Internal problem ID [7260]
Internal file name [OUTPUT/6246_Sunday_June_05_2022_04_35_21_PM_68480318/index.tex]
Book: Own collection of miscellaneous problems
Section: section 4.0
Problem number: 36.
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]]
\[ \boxed {x \left (x -1\right ) y^{\prime \prime }+3 x y^{\prime }+y=0} \] 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. \[ \left (x^{2}-x \right ) y^{\prime \prime }+3 x y^{\prime }+y = 0 \] 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 {3}{x -1}\\ q(x) &= \frac {1}{x \left (x -1\right )}\\ \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 \[ x \left (x -1\right ) y^{\prime \prime }+3 x y^{\prime }+y = 0 \] 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} x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 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 (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}3 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ -x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ -x^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ -x^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r \left (-1+r \right ) = 0 \] 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 \[ -x^{-1+r} r \left (-1+r \right ) = 0 \] 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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -1} \left (n +r -1\right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {\left (n +r \right ) a_{n -1}}{n +r -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (n +1\right ) a_{n -1}}{n}\tag {5} \] 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 \[ a_{1}=\frac {1+r}{r} \] Which for the root \(r = 1\) becomes \[ a_{1}=2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {2+r}{r} \] Which for the root \(r = 1\) becomes \[ a_{2}=3 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {3+r}{r} \] Which for the root \(r = 1\) becomes \[ a_{3}=4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
| \(a_{3}\) | \(\frac {3+r}{r}\) | \(4\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {4+r}{r} \] Which for the root \(r = 1\) becomes \[ a_{4}=5 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
| \(a_{3}\) | \(\frac {3+r}{r}\) | \(4\) |
| \(a_{4}\) | \(\frac {4+r}{r}\) | \(5\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {5+r}{r} \] Which for the root \(r = 1\) becomes \[ a_{5}=6 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1+r}{r}\) | \(2\) |
| \(a_{2}\) | \(\frac {2+r}{r}\) | \(3\) |
| \(a_{3}\) | \(\frac {3+r}{r}\) | \(4\) |
| \(a_{4}\) | \(\frac {4+r}{r}\) | \(5\) |
| \(a_{5}\) | \(\frac {5+r}{r}\) | \(6\) |
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+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+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 {1+r}{r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1+r}{r}&= \lim _{r\rightarrow 0}\frac {1+r}{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 \(x \left (x -1\right ) y^{\prime \prime }+3 x y^{\prime }+y = 0\) gives \[ x \left (x -1\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+3 x \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x \left (x -1\right ) y_{1}^{\prime \prime }\left (x \right )+3 y_{1}^{\prime }\left (x \right ) x +y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (x -1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+3 y_{1}\left (x \right )\right ) C +x \left (x -1\right ) \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 )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{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 \[ x \left (x -1\right ) y_{1}^{\prime \prime }\left (x \right )+3 y_{1}^{\prime }\left (x \right ) x +y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (x -1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+3 y_{1}\left (x \right )\right ) C +x \left (x -1\right ) \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 )+3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{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} \frac {\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 (1+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {x^{2} \left (x -1\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 )+3 \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 (2 x \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} a_{n} \left (n +1\right )\right )+\left (1+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +1}\right )\right ) C}{x}+\frac {x^{2} \left (x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right )+3 \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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} \left (n +1\right )\right )+\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 }}C a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}2 C \,x^{n +1} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} \left (n -1\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n} a_{n} \left (n +1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}C a_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -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 \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} \left (n -1\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 C a_{n -1} n \,x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}C a_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{-2+n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} \left (-2+n \right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-n \,x^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 \[ -C +1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=1 \] 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 \[ -2-2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-1 \] For \(n=3\), Eq (2B) gives \[ \left (6 a_{1}-5 a_{2}\right ) C +9 b_{2}-6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -12-6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-2 \] For \(n=4\), Eq (2B) gives \[ \left (8 a_{2}-7 a_{3}\right ) C +16 b_{3}-12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -36-12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-3 \] For \(n=5\), Eq (2B) gives \[ \left (10 a_{3}-9 a_{4}\right ) C +25 b_{4}-20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -80-20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-4 \] 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 \(C=1\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= 1\eslowast \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right ) \] 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+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) + c_{2} \left (1\eslowast \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+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+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+1-x^{2}-2 x^{3}-3 x^{4}-4 x^{5}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) y^{\prime \prime }+3 x y^{\prime }+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 {y}{x \left (x -1\right )}-\frac {3 y^{\prime }}{x -1} \\ \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 {3 y^{\prime }}{x -1}+\frac {y}{x \left (x -1\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 {3}{x -1}, P_{3}\left (x \right )=\frac {1}{x \left (x -1\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}}}=0 \\ {} & \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} \\ {} & {} & x \left (x -1\right ) y^{\prime \prime }+3 x y^{\prime }+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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +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} \\ {} & {} & -a_{0} r \left (-1+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +r +1\right ) \left (k +r \right )+a_{k} \left (k +r +1\right )^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r +1\right ) \left (-a_{k +1} \left (k +r \right )+a_{k} \left (k +r +1\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r +1\right )}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +1\right )}{k} \\ \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} \left (k +1\right )}{k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1} \\ \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} \left (k +2\right )}{k +1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1}\right ), a_{k +1}=\frac {a_{k} \left (k +1\right )}{k}, b_{k +1}=\frac {b_{k} \left (k +2\right )}{k +1}\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: 0.0 (sec). Leaf size: 60
Order:=6; dsolve(x*(x-1)*diff(y(x), x$2) +3*x*diff(y(x),x)+y(x) = 0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (x +2 x^{2}+3 x^{3}+4 x^{4}+5 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \ln \left (x \right ) c_{2} +\left (1+3 x +5 x^{2}+7 x^{3}+9 x^{4}+11 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.039 (sec). Leaf size: 63
AsymptoticDSolveValue[x*(x-1)*y''[x] +3*x*y'[x]+y[x] == 0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (x^4+x^3+x^2+\left (4 x^3+3 x^2+2 x+1\right ) x \log (x)+x+1\right )+c_2 \left (5 x^5+4 x^4+3 x^3+2 x^2+x\right ) \]