Internal problem ID [13981]
Internal file name [OUTPUT/13153_Friday_February_23_2024_06_57_56_AM_31825060/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 35. Modified Power series solutions and basic method of Frobenius. Additional
Exercises. page 715
Problem number: 35.3 (d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+\frac {y^{\prime }}{x -3}+\frac {y}{x -4}=0} \] With the expansion point for the power series method at \(x = 3\).
The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let \[ t = x -3 \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \frac {d^{2}}{d t^{2}}y \left (t \right )+\frac {\frac {d}{d t}y \left (t \right )}{t}+\frac {y \left (t \right )}{t -1} = 0 \] With its expansion point and initial conditions now at \(t = 0\). The transformed ODE is now solved. The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \frac {d^{2}}{d t^{2}}y \left (t \right )+\frac {\frac {d}{d t}y \left (t \right )}{t}+\frac {y \left (t \right )}{t -1} = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} \frac {d^{2}}{d t^{2}}y \left (t \right )+p(t) \frac {d}{d t}y \left (t \right ) + q(t) y \left (t \right ) &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {1}{t}\\ q(t) &= \frac {1}{t -1}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t \left (t -1\right )+\left (\frac {d}{d t}y \left (t \right )\right ) \left (t -1\right )+y \left (t \right ) t = 0 \] Let the solution be represented as Frobenius power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{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} t^{n +r -2}\right ) t \left (t -1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right ) \left (t -1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) t = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}t^{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 ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} t^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) 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 ) t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -t^{-1+r} a_{0} r \left (-1+r \right )-r a_{0} t^{-1+r} = 0 \] Or \[ \left (-t^{-1+r} r \left (-1+r \right )-r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ -t^{-1+r} r^{2} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ -r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ -t^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y \left (t \right ) = c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution \(y_{1}\left (t \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {r^{2}}{\left (1+r \right )^{2}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -1}+2 n r a_{n -1}+r^{2} a_{n -1}-2 n a_{n -1}-2 r a_{n -1}+a_{n -2}+a_{n -1}}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (n -1\right )^{2} a_{n -1}+a_{n -2}}{n^{2}}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}}{\left (1+r \right )^{2}}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2}+1}{\left (r +2\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2}+1}{\left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {\left (r^{2}+r +1\right )^{2}}{\left (3+r \right )^{2} \left (1+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2}+1}{\left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}+r +1\right )^{2}}{\left (3+r \right )^{2} \left (1+r \right )^{2}}\) | \(\frac {1}{9}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{6}+6 r^{5}+16 r^{4}+24 r^{3}+23 r^{2}+14 r +5}{\left (r +4\right )^{2} \left (r +2\right )^{2} \left (1+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {5}{64}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2}+1}{\left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}+r +1\right )^{2}}{\left (3+r \right )^{2} \left (1+r \right )^{2}}\) | \(\frac {1}{9}\) |
| \(a_{4}\) | \(\frac {r^{6}+6 r^{5}+16 r^{4}+24 r^{3}+23 r^{2}+14 r +5}{\left (r +4\right )^{2} \left (r +2\right )^{2} \left (1+r \right )^{2}}\) | \(\frac {5}{64}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (r^{4}+6 r^{3}+12 r^{2}+9 r +7\right ) \left (r^{4}+6 r^{3}+14 r^{2}+15 r +7\right )}{\left (r +5\right )^{2} \left (3+r \right )^{2} \left (1+r \right )^{2} \left (r +2\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {49}{900}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2}+1}{\left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}+r +1\right )^{2}}{\left (3+r \right )^{2} \left (1+r \right )^{2}}\) | \(\frac {1}{9}\) |
| \(a_{4}\) | \(\frac {r^{6}+6 r^{5}+16 r^{4}+24 r^{3}+23 r^{2}+14 r +5}{\left (r +4\right )^{2} \left (r +2\right )^{2} \left (1+r \right )^{2}}\) | \(\frac {5}{64}\) |
| \(a_{5}\) | \(\frac {\left (r^{4}+6 r^{3}+12 r^{2}+9 r +7\right ) \left (r^{4}+6 r^{3}+14 r^{2}+15 r +7\right )}{\left (r +5\right )^{2} \left (3+r \right )^{2} \left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {49}{900}\) |
Using the above table, then the first solution \(y_{1}\left (t \right )\) becomes \begin{align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \\ &= 1+\frac {t^{2}}{4}+\frac {t^{3}}{9}+\frac {5 t^{4}}{64}+\frac {49 t^{5}}{900}+O\left (t^{6}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (t \right ) = y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {r^{2}}{\left (1+r \right )^{2}}\) | \(0\) | \(\frac {2 r}{\left (1+r \right )^{3}}\) | \(0\) |
| \(b_{2}\) | \(\frac {r^{2}+1}{\left (r +2\right )^{2}}\) | \(\frac {1}{4}\) | \(\frac {4 r -2}{\left (r +2\right )^{3}}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {\left (r^{2}+r +1\right )^{2}}{\left (3+r \right )^{2} \left (1+r \right )^{2}}\) | \(\frac {1}{9}\) | \(\frac {6 r^{4}+14 r^{3}+12 r^{2}+6 r -2}{\left (3+r \right )^{3} \left (1+r \right )^{3}}\) | \(-{\frac {2}{27}}\) |
| \(b_{4}\) | \(\frac {r^{6}+6 r^{5}+16 r^{4}+24 r^{3}+23 r^{2}+14 r +5}{\left (r +4\right )^{2} \left (r +2\right )^{2} \left (1+r \right )^{2}}\) | \(\frac {5}{64}\) | \(\frac {8 r^{7}+66 r^{6}+228 r^{5}+428 r^{4}+456 r^{3}+252 r^{2}+32 r -28}{\left (r +4\right )^{3} \left (r +2\right )^{3} \left (1+r \right )^{3}}\) | \(-{\frac {7}{128}}\) |
| \(b_{5}\) | \(\frac {\left (r^{4}+6 r^{3}+12 r^{2}+9 r +7\right ) \left (r^{4}+6 r^{3}+14 r^{2}+15 r +7\right )}{\left (r +5\right )^{2} \left (3+r \right )^{2} \left (1+r \right )^{2} \left (r +2\right )^{2}}\) | \(\frac {49}{900}\) | \(\frac {10 r^{10}+172 r^{9}+1302 r^{8}+5700 r^{7}+15906 r^{6}+29328 r^{5}+35658 r^{4}+27284 r^{3}+11202 r^{2}+736 r -938}{\left (r +5\right )^{3} \left (3+r \right )^{3} \left (1+r \right )^{3} \left (r +2\right )^{3}}\) | \(-{\frac {469}{13500}}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (t \right )&=y_{1}\left (t \right ) \ln \left (t \right )+b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}\dots \\ &= \left (1+\frac {t^{2}}{4}+\frac {t^{3}}{9}+\frac {5 t^{4}}{64}+\frac {49 t^{5}}{900}+O\left (t^{6}\right )\right ) \ln \left (t \right )-\frac {t^{2}}{4}-\frac {2 t^{3}}{27}-\frac {7 t^{4}}{128}-\frac {469 t^{5}}{13500}+O\left (t^{6}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (1+\frac {t^{2}}{4}+\frac {t^{3}}{9}+\frac {5 t^{4}}{64}+\frac {49 t^{5}}{900}+O\left (t^{6}\right )\right ) + c_{2} \left (\left (1+\frac {t^{2}}{4}+\frac {t^{3}}{9}+\frac {5 t^{4}}{64}+\frac {49 t^{5}}{900}+O\left (t^{6}\right )\right ) \ln \left (t \right )-\frac {t^{2}}{4}-\frac {2 t^{3}}{27}-\frac {7 t^{4}}{128}-\frac {469 t^{5}}{13500}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y \left (t \right ) &= y_h \\ &= c_{1} \left (1+\frac {t^{2}}{4}+\frac {t^{3}}{9}+\frac {5 t^{4}}{64}+\frac {49 t^{5}}{900}+O\left (t^{6}\right )\right )+c_{2} \left (\left (1+\frac {t^{2}}{4}+\frac {t^{3}}{9}+\frac {5 t^{4}}{64}+\frac {49 t^{5}}{900}+O\left (t^{6}\right )\right ) \ln \left (t \right )-\frac {t^{2}}{4}-\frac {2 t^{3}}{27}-\frac {7 t^{4}}{128}-\frac {469 t^{5}}{13500}+O\left (t^{6}\right )\right ) \\ \end{align*} Replacing \(t\) in the above with the original independent variable \(xs\)using \(t = x -3\) results in \[ y = c_{1} \left (1+\frac {\left (x -3\right )^{2}}{4}+\frac {\left (x -3\right )^{3}}{9}+\frac {5 \left (x -3\right )^{4}}{64}+\frac {49 \left (x -3\right )^{5}}{900}+O\left (\left (x -3\right )^{6}\right )\right )+c_{2} \left (\left (1+\frac {\left (x -3\right )^{2}}{4}+\frac {\left (x -3\right )^{3}}{9}+\frac {5 \left (x -3\right )^{4}}{64}+\frac {49 \left (x -3\right )^{5}}{900}+O\left (\left (x -3\right )^{6}\right )\right ) \ln \left (x -3\right )-\frac {\left (x -3\right )^{2}}{4}-\frac {2 \left (x -3\right )^{3}}{27}-\frac {7 \left (x -3\right )^{4}}{128}-\frac {469 \left (x -3\right )^{5}}{13500}+O\left (\left (x -3\right )^{6}\right )\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {\left (x -3\right )^{2}}{4}+\frac {\left (x -3\right )^{3}}{9}+\frac {5 \left (x -3\right )^{4}}{64}+\frac {49 \left (x -3\right )^{5}}{900}+O\left (\left (x -3\right )^{6}\right )\right )+c_{2} \left (\left (1+\frac {\left (x -3\right )^{2}}{4}+\frac {\left (x -3\right )^{3}}{9}+\frac {5 \left (x -3\right )^{4}}{64}+\frac {49 \left (x -3\right )^{5}}{900}+O\left (\left (x -3\right )^{6}\right )\right ) \ln \left (x -3\right )-\frac {\left (x -3\right )^{2}}{4}-\frac {2 \left (x -3\right )^{3}}{27}-\frac {7 \left (x -3\right )^{4}}{128}-\frac {469 \left (x -3\right )^{5}}{13500}+O\left (\left (x -3\right )^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {\left (x -3\right )^{2}}{4}+\frac {\left (x -3\right )^{3}}{9}+\frac {5 \left (x -3\right )^{4}}{64}+\frac {49 \left (x -3\right )^{5}}{900}+O\left (\left (x -3\right )^{6}\right )\right )+c_{2} \left (\left (1+\frac {\left (x -3\right )^{2}}{4}+\frac {\left (x -3\right )^{3}}{9}+\frac {5 \left (x -3\right )^{4}}{64}+\frac {49 \left (x -3\right )^{5}}{900}+O\left (\left (x -3\right )^{6}\right )\right ) \ln \left (x -3\right )-\frac {\left (x -3\right )^{2}}{4}-\frac {2 \left (x -3\right )^{3}}{27}-\frac {7 \left (x -3\right )^{4}}{128}-\frac {469 \left (x -3\right )^{5}}{13500}+O\left (\left (x -3\right )^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{x -3}+\frac {y}{x -4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {1}{x -3}, P_{3}\left (x \right )=\frac {1}{x -4}\right ] \\ {} & \circ & \left (x -3\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =3 \\ {} & {} & \left (\left (x -3\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}3}}}=1 \\ {} & \circ & \left (x -3\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =3 \\ {} & {} & \left (\left (x -3\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}3}}}=0 \\ {} & \circ & x =3\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}=3 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } \left (x -3\right ) \left (x -4\right )+y^{\prime } \left (x -4\right )+y \left (x -3\right )=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +3\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (u -1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+u y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & u \cdot y \left (u \right )=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r^{2} u^{-1+r}+\left (-a_{1} \left (1+r \right )^{2}+a_{0} r^{2}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k +1} \left (k +r +1\right )^{2}+a_{k} \left (k +r \right )^{2}+a_{k -1}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (1+r \right )^{2}+a_{0} r^{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1\right )^{2}+a_{k} k^{2}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -a_{k +2} \left (k +2\right )^{2}+a_{k +1} \left (k +1\right )^{2}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+2 k a_{k +1}+a_{k}+a_{k +1}}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k +1}+2 k a_{k +1}+a_{k}+a_{k +1}}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=\frac {k^{2} a_{k +1}+2 k a_{k +1}+a_{k}+a_{k +1}}{\left (k +2\right )^{2}}, -a_{1}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -3\right )^{k}, a_{k +2}=\frac {k^{2} a_{k +1}+2 k a_{k +1}+a_{k}+a_{k +1}}{\left (k +2\right )^{2}}, -a_{1}=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 -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 55
Order:=6; dsolve(diff(y(x),x$2)+1/(x-3)*diff(y(x),x)+1/(x-4)*y(x)=0,y(x),type='series',x=3);
\[ y \left (x \right ) = \left (\ln \left (-3+x \right ) c_{2} +c_{1} \right ) \left (1+\frac {1}{4} \left (-3+x \right )^{2}+\frac {1}{9} \left (-3+x \right )^{3}+\frac {5}{64} \left (-3+x \right )^{4}+\frac {49}{900} \left (-3+x \right )^{5}+\operatorname {O}\left (\left (-3+x \right )^{6}\right )\right )+\left (-\frac {1}{4} \left (-3+x \right )^{2}-\frac {2}{27} \left (-3+x \right )^{3}-\frac {7}{128} \left (-3+x \right )^{4}-\frac {469}{13500} \left (-3+x \right )^{5}+\operatorname {O}\left (\left (-3+x \right )^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 128
AsymptoticDSolveValue[y''[x]+1/(x-3)*y'[x]+1/(x-4)*y[x]==0,y[x],{x,3,5}]
\[ y(x)\to c_1 \left (\frac {49}{900} (x-3)^5+\frac {5}{64} (x-3)^4+\frac {1}{9} (x-3)^3+\frac {1}{4} (x-3)^2+1\right )+c_2 \left (-\frac {469 (x-3)^5}{13500}-\frac {7}{128} (x-3)^4-\frac {2}{27} (x-3)^3-\frac {1}{4} (x-3)^2+\left (\frac {49}{900} (x-3)^5+\frac {5}{64} (x-3)^4+\frac {1}{9} (x-3)^3+\frac {1}{4} (x-3)^2+1\right ) \log (x-3)\right ) \]