Internal problem ID [4726]
Internal file name [OUTPUT/4219_Sunday_June_05_2022_12_42_52_PM_2605663/index.tex]
Book: A treatise on ordinary and partial differential equations by William Woolsey Johnson.
1913
Section: Chapter VII, Solutions in series. Examples XVI. page 220
Problem number: 6.
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]]
\[ \boxed {2 x \left (1-x \right ) y^{\prime \prime }+y^{\prime } x -y=0} \] 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. \[ \left (-2 x^{2}+2 x \right ) y^{\prime \prime }+y^{\prime } x -y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {1}{2 \left (x -1\right )}\\ q(x) &= \frac {1}{2 \left (x -1\right ) x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -2 y^{\prime \prime } x \left (x -1\right )+y^{\prime } x -y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x \left (x -1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x -\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 (-2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\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 }}\left (-2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{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 -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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{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 ) 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 \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ 2 x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 2 x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 2 x^{-1+r} r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).
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 ) &= x \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 )+\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 +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}\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} -2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (2 n^{2}+4 n r +2 r^{2}-7 n -7 r +6\right )}{2 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {a_{n -1} \left (2 n^{2}-3 n +1\right )}{2 \left (n +1\right ) n}\tag {5} \] 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 \[ a_{1}=\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r} \] Which for the root \(r = 1\) becomes \[ a_{1}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{3}-4 r^{2}-r +1}{4 \left (1+r \right )^{2} \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {4 r^{3}-4 r^{2}-r +1}{4 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8 r^{4}+4 r^{3}-14 r^{2}-r +3}{8 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {4 r^{3}-4 r^{2}-r +1}{4 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 r^{4}+4 r^{3}-14 r^{2}-r +3}{8 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r^{5}+48 r^{4}-8 r^{3}-72 r^{2}+r +15}{16 \left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {4 r^{3}-4 r^{2}-r +1}{4 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 r^{4}+4 r^{3}-14 r^{2}-r +3}{8 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {16 r^{5}+48 r^{4}-8 r^{3}-72 r^{2}+r +15}{16 \left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )^{2}}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {32 r^{6}+208 r^{5}+320 r^{4}-200 r^{3}-502 r^{2}+37 r +105}{32 \left (5+r \right ) \left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {4 r^{3}-4 r^{2}-r +1}{4 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 r^{4}+4 r^{3}-14 r^{2}-r +3}{8 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {16 r^{5}+48 r^{4}-8 r^{3}-72 r^{2}+r +15}{16 \left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )^{2}}\) | \(0\) |
| \(a_{5}\) | \(\frac {32 r^{6}+208 r^{5}+320 r^{4}-200 r^{3}-502 r^{2}+37 r +105}{32 \left (5+r \right ) \left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )^{2}}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 ) \\ &= x \left (1+O\left (x^{6}\right )\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 {2 r^{2}-3 r +1}{2 \left (1+r \right ) r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}&= \lim _{r\rightarrow 0}\frac {2 r^{2}-3 r +1}{2 \left (1+r \right ) r}\\ &= \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 \(-2 y^{\prime \prime } x \left (x -1\right )+y^{\prime } x -y = 0\) gives \[ -2 \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right ) x \left (x -1\right )+\left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) x -C y_{1}\left (x \right ) \ln \left (x \right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (-2 y_{1}^{\prime \prime }\left (x \right ) x \left (x -1\right )+y_{1}^{\prime }\left (x \right ) x -y_{1}\left (x \right )\right ) \ln \left (x \right )-2 \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x \left (x -1\right )+y_{1}\left (x \right )\right ) C -2 \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 \left (x -1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x -\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 \[ -2 y_{1}^{\prime \prime }\left (x \right ) x \left (x -1\right )+y_{1}^{\prime }\left (x \right ) x -y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (-2 \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) x \left (x -1\right )+y_{1}\left (x \right )\right ) C -2 \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 \left (x -1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x -\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} \frac {\left (-4 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 (3 x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {2 \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 (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x^{2}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (-4 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (n +1\right )\right )+\left (3 x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\right )\right ) C}{x}+\frac {2 \left (-x^{3}+x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right ) x^{2}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) x}{x} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +1} a_{n} \left (n +1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 C \,x^{n} a_{n} \left (n +1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n +1} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C \,x^{n +1} a_{n} \left (n +1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 C a_{-2+n} \left (n -1\right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 C \,x^{n} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 C a_{n -1} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 C \,x^{n +1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}3 C a_{-2+n} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C a_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} n \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 \left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -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 \(n -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 C a_{-2+n} \left (n -1\right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 C a_{n -1} n \,x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 C a_{-2+n} x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 \left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} x^{n -1}\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 \[ 2 C -1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C={\frac {1}{2}} \] For \(n=2\), Eq (2B) gives \[ \left (-a_{0}+6 a_{1}\right ) C +4 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {1}{2}+4 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {1}{8}} \] For \(n=3\), Eq (2B) gives \[ \left (-5 a_{1}+10 a_{2}\right ) C -3 b_{2}+12 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {3}{8}+12 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {1}{32}} \] For \(n=4\), Eq (2B) gives \[ \left (-9 a_{2}+14 a_{3}\right ) C -10 b_{3}+24 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {5}{16}+24 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {5}{384}} \] For \(n=5\), Eq (2B) gives \[ \left (-13 a_{3}+18 a_{4}\right ) C -21 b_{4}+40 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {35}{128}+40 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {7}{1024}} \] 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 \(C={\frac {1}{2}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \frac {1}{2}\eslowast \left (x \left (1+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1+\frac {x^{2}}{8}+\frac {x^{3}}{32}+\frac {5 x^{4}}{384}+\frac {7 x^{5}}{1024}+O\left (x^{6}\right ) \] Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x \left (1+O\left (x^{6}\right )\right ) + c_{2} \left (\frac {1}{2}\eslowast \left (x \left (1+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1+\frac {x^{2}}{8}+\frac {x^{3}}{32}+\frac {5 x^{4}}{384}+\frac {7 x^{5}}{1024}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1+O\left (x^{6}\right )\right )+c_{2} \left (\frac {x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1+\frac {x^{2}}{8}+\frac {x^{3}}{32}+\frac {5 x^{4}}{384}+\frac {7 x^{5}}{1024}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1+O\left (x^{6}\right )\right )+c_{2} \left (\frac {x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1+\frac {x^{2}}{8}+\frac {x^{3}}{32}+\frac {5 x^{4}}{384}+\frac {7 x^{5}}{1024}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1+O\left (x^{6}\right )\right )+c_{2} \left (\frac {x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+1+\frac {x^{2}}{8}+\frac {x^{3}}{32}+\frac {5 x^{4}}{384}+\frac {7 x^{5}}{1024}+O\left (x^{6}\right )\right ) \] 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: 0.016 (sec). Leaf size: 42
Order:=6; dsolve(2*x*(1-x)*diff(y(x),x$2)+x*diff(y(x),x)-y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\frac {1}{2} x +\operatorname {O}\left (x^{6}\right )\right ) \ln \left (x \right ) c_{2} +c_{1} x \left (1+\operatorname {O}\left (x^{6}\right )\right )+\left (1-\frac {1}{2} x +\frac {1}{8} x^{2}+\frac {1}{32} x^{3}+\frac {5}{384} x^{4}+\frac {7}{1024} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.051 (sec). Leaf size: 43
AsymptoticDSolveValue[2*x*(1-x)*y''[x]+x*y'[x]-y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{384} \left (5 x^4+12 x^3+48 x^2-768 x+384\right )+\frac {1}{2} x \log (x)\right )+c_2 x \]