Internal problem ID [14807]
Internal file name [OUTPUT/14487_Monday_April_08_2024_06_25_53_AM_6070828/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.9, page 215
Problem number: 7.
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]]
\[ \boxed {9 x y^{\prime \prime }+14 y^{\prime }+y \left (x -1\right )=0} \] 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. \[ 9 x y^{\prime \prime }+14 y^{\prime }+y \left (x -1\right ) = 0 \] 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 {14}{9 x}\\ q(x) &= \frac {x -1}{9 x}\\ \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 \[ 9 x y^{\prime \prime }+14 y^{\prime }+y \left (x -1\right ) = 0 \] 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} 9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x +14 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) \left (x -1\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) 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 =2}{\sum }}a_{n -2} x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\right ) \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+14 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{-1+r} a_{0} r \left (-1+r \right )+14 r a_{0} x^{-1+r} = 0 \] Or \[ \left (9 x^{-1+r} r \left (-1+r \right )+14 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (5+9 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 9 r^{2}+5 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {5}{9}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (5+9 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {5}{9}}\right ]\).
Since \(r_1 - r_2 = {\frac {5}{9}}\) 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}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {5}{9}} \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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {1}{9 r^{2}+23 r +14} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 9 a_{n} \left (n +r \right ) \left (n +r -1\right )+14 a_{n} \left (n +r \right )+a_{n -2}-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -2}-a_{n -1}}{9 n^{2}+18 n r +9 r^{2}+5 n +5 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {-a_{n -2}+a_{n -1}}{n \left (9 n +5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{14}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )} \] Which for the root \(r = 0\) becomes \[ a_{2}=-{\frac {13}{644}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{14}\) |
| \(a_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {13}{644}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {59}{61824}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{14}\) |
| \(a_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {13}{644}}\) |
| \(a_{3}\) | \(\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )}\) | \(-{\frac {59}{61824}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r^{4}+738 r^{3}+2320 r^{2}+2911 r +1189}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {29}{247296}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{14}\) |
| \(a_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {13}{644}}\) |
| \(a_{3}\) | \(\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )}\) | \(-{\frac {59}{61824}}\) |
| \(a_{4}\) | \(\frac {81 r^{4}+738 r^{3}+2320 r^{2}+2911 r +1189}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right )}\) | \(\frac {29}{247296}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {243 r^{4}+2700 r^{3}+10731 r^{2}+17950 r +10865}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right ) \left (9 r^{2}+95 r +250\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {53}{12364800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{14}\) |
| \(a_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {13}{644}}\) |
| \(a_{3}\) | \(\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )}\) | \(-{\frac {59}{61824}}\) |
| \(a_{4}\) | \(\frac {81 r^{4}+738 r^{3}+2320 r^{2}+2911 r +1189}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right )}\) | \(\frac {29}{247296}\) |
| \(a_{5}\) | \(\frac {243 r^{4}+2700 r^{3}+10731 r^{2}+17950 r +10865}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right ) \left (9 r^{2}+95 r +250\right )}\) | \(\frac {53}{12364800}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1+\frac {x}{14}-\frac {13 x^{2}}{644}-\frac {59 x^{3}}{61824}+\frac {29 x^{4}}{247296}+\frac {53 x^{5}}{12364800}+O\left (x^{6}\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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {1}{9 r^{2}+23 r +14} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 9 b_{n} \left (n +r \right ) \left (n +r -1\right )+14 \left (n +r \right ) b_{n}+b_{n -2}-b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -2}-b_{n -1}}{9 n^{2}+18 n r +9 r^{2}+5 n +5 r}\tag {4} \] Which for the root \(r = -{\frac {5}{9}}\) becomes \[ b_{n} = \frac {-b_{n -2}+b_{n -1}}{n \left (9 n -5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {5}{9}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{4}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )} \] Which for the root \(r = -{\frac {5}{9}}\) becomes \[ b_{2}=-{\frac {3}{104}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {3}{104}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )} \] Which for the root \(r = -{\frac {5}{9}}\) becomes \[ b_{3}=-{\frac {29}{6864}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {3}{104}}\) |
| \(b_{3}\) | \(\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )}\) | \(-{\frac {29}{6864}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {81 r^{4}+738 r^{3}+2320 r^{2}+2911 r +1189}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right )} \] Which for the root \(r = -{\frac {5}{9}}\) becomes \[ b_{4}={\frac {13}{65472}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {3}{104}}\) |
| \(b_{3}\) | \(\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )}\) | \(-{\frac {29}{6864}}\) |
| \(b_{4}\) | \(\frac {81 r^{4}+738 r^{3}+2320 r^{2}+2911 r +1189}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right )}\) | \(\frac {13}{65472}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {243 r^{4}+2700 r^{3}+10731 r^{2}+17950 r +10865}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right ) \left (9 r^{2}+95 r +250\right )} \] Which for the root \(r = -{\frac {5}{9}}\) becomes \[ b_{5}={\frac {251}{11348480}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{9 r^{2}+23 r +14}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {-9 r^{2}-23 r -13}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right )}\) | \(-{\frac {3}{104}}\) |
| \(b_{3}\) | \(\frac {-18 r^{2}-64 r -59}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right )}\) | \(-{\frac {29}{6864}}\) |
| \(b_{4}\) | \(\frac {81 r^{4}+738 r^{3}+2320 r^{2}+2911 r +1189}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right )}\) | \(\frac {13}{65472}\) |
| \(b_{5}\) | \(\frac {243 r^{4}+2700 r^{3}+10731 r^{2}+17950 r +10865}{\left (9 r^{2}+23 r +14\right ) \left (9 r^{2}+41 r +46\right ) \left (9 r^{2}+59 r +96\right ) \left (9 r^{2}+77 r +164\right ) \left (9 r^{2}+95 r +250\right )}\) | \(\frac {251}{11348480}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \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 ) \\ &= \frac {1+\frac {x}{4}-\frac {3 x^{2}}{104}-\frac {29 x^{3}}{6864}+\frac {13 x^{4}}{65472}+\frac {251 x^{5}}{11348480}+O\left (x^{6}\right )}{x^{\frac {5}{9}}} \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} \left (1+\frac {x}{14}-\frac {13 x^{2}}{644}-\frac {59 x^{3}}{61824}+\frac {29 x^{4}}{247296}+\frac {53 x^{5}}{12364800}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {x}{4}-\frac {3 x^{2}}{104}-\frac {29 x^{3}}{6864}+\frac {13 x^{4}}{65472}+\frac {251 x^{5}}{11348480}+O\left (x^{6}\right )\right )}{x^{\frac {5}{9}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+\frac {x}{14}-\frac {13 x^{2}}{644}-\frac {59 x^{3}}{61824}+\frac {29 x^{4}}{247296}+\frac {53 x^{5}}{12364800}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{4}-\frac {3 x^{2}}{104}-\frac {29 x^{3}}{6864}+\frac {13 x^{4}}{65472}+\frac {251 x^{5}}{11348480}+O\left (x^{6}\right )\right )}{x^{\frac {5}{9}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {x}{14}-\frac {13 x^{2}}{644}-\frac {59 x^{3}}{61824}+\frac {29 x^{4}}{247296}+\frac {53 x^{5}}{12364800}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{4}-\frac {3 x^{2}}{104}-\frac {29 x^{3}}{6864}+\frac {13 x^{4}}{65472}+\frac {251 x^{5}}{11348480}+O\left (x^{6}\right )\right )}{x^{\frac {5}{9}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {x}{14}-\frac {13 x^{2}}{644}-\frac {59 x^{3}}{61824}+\frac {29 x^{4}}{247296}+\frac {53 x^{5}}{12364800}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{4}-\frac {3 x^{2}}{104}-\frac {29 x^{3}}{6864}+\frac {13 x^{4}}{65472}+\frac {251 x^{5}}{11348480}+O\left (x^{6}\right )\right )}{x^{\frac {5}{9}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 y^{\prime \prime } x +14 y^{\prime }+y \left (x -1\right )=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 {\left (x -1\right ) y}{9 x}-\frac {14 y^{\prime }}{9 x} \\ \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 {14 y^{\prime }}{9 x}+\frac {\left (x -1\right ) y}{9 x}=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 {14}{9 x}, P_{3}\left (x \right )=\frac {x -1}{9 x}\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 {14}{9} \\ {} & \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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 9 y^{\prime \prime } x +14 y^{\prime }+y \left (x -1\right )=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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\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 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (5+9 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (14+9 r \right )-a_{0}\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (9 k +14+9 r \right )-a_{k}+a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (5+9 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {5}{9}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (14+9 r \right )-a_{0}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (k +1+r \right ) \left (k +\frac {14}{9}+r \right ) a_{k +1}-a_{k}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 9 \left (k +2+r \right ) \left (k +\frac {23}{9}+r \right ) a_{k +2}-a_{k +1}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {-a_{k +1}+a_{k}}{\left (k +2+r \right ) \left (9 k +23+9 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {-a_{k +1}+a_{k}}{\left (k +2\right ) \left (9 k +23\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {-a_{k +1}+a_{k}}{\left (k +2\right ) \left (9 k +23\right )}, 14 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {5}{9} \\ {} & {} & a_{k +2}=-\frac {-a_{k +1}+a_{k}}{\left (k +\frac {13}{9}\right ) \left (9 k +18\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {5}{9} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {5}{9}}, a_{k +2}=-\frac {-a_{k +1}+a_{k}}{\left (k +\frac {13}{9}\right ) \left (9 k +18\right )}, 4 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {5}{9}}\right ), a_{k +2}=-\frac {-a_{k +1}+a_{k}}{\left (k +2\right ) \left (9 k +23\right )}, 14 a_{1}-a_{0}=0, b_{k +2}=-\frac {-b_{k +1}+b_{k}}{\left (k +\frac {13}{9}\right ) \left (9 k +18\right )}, 4 b_{1}-b_{0}=0\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 <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 44
Order:=6; dsolve(9*x*diff(y(x),x$2)+14*diff(y(x),x)+(x-1)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1+\frac {1}{4} x -\frac {3}{104} x^{2}-\frac {29}{6864} x^{3}+\frac {13}{65472} x^{4}+\frac {251}{11348480} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {5}{9}}}+c_{2} \left (1+\frac {1}{14} x -\frac {13}{644} x^{2}-\frac {59}{61824} x^{3}+\frac {29}{247296} x^{4}+\frac {53}{12364800} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 85
AsymptoticDSolveValue[9*x*y''[x]+14*y'[x]+(x-1)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {53 x^5}{12364800}+\frac {29 x^4}{247296}-\frac {59 x^3}{61824}-\frac {13 x^2}{644}+\frac {x}{14}+1\right )+\frac {c_2 \left (\frac {251 x^5}{11348480}+\frac {13 x^4}{65472}-\frac {29 x^3}{6864}-\frac {3 x^2}{104}+\frac {x}{4}+1\right )}{x^{5/9}} \]