Internal problem ID [7004]
Internal file name [OUTPUT/6247_Thursday_August_18_2022_07_11_39_AM_54095803/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. Miscellaneous Exercises. page 394
Problem number: 2.
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
[_Laguerre]
\[ \boxed {x y^{\prime \prime }-\left (x +2\right ) y^{\prime }-2 y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x y^{\prime \prime }+\left (-2-x \right ) y^{\prime }-2 y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {x +2}{x}\\ q(x) &= -\frac {2}{x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }+\left (-2-x \right ) y^{\prime }-2 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-2-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} 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 ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )-2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )-2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-3+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, 0]\).
Since \(r_1 - r_2 = 3\) 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^{3} \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 +3}\\ 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} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )-2 a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r +1\right )}{n^{2}+2 n r +r^{2}-3 n -3 r}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = \frac {a_{n -1} \left (n +4\right )}{n \left (n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 3\) 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 \[ a_{1}=\frac {2+r}{r^{2}-r -2} \] Which for the root \(r = 3\) becomes \[ a_{1}={\frac {5}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3+r}{r^{3}-2 r^{2}-r +2} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {3}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
| \(a_{2}\) | \(\frac {3+r}{r^{3}-2 r^{2}-r +2}\) | \(\frac {3}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )} \] Which for the root \(r = 3\) becomes \[ a_{3}={\frac {7}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
| \(a_{2}\) | \(\frac {3+r}{r^{3}-2 r^{2}-r +2}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}\) | \(\frac {7}{24}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r -2\right )} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {1}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
| \(a_{2}\) | \(\frac {3+r}{r^{3}-2 r^{2}-r +2}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}\) | \(\frac {7}{24}\) |
| \(a_{4}\) | \(\frac {5+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r -2\right )}\) | \(\frac {1}{12}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {6+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )} \] Which for the root \(r = 3\) becomes \[ a_{5}={\frac {3}{160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
| \(a_{2}\) | \(\frac {3+r}{r^{3}-2 r^{2}-r +2}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}\) | \(\frac {7}{24}\) |
| \(a_{4}\) | \(\frac {5+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r -2\right )}\) | \(\frac {1}{12}\) |
| \(a_{5}\) | \(\frac {6+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )}\) | \(\frac {3}{160}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {7+r}{\left (3+r \right ) \left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )} \] Which for the root \(r = 3\) becomes \[ a_{6}={\frac {1}{288}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
| \(a_{2}\) | \(\frac {3+r}{r^{3}-2 r^{2}-r +2}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}\) | \(\frac {7}{24}\) |
| \(a_{4}\) | \(\frac {5+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r -2\right )}\) | \(\frac {1}{12}\) |
| \(a_{5}\) | \(\frac {6+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )}\) | \(\frac {3}{160}\) |
| \(a_{6}\) | \(\frac {7+r}{\left (3+r \right ) \left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )}\) | \(\frac {1}{288}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {8+r}{\left (4+r \right ) \left (3+r \right ) \left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )} \] Which for the root \(r = 3\) becomes \[ a_{7}={\frac {11}{20160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2+r}{r^{2}-r -2}\) | \(\frac {5}{4}\) |
| \(a_{2}\) | \(\frac {3+r}{r^{3}-2 r^{2}-r +2}\) | \(\frac {3}{4}\) |
| \(a_{3}\) | \(\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}\) | \(\frac {7}{24}\) |
| \(a_{4}\) | \(\frac {5+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r -2\right )}\) | \(\frac {1}{12}\) |
| \(a_{5}\) | \(\frac {6+r}{\left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )}\) | \(\frac {3}{160}\) |
| \(a_{6}\) | \(\frac {7+r}{\left (3+r \right ) \left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )}\) | \(\frac {1}{288}\) |
| \(a_{7}\) | \(\frac {8+r}{\left (4+r \right ) \left (3+r \right ) \left (r +1\right )^{2} r \left (-1+r \right ) \left (r^{2}-4\right )}\) | \(\frac {11}{20160}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{3} \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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\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 \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). 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_{3} \\ &= \frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}&= \lim _{r\rightarrow 0}\frac {4+r}{r \left (r^{3}-2 r^{2}-r +2\right )}\\ &= \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 \(x y^{\prime \prime }+\left (-2-x \right ) y^{\prime }-2 y = 0\) gives \[ x \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-2-x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )-2 C y_{1}\left (x \right ) \ln \left (x \right )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x y_{1}^{\prime \prime }\left (x \right )+\left (-2-x \right ) y_{1}^{\prime }\left (x \right )-2 y_{1}\left (x \right )\right ) \ln \left (x \right )+x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-2-x \right ) y_{1}\left (x \right )}{x}\right ) C +x \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 )+\left (-2-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )-2 \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 \[ x y_{1}^{\prime \prime }\left (x \right )+\left (-2-x \right ) y_{1}^{\prime }\left (x \right )-2 y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-2-x \right ) y_{1}\left (x \right )}{x}\right ) C +x \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 )+\left (-2-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )-2 \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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -\left (x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\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 ) x^{2}+\left (-x^{2}-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x}{x} = 0 \end{equation} Since \(r_{1} = 3\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{2+n} a_{n} \left (n +3\right )\right ) x -\left (x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +3}\right )\right ) C}{x}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} n \left (n -1\right )\right ) x^{2}+\left (-x^{2}-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} n \right )-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^{2+n} a_{n} \left (n +3\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +3} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 C \,x^{2+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} b_{n} n \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n -1} b_{n} n \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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^{2+n} a_{n} \left (n +3\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} n \,x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +3} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-C a_{n -4} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 C \,x^{2+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-3 C a_{n -3} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n} b_{n} n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 b_{n} x^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 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} \left (\moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} n \,x^{n -1}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-C a_{n -4} x^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-3 C a_{n -3} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n -1} b_{n} \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\left (n -1\right ) b_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n -1} b_{n} n \right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 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 \(n=1\), Eq (2B) gives \[ -2 b_{1}-2 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -2 b_{1}-2 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=-1 \] For \(n=2\), Eq (2B) gives \[ -2 b_{2}-3 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -2 b_{2}+3 = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {3}{2}} \] For \(n=N\), where \(N=3\) which is the difference between the two roots, we are free to choose \(b_{3} = 0\). Hence for \(n=3\), Eq (2B) gives \[ 3 C -6 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=2 \] For \(n=4\), Eq (2B) gives \[ \left (-a_{0}+5 a_{1}\right ) C -5 b_{3}+4 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {21}{2}+4 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {21}{8}} \] For \(n=5\), Eq (2B) gives \[ \left (-a_{1}+7 a_{2}\right ) C -6 b_{4}+10 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {95}{4}+10 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {19}{8}} \] For \(n=6\), Eq (2B) gives \[ \left (-a_{2}+9 a_{3}\right ) C -7 b_{5}+18 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {163}{8}+18 b_{6} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}=-{\frac {163}{144}} \] For \(n=7\), Eq (2B) gives \[ \left (-a_{3}+11 a_{4}\right ) C -8 b_{6}+28 b_{7} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {371}{36}+28 b_{7} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}=-{\frac {53}{144}} \] 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 \(C=2\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= 2\eslowast \left (x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}-\frac {163 x^{6}}{144}-\frac {53 x^{7}}{144}+O\left (x^{8}\right ) \] 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^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right ) + c_{2} \left (2\eslowast \left (x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}-\frac {163 x^{6}}{144}-\frac {53 x^{7}}{144}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right )+c_{2} \left (2 x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}-\frac {163 x^{6}}{144}-\frac {53 x^{7}}{144}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right )+c_{2} \left (2 x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}-\frac {163 x^{6}}{144}-\frac {53 x^{7}}{144}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right )+c_{2} \left (2 x^{3} \left (1+\frac {5 x}{4}+\frac {3 x^{2}}{4}+\frac {7 x^{3}}{24}+\frac {x^{4}}{12}+\frac {3 x^{5}}{160}+\frac {x^{6}}{288}+\frac {11 x^{7}}{20160}+O\left (x^{8}\right )\right ) \ln \left (x \right )+1-x +\frac {3 x^{2}}{2}-\frac {21 x^{4}}{8}-\frac {19 x^{5}}{8}-\frac {163 x^{6}}{144}-\frac {53 x^{7}}{144}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (-2-x \right ) y^{\prime }-2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {2 y}{x}+\frac {\left (x +2\right ) y^{\prime }}{x} \\ \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 {\left (x +2\right ) y^{\prime }}{x}-\frac {2 y}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {x +2}{x}, P_{3}\left (x \right )=-\frac {2}{x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-2 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (-2-x \right ) y^{\prime }-2 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-3+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k -2+r \right )-a_{k} \left (k +r +2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k -2+r \right )-a_{k} \left (k +r +2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r +2\right )}{\left (k +1+r \right ) \left (k -2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) \left (k -2\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) \left (k -2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +5\right )}{\left (k +4\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +3}, a_{k +1}=\frac {a_{k} \left (k +5\right )}{\left (k +4\right ) \left (k +1\right )}\right ] \end {array} \]
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: 0.015 (sec). Leaf size: 70
Order:=8; dsolve(x*diff(y(x),x$2)-(2+x)*diff(y(x),x)-2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{3} \left (1+\frac {5}{4} x +\frac {3}{4} x^{2}+\frac {7}{24} x^{3}+\frac {1}{12} x^{4}+\frac {3}{160} x^{5}+\frac {1}{288} x^{6}+\frac {11}{20160} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (24 x^{3}+30 x^{4}+18 x^{5}+7 x^{6}+2 x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (12-12 x +18 x^{2}+26 x^{3}+x^{4}-9 x^{5}-6 x^{6}-\frac {9}{4} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.103 (sec). Leaf size: 115
AsymptoticDSolveValue[x*y''[x]-(2+x)*y'[x]-2*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {1}{12} \left (7 x^3+18 x^2+30 x+24\right ) x^3 \log (x)+\frac {1}{36} \left (-25 x^6-45 x^5-27 x^4+54 x^3+54 x^2-36 x+36\right )\right )+c_2 \left (\frac {x^9}{288}+\frac {3 x^8}{160}+\frac {x^7}{12}+\frac {7 x^6}{24}+\frac {3 x^5}{4}+\frac {5 x^4}{4}+x^3\right ) \]