Internal problem ID [1199]
Internal file name [OUTPUT/1200_Sunday_June_05_2022_02_04_55_AM_17975047/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.1 Exercises. Page
318
Problem number: 21.
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, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (1-x \right ) y^{\prime \prime }+x \left (4+x \right ) y^{\prime }+\left (2-x \right ) y=0} \] 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. \[ \left (-x^{3}+x^{2}\right ) y^{\prime \prime }+\left (x^{2}+4 x \right ) y^{\prime }+\left (2-x \right ) y = 0 \] 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 {4+x}{x \left (x -1\right )}\\ q(x) &= \frac {x -2}{x^{2} \left (x -1\right )}\\ \end {align*}
| \(p(x)=-\frac {4+x}{x \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x -2}{x^{2} \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -y^{\prime \prime } x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) y^{\prime }+\left (2-x \right ) y = 0 \] 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} -\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} \left (x -1\right )+\left (x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (2-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+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 ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\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 }}\left (-x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\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 \(n +r\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} 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 ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+4 x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+3 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+3 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -1\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+3 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-1, -2]\).
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 ) &= \frac {\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}}{x}\\ 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^{2}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\\ 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 -2}\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 )+a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+4 a_{n} \left (n +r \right )+2 a_{n}-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-4 n -4 r +4\right )}{n^{2}+2 n r +r^{2}+3 n +3 r +2}\tag {4} \] Which for the root \(r = -1\) becomes \[ a_{n} = \frac {a_{n -1} \left (n -3\right )^{2}}{n \left (1+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -1\) 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 \[ a_{1}=\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6} \] Which for the root \(r = -1\) becomes \[ a_{1}=2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2} \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )} \] Which for the root \(r = -1\) becomes \[ a_{2}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}\) | \(2\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )}\) | \(\frac {1}{3}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {\left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right )} \] Which for the root \(r = -1\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}\) | \(2\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )}\) | \(\frac {1}{3}\) |
| \(a_{3}\) | \(\frac {\left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r +2\right ) \left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +6\right ) \left (r +5\right )^{2} \left (r +4\right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}\) | \(2\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )}\) | \(\frac {1}{3}\) |
| \(a_{3}\) | \(\frac {\left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (r +2\right ) \left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +6\right ) \left (r +5\right )^{2} \left (r +4\right )^{2} \left (r +3\right )^{2}}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (r +2\right ) \left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +7\right ) \left (r +6\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = -1\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}\) | \(2\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )}\) | \(\frac {1}{3}\) |
| \(a_{3}\) | \(\frac {\left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (r +2\right ) \left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +6\right ) \left (r +5\right )^{2} \left (r +4\right )^{2} \left (r +3\right )^{2}}\) | \(0\) |
| \(a_{5}\) | \(\frac {\left (r +2\right ) \left (-1+r \right )^{2} r^{2} \left (r +1\right )^{2}}{\left (r +7\right ) \left (r +6\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \frac {1}{x} \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 ) \\ &= \frac {1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )}{x} \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 {\left (-1+r \right )^{2}}{r^{2}+5 r +6} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}&= \lim _{r\rightarrow -2}\frac {\left (-1+r \right )^{2}}{r^{2}+5 r +6}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(-y^{\prime \prime } x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) y^{\prime }+\left (2-x \right ) y = 0\) gives \[ -\left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (2-x \right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (-y_{1}^{\prime \prime }\left (x \right ) x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) y_{1}^{\prime }\left (x \right )+\left (2-x \right ) y_{1}\left (x \right )\right ) \ln \left (x \right )-\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x^{2} \left (x -1\right )+\frac {\left (x^{2}+4 x \right ) y_{1}\left (x \right )}{x}\right ) C -\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (2-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ -y_{1}^{\prime \prime }\left (x \right ) x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) y_{1}^{\prime }\left (x \right )+\left (2-x \right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (-\left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x^{2} \left (x -1\right )+\frac {\left (x^{2}+4 x \right ) y_{1}\left (x \right )}{x}\right ) C -\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x^{2} \left (x -1\right )+\left (x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (2-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \left (-2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (2 x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\left (-x^{3}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right )+\left (x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )-\left (x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = -1\) and \(r_{2} = -2\) then the above becomes \begin{equation} \tag{10} \left (-2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} a_{n} \left (n -1\right )\right )+\left (2 x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\right )\right ) C +\left (-x^{3}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-4+n} b_{n} \left (n -2\right ) \left (n -3\right )\right )+\left (x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -3} b_{n} \left (n -2\right )\right )-\left (x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n} a_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n -1} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n -1} \left (n -2\right ) \left (n -3\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n -2} b_{n} \left (n -2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n -1} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 b_{n} x^{n -2}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -2\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -2}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n} a_{n} \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} \left (n -3\right ) x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n -1} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} \left (n -2\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C a_{n} x^{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n -1} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 C a_{n -1} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n -1} \left (n -2\right ) \left (n -3\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} \left (-4+n \right ) \left (n -3\right ) x^{n -2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (n -2\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (n -3\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n -1} b_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -2}\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 \(n -2\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{n -2} \left (n -3\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} \left (n -2\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 C a_{n -1} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} \left (-4+n \right ) \left (n -3\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (n -3\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n -2} b_{n} \left (n -2\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 b_{n} x^{n -2}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives \[ C -9 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=9 \] For \(n=2\), Eq (2B) gives \[ \left (4 a_{0}+3 a_{1}\right ) C -4 b_{1}+2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 90+2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-45 \] For \(n=3\), Eq (2B) gives \[ \left (2 a_{1}+5 a_{2}\right ) C -b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 96+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-16 \] For \(n=4\), Eq (2B) gives \[ 7 C a_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=0 \] For \(n=5\), Eq (2B) gives \[ \left (-2 a_{3}+9 a_{4}\right ) C -b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=0 \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Using the above value found for \(C=9\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= 9\eslowast \left (\frac {1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )}{x}\right ) \ln \left (x \right )+\frac {1-45 x^{2}-16 x^{3}+O\left (x^{6}\right )}{x^{2}} \] Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \frac {c_{1} \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right )}{x} + c_{2} \left (9\eslowast \left (\frac {1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )}{x}\right ) \ln \left (x \right )+\frac {1-45 x^{2}-16 x^{3}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right )}{x}+c_{2} \left (\frac {9 \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x}+\frac {1-45 x^{2}-16 x^{3}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right )}{x}+c_{2} \left (\frac {9 \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x}+\frac {1-45 x^{2}-16 x^{3}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right )}{x}+c_{2} \left (\frac {9 \left (1+2 x +\frac {x^{2}}{3}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x}+\frac {1-45 x^{2}-16 x^{3}+O\left (x^{6}\right )}{x^{2}}\right ) \] Verified OK.
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 55
Order:=6; dsolve(x^2*(1-x)*diff(y(x),x$2)+x*(4+x)*diff(y(x),x)+(2-x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\ln \left (x \right ) \left (9 x +18 x^{2}+3 x^{3}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} +c_{1} \left (1+2 x +\frac {1}{3} x^{2}+\operatorname {O}\left (x^{6}\right )\right ) x +\left (1-5 x -55 x^{2}-\frac {53}{3} x^{3}+\operatorname {O}\left (x^{6}\right )\right ) c_{2}}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.049 (sec). Leaf size: 56
AsymptoticDSolveValue[x^2*(1-x)*y''[x]+x*(4+x)*y'[x]+(2-x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {3 \left (x^2+6 x+3\right ) \log (x)}{x}-\frac {21 x^3+75 x^2+15 x-1}{x^2}\right )+c_2 \left (\frac {x}{3}+\frac {1}{x}+2\right ) \]