Internal problem ID [7263]
Internal file name [OUTPUT/6249_Sunday_June_05_2022_04_35_32_PM_87536086/index.tex]
Book: Own collection of miscellaneous problems
Section: section 4.0
Problem number: 39.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {2 y^{\prime \prime } x^{2}+x y^{\prime }+\left (x -5\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ 2 y^{\prime \prime } x^{2}+x y^{\prime }+\left (x -5\right ) y = 0 \] 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*}
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 \[ 2 y^{\prime \prime } x^{2}+x y^{\prime }+\left (x -5\right ) y = 0 \] 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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2}+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*}
The solution(s) found are the following \begin{align*} \tag{1} y &= 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*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{4}+\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2-\sqrt {41}}+\frac {x^{2}}{2 \left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )}-\frac {x^{3}}{3240+510 \sqrt {41}}+\frac {x^{4}}{187320+29280 \sqrt {41}}-\frac {x^{5}}{600 \left (1561+244 \sqrt {41}\right ) \left (10+\sqrt {41}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}-\frac {\sqrt {41}}{4}} \left (1+\frac {x}{-2+\sqrt {41}}+\frac {x^{2}}{2 \left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )}+\frac {x^{3}}{-3240+510 \sqrt {41}}+\frac {x^{4}}{187320-29280 \sqrt {41}}-\frac {x^{5}}{600 \left (-1561+244 \sqrt {41}\right ) \left (-10+\sqrt {41}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime } x^{2}+x y^{\prime }+\left (x -5\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {y^{\prime }}{2 x}-\frac {\left (x -5\right ) y}{2 x^{2}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{2 x}+\frac {\left (x -5\right ) y}{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 y^{\prime \prime } x^{2}+x y^{\prime }+\left (x -5\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\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=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\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=\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=\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 (\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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 665
Order:=6; dsolve(2*x^2*diff(y(x), x, x) +x*diff(y(x), x) +(x-5)*y(x) = 0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {1}{4}} \left (c_{1} x^{-\frac {\sqrt {41}}{4}} \left (1+\frac {1}{-2+\sqrt {41}} x +\frac {1}{2} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right )} x^{2}+\frac {1}{6} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right ) \left (-6+\sqrt {41}\right )} x^{3}+\frac {1}{24} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right ) \left (-6+\sqrt {41}\right ) \left (-8+\sqrt {41}\right )} x^{4}+\frac {1}{120} \frac {1}{\left (-2+\sqrt {41}\right ) \left (-4+\sqrt {41}\right ) \left (-6+\sqrt {41}\right ) \left (-8+\sqrt {41}\right ) \left (-10+\sqrt {41}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\frac {\sqrt {41}}{4}} \left (1+\frac {1}{-2-\sqrt {41}} x +\frac {1}{2} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right )} x^{2}-\frac {1}{6} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right ) \left (6+\sqrt {41}\right )} x^{3}+\frac {1}{24} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right ) \left (6+\sqrt {41}\right ) \left (8+\sqrt {41}\right )} x^{4}-\frac {1}{120} \frac {1}{\left (2+\sqrt {41}\right ) \left (4+\sqrt {41}\right ) \left (6+\sqrt {41}\right ) \left (8+\sqrt {41}\right ) \left (10+\sqrt {41}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 1668
AsymptoticDSolveValue[2*x^2*y''[x]+x*y'[x]+(x-5)*y[x]==0,y[x],{x,0,5}]
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