {1}{4}-\frac {\sqrt {41}}{4}\right )+2 \left (\frac {1}{4}-\frac {\sqrt {41}}{4}\right )^{2}+3 k -\frac {13}{4}-\frac {3 \sqrt {41}}{4}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4}+\frac {\sqrt {41}}{4} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 k^{2}+4 k \left (\frac {1}{4}+\frac {\sqrt {41}}{4}\right )+2 \left (\frac {1}{4}+\frac {\sqrt {41}}{4}\right )^{2}+3 k -\frac {13}{4}+\frac {3 \sqrt {41}}{4}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{4}+\frac {\sqrt {41}}{4} \\ {} & {} & \left [y \left (x \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{4}+\frac {\sqrt {41}}{4}}, a_{k +1}=-\frac {a_{k}}{2 k^{2}+4 k \left (\frac {1}{4}+\frac {\sqrt {41}}{4}\right )+2 \left (\frac {1}{4}+\frac {\sqrt {41}}{4}\right )^{2}+3 k -\frac {13}{4}+\frac {3 \sqrt {41}}{4}}\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 +\frac {1}{4}-\frac {\sqrt {41}}{4}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{4}+\frac {\sqrt {41}}{4}}\right ), a_{k +1}=-\frac {a_{k}}{2 k^{2}+4 k \left (\frac {1}{4}-\frac {\sqrt {41}}{4}\right )+2 \left (\frac {1}{4}-\frac {\sqrt {41}}{4}\right )^{2}+3 k -\frac {13}{4}-\frac {3 \sqrt {41}}{4}}, b_{k +1}=-\frac {b_{k}}{2 k^{2}+4 k \left (\frac {1}{4}+\frac {\sqrt {41}}{4}\right )+2 \left (\frac {1}{4}+\frac {\sqrt {41}}{4}\right )^{2}+3 k -\frac {13}{4}+\frac {3 \sqrt {41}}{4}}\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
<- Bessel successful
<- special function solution successful`
Maple dsolve solution
Solving time : 0.021 (sec)
Leaf size : 277
dsolve(2*x^2*diff(diff(y(x),x),x)+diff(y(x),x)*x+(x-5)*y(x) = 0,y(x),
series,x=0)
\[
y = x^{{1}/{4}} \left (c_{1} x^{-\frac {\sqrt {41}}{4}} \left (1+\frac {1}{-2+\sqrt {41}} x +\frac {1}{2} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )} x^{2}+\frac {1}{6} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right ) \left (-6+\sqrt {41}\right )} x^{3}+\frac {1}{24} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right ) \left (-6+\sqrt {41}\right ) \left (-8+\sqrt {41}\right )} x^{4}+\frac {1}{120} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right ) \left (-6+\sqrt {41}\right ) \left (-8+\sqrt {41}\right ) \left (-10+\sqrt {41}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\frac {\sqrt {41}}{4}} \left (1+\frac {1}{-2-\sqrt {41}} x +\frac {1}{2} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )} x^{2}-\frac {1}{6} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right ) \left (6+\sqrt {41}\right )} x^{3}+\frac {1}{24} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right ) \left (6+\sqrt {41}\right ) \left (8+\sqrt {41}\right )} x^{4}-\frac {1}{120} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right ) \left (6+\sqrt {41}\right ) \left (8+\sqrt {41}\right ) \left (10+\sqrt {41}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )
\]
Mathematica DSolve solution
Solving time : 0.006 (sec)
Leaf size : 1668
AsymptoticDSolveValue[{2*x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x-5)*y[x]==0,{}},
y[x],{x,0,5}]
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