Internal problem ID [13929]
Internal file name [OUTPUT/13101_Friday_February_23_2024_06_54_45_AM_88844513/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 33. Power series solutions I: Basic computational methods. Additional Exercises.
page 641
Problem number: 33.5 (h).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _homogeneous]]
\[ \boxed {\left (x^{2}-6 x \right ) y^{\prime \prime }+4 \left (x -3\right ) y^{\prime }+2 y=0} \] 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. \[ \left (x^{2}-6 x \right ) y^{\prime \prime }+\left (4 x -12\right ) y^{\prime }+2 y = 0 \] 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 {4 x -12}{x \left (x -6\right )}\\ q(x) &= \frac {2}{x \left (x -6\right )}\\ \end {align*}
| \(p(x)=\frac {4 x -12}{x \left (x -6\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 6\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2}{x \left (x -6\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 6\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 6, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x \left (x -6\right )+\left (4 x -12\right ) y^{\prime }+2 y = 0 \] 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} \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 (x -6\right )+\left (4 x -12\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-12 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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^{n +r} a_{n} \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 ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r -1} \\ \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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-12 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -6 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-12 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -6 x^{-1+r} a_{0} r \left (-1+r \right )-12 r a_{0} x^{-1+r} = 0 \] Or \[ \left (-6 x^{-1+r} r \left (-1+r \right )-12 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ -6 r \,x^{-1+r} \left (1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -6 r \left (1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ -6 r \,x^{-1+r} \left (1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -1]\).
Since \(r_1 - r_2 = 1\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x} \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 ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-6 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -1} \left (n +r -1\right )-12 a_{n} \left (n +r \right )+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{6}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {a_{n -1}}{6}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}={\frac {1}{6}} \] Which for the root \(r = 0\) becomes \[ a_{1}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}={\frac {1}{36}} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{36}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(a_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}={\frac {1}{216}} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{216}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(a_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
| \(a_{3}\) | \(\frac {1}{216}\) | \(\frac {1}{216}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}={\frac {1}{1296}} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{1296}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(a_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
| \(a_{3}\) | \(\frac {1}{216}\) | \(\frac {1}{216}\) |
| \(a_{4}\) | \(\frac {1}{1296}\) | \(\frac {1}{1296}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}={\frac {1}{7776}} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {1}{7776}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(a_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
| \(a_{3}\) | \(\frac {1}{216}\) | \(\frac {1}{216}\) |
| \(a_{4}\) | \(\frac {1}{1296}\) | \(\frac {1}{1296}\) |
| \(a_{5}\) | \(\frac {1}{7776}\) | \(\frac {1}{7776}\) |
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}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= {\frac {1}{6}} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}{\frac {1}{6}}&= \lim _{r\rightarrow -1}{\frac {1}{6}}\\ &= {\frac {1}{6}} \end {align*}
The limit is \(\frac {1}{6}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \end {align*}
Eq (3) derived above is 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{4} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-6 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -1} \left (n +r -1\right )-12 \left (n +r \right ) b_{n}+2 b_{n -1} = 0 \end{equation} Which for for the root \(r = -1\) becomes \begin{equation} \tag{4A} b_{n -1} \left (n -2\right ) \left (n -3\right )-6 b_{n} \left (n -1\right ) \left (n -2\right )+4 b_{n -1} \left (n -2\right )-12 \left (n -1\right ) b_{n}+2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {b_{n -1}}{6}\tag {5} \] Which for the root \(r = -1\) becomes \[ b_{n} = \frac {b_{n -1}}{6}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) 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 \[ b_{1}={\frac {1}{6}} \] Which for the root \(r = -1\) becomes \[ b_{1}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}={\frac {1}{36}} \] Which for the root \(r = -1\) becomes \[ b_{2}={\frac {1}{36}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(b_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}={\frac {1}{216}} \] Which for the root \(r = -1\) becomes \[ b_{3}={\frac {1}{216}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(b_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
| \(b_{3}\) | \(\frac {1}{216}\) | \(\frac {1}{216}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}={\frac {1}{1296}} \] Which for the root \(r = -1\) becomes \[ b_{4}={\frac {1}{1296}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(b_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
| \(b_{3}\) | \(\frac {1}{216}\) | \(\frac {1}{216}\) |
| \(b_{4}\) | \(\frac {1}{1296}\) | \(\frac {1}{1296}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}={\frac {1}{7776}} \] Which for the root \(r = -1\) becomes \[ b_{5}={\frac {1}{7776}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{6}\) | \(\frac {1}{6}\) |
| \(b_{2}\) | \(\frac {1}{36}\) | \(\frac {1}{36}\) |
| \(b_{3}\) | \(\frac {1}{216}\) | \(\frac {1}{216}\) |
| \(b_{4}\) | \(\frac {1}{1296}\) | \(\frac {1}{1296}\) |
| \(b_{5}\) | \(\frac {1}{7776}\) | \(\frac {1}{7776}\) |
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}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )}{x} \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}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{6}+\frac {x^{2}}{36}+\frac {x^{3}}{216}+\frac {x^{4}}{1296}+\frac {x^{5}}{7776}+O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x \left (x -6\right )+\left (4 x -12\right ) y^{\prime }+2 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 {2 y}{x \left (x -6\right )}-\frac {4 \left (x -3\right ) y^{\prime }}{x \left (x -6\right )} \\ \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 {4 \left (x -3\right ) y^{\prime }}{x \left (x -6\right )}+\frac {2 y}{x \left (x -6\right )}=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 {4 \left (x -3\right )}{x \left (x -6\right )}, P_{3}\left (x \right )=\frac {2}{x \left (x -6\right )}\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}}}=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} \\ {} & {} & y^{\prime \prime } x \left (x -6\right )+\left (4 x -12\right ) y^{\prime }+2 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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\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 y^{\prime \prime }=\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} \\ {} & {} & -6 a_{0} r \left (1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-6 a_{k +1} \left (k +r +1\right ) \left (k +r +2\right )+a_{k} \left (k +r +2\right ) \left (k +r +1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -6 r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r +2\right ) \left (k +r +1\right ) \left (-6 a_{k +1}+a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{6} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{6} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +1}=\frac {a_{k}}{6}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k}}{6} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {a_{k}}{6}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +1}=\frac {a_{k}}{6}, b_{k +1}=\frac {b_{k}}{6}\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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 44
Order:=6; dsolve((x^2-6*x)*diff(y(x),x$2)+4*(x-3)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1+\frac {1}{6} x +\frac {1}{36} x^{2}+\frac {1}{216} x^{3}+\frac {1}{1296} x^{4}+\frac {1}{7776} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {1}{3} x +\frac {1}{18} x^{2}+\frac {1}{108} x^{3}+\frac {1}{648} x^{4}+\frac {1}{3888} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.034 (sec). Leaf size: 64
AsymptoticDSolveValue[(x^2-6*x)*y''[x]+4*(x-3)*y'[x]+2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {x^3}{1296}+\frac {x^2}{216}+\frac {x}{36}+\frac {1}{x}+\frac {1}{6}\right )+c_2 \left (\frac {x^4}{1296}+\frac {x^3}{216}+\frac {x^2}{36}+\frac {x}{6}+1\right ) \]