Internal problem ID [1201]
Internal file name [OUTPUT/1202_Sunday_June_05_2022_02_05_02_AM_53203217/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: 23.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (x +1\right ) y^{\prime \prime }-x \left (-x^{2}-6 x +1\right ) y^{\prime }+\left (x^{2}+6 x +1\right ) y=0} \] 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. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (x^{3}+6 x^{2}-x \right ) y^{\prime }+\left (x^{2}+6 x +1\right ) y = 0 \] 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 {x^{2}+6 x -1}{x \left (x +1\right )}\\ q(x) &= \frac {x^{2}+6 x +1}{x^{2} \left (x +1\right )}\\ \end {align*}
| \(p(x)=\frac {x^{2}+6 x -1}{x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x^{2}+6 x +1}{x^{2} \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x +1\right ) y^{\prime \prime }+\left (x^{3}+6 x^{2}-x \right ) y^{\prime }+\left (x^{2}+6 x +1\right ) y = 0 \] 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} x^{2} \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 )+\left (x^{3}+6 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x^{2}+6 x +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^{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^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n}\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 \(n +r\) 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 }}x^{1+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} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} x^{n +r} \\ \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} \left (\moverset {\infty }{\munderset {n =1}{\sum }}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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-1+r \right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (-1+r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-1+r \right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 1]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = 1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{1+n}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {-r^{2}-5 r -6}{r^{2}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -2} \left (n +r -2\right )+6 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n -2}+6 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -1}+2 n r a_{n -1}+r^{2} a_{n -1}+n a_{n -2}+3 n a_{n -1}+r a_{n -2}+3 r a_{n -1}-a_{n -2}+2 a_{n -1}}{n^{2}+2 n r +r^{2}-2 n -2 r +1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-n^{2} a_{n -1}+\left (-a_{n -2}-5 a_{n -1}\right ) n -6 a_{n -1}}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-5 r -6}{r^{2}}\) | \(-12\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{4}+11 r^{3}+52 r^{2}+102 r +72}{r^{2} \left (1+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {119}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-5 r -6}{r^{2}}\) | \(-12\) |
| \(a_{2}\) | \(\frac {r^{4}+11 r^{3}+52 r^{2}+102 r +72}{r^{2} \left (1+r \right )^{2}}\) | \(\frac {119}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{6}-19 r^{5}-162 r^{4}-759 r^{3}-1979 r^{2}-2648 r -1428}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {583}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-5 r -6}{r^{2}}\) | \(-12\) |
| \(a_{2}\) | \(\frac {r^{4}+11 r^{3}+52 r^{2}+102 r +72}{r^{2} \left (1+r \right )^{2}}\) | \(\frac {119}{2}\) |
| \(a_{3}\) | \(\frac {-r^{6}-19 r^{5}-162 r^{4}-759 r^{3}-1979 r^{2}-2648 r -1428}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {583}{3}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{8}+29 r^{7}+383 r^{6}+2966 r^{5}+14534 r^{4}+45437 r^{3}+87166 r^{2}+92772 r +41976}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1981}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-5 r -6}{r^{2}}\) | \(-12\) |
| \(a_{2}\) | \(\frac {r^{4}+11 r^{3}+52 r^{2}+102 r +72}{r^{2} \left (1+r \right )^{2}}\) | \(\frac {119}{2}\) |
| \(a_{3}\) | \(\frac {-r^{6}-19 r^{5}-162 r^{4}-759 r^{3}-1979 r^{2}-2648 r -1428}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {583}{3}}\) |
| \(a_{4}\) | \(\frac {r^{8}+29 r^{7}+383 r^{6}+2966 r^{5}+14534 r^{4}+45437 r^{3}+87166 r^{2}+92772 r +41976}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1981}{4}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r^{10}-41 r^{9}-773 r^{8}-8778 r^{7}-66136 r^{6}-343352 r^{5}-1234958 r^{4}-3013745 r^{3}-4736076 r^{2}-4299660 r -1711584}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {80287}{75}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-5 r -6}{r^{2}}\) | \(-12\) |
| \(a_{2}\) | \(\frac {r^{4}+11 r^{3}+52 r^{2}+102 r +72}{r^{2} \left (1+r \right )^{2}}\) | \(\frac {119}{2}\) |
| \(a_{3}\) | \(\frac {-r^{6}-19 r^{5}-162 r^{4}-759 r^{3}-1979 r^{2}-2648 r -1428}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {583}{3}}\) |
| \(a_{4}\) | \(\frac {r^{8}+29 r^{7}+383 r^{6}+2966 r^{5}+14534 r^{4}+45437 r^{3}+87166 r^{2}+92772 r +41976}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1981}{4}\) |
| \(a_{5}\) | \(\frac {-r^{10}-41 r^{9}-773 r^{8}-8778 r^{7}-66136 r^{6}-343352 r^{5}-1234958 r^{4}-3013745 r^{3}-4736076 r^{2}-4299660 r -1711584}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {80287}{75}}\) |
Using the above table, then the first 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 (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-r^{2}-5 r -6}{r^{2}}\) | \(-12\) | \(\frac {5 r +12}{r^{3}}\) | \(17\) |
| \(b_{2}\) | \(\frac {r^{4}+11 r^{3}+52 r^{2}+102 r +72}{r^{2} \left (1+r \right )^{2}}\) | \(\frac {119}{2}\) | \(\frac {-9 r^{4}-93 r^{3}-306 r^{2}-390 r -144}{r^{3} \left (1+r \right )^{3}}\) | \(-{\frac {471}{4}}\) |
| \(b_{3}\) | \(\frac {-r^{6}-19 r^{5}-162 r^{4}-759 r^{3}-1979 r^{2}-2648 r -1428}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {583}{3}}\) | \(\frac {13 r^{7}+259 r^{6}+2163 r^{5}+9545 r^{4}+23596 r^{3}+32400 r^{2}+22432 r +5712}{r^{3} \left (1+r \right )^{3} \left (2+r \right )^{3}}\) | \(445\) |
| \(b_{4}\) | \(\frac {r^{8}+29 r^{7}+383 r^{6}+2966 r^{5}+14534 r^{4}+45437 r^{3}+87166 r^{2}+92772 r +41976}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {1981}{4}\) | \(\frac {-17 r^{10}-548 r^{9}-7905 r^{8}-66636 r^{7}-359775 r^{6}-1287474 r^{5}-3066787 r^{4}-4763998 r^{3}-4572612 r^{2}-2403576 r -503712}{r^{3} \left (1+r \right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3}}\) | \(-{\frac {118285}{96}}\) |
| \(b_{5}\) | \(\frac {-r^{10}-41 r^{9}-773 r^{8}-8778 r^{7}-66136 r^{6}-343352 r^{5}-1234958 r^{4}-3013745 r^{3}-4736076 r^{2}-4299660 r -1711584}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {80287}{75}}\) | \(\frac {21 r^{13}+996 r^{12}+21729 r^{11}+287772 r^{10}+2570762 r^{9}+16282296 r^{8}+74844895 r^{7}+251506262 r^{6}+614583381 r^{5}+1072551946 r^{4}+1290644940 r^{3}+1004381640 r^{2}+445508640 r +82156032}{r^{3} \left (1+r \right )^{3} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(\frac {702451}{250}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (17 x -\frac {471 x^{2}}{4}+445 x^{3}-\frac {118285 x^{4}}{96}+\frac {702451 x^{5}}{250}+O\left (x^{6}\right )\right ) \\ \end{align*} 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 (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) + c_{2} \left (x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (17 x -\frac {471 x^{2}}{4}+445 x^{3}-\frac {118285 x^{4}}{96}+\frac {702451 x^{5}}{250}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (17 x -\frac {471 x^{2}}{4}+445 x^{3}-\frac {118285 x^{4}}{96}+\frac {702451 x^{5}}{250}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (17 x -\frac {471 x^{2}}{4}+445 x^{3}-\frac {118285 x^{4}}{96}+\frac {702451 x^{5}}{250}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1-12 x +\frac {119 x^{2}}{2}-\frac {583 x^{3}}{3}+\frac {1981 x^{4}}{4}-\frac {80287 x^{5}}{75}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (17 x -\frac {471 x^{2}}{4}+445 x^{3}-\frac {118285 x^{4}}{96}+\frac {702451 x^{5}}{250}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) y^{\prime \prime }+\left (x^{3}+6 x^{2}-x \right ) y^{\prime }+\left (x^{2}+6 x +1\right ) 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 {\left (x^{2}+6 x +1\right ) y}{x^{2} \left (x +1\right )}-\frac {\left (x^{2}+6 x -1\right ) y^{\prime }}{x \left (x +1\right )} \\ \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}+6 x -1\right ) y^{\prime }}{x \left (x +1\right )}+\frac {\left (x^{2}+6 x +1\right ) y}{x^{2} \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 {x^{2}+6 x -1}{x \left (x +1\right )}, P_{3}\left (x \right )=\frac {x^{2}+6 x +1}{x^{2} \left (x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=6 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) y^{\prime \prime }+x \left (x^{2}+6 x -1\right ) y^{\prime }+\left (x^{2}+6 x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-2 u^{2}+u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (u^{3}+3 u^{2}-10 u +6\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u^{2}+4 u -4\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (5+r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (6+r \right )-2 a_{0} \left (r^{2}+4 r +2\right )\right ) u^{r}+\left (a_{2} \left (2+r \right ) \left (7+r \right )-2 a_{1} \left (r^{2}+6 r +7\right )+a_{0} \left (r^{2}+2 r +4\right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k +6+r \right )-2 a_{k} \left (k^{2}+2 k r +r^{2}+4 k +4 r +2\right )+a_{k -1} \left (\left (k -1\right )^{2}+2 \left (k -1\right ) r +r^{2}+2 k +2+2 r \right )+a_{k -2} \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (5+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-5, 0\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (6+r \right )-2 a_{0} \left (r^{2}+4 r +2\right )=0, a_{2} \left (2+r \right ) \left (7+r \right )-2 a_{1} \left (r^{2}+6 r +7\right )+a_{0} \left (r^{2}+2 r +4\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {2 a_{0} \left (r^{2}+4 r +2\right )}{r^{2}+7 r +6}, a_{2}=\frac {a_{0} \left (3 r^{4}+31 r^{3}+108 r^{2}+120 r +32\right )}{r^{4}+16 r^{3}+83 r^{2}+152 r +84}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+\left (\left (-4 a_{k}+2 a_{k -1}+2 a_{k +1}\right ) r -8 a_{k}+a_{k -2}+7 a_{k +1}\right ) k +\left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+\left (-8 a_{k}+a_{k -2}+7 a_{k +1}\right ) r -4 a_{k}-a_{k -2}+3 a_{k -1}+6 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) \left (k +2\right )^{2}+\left (\left (-4 a_{k +2}+2 a_{k +1}+2 a_{k +3}\right ) r -8 a_{k +2}+a_{k}+7 a_{k +3}\right ) \left (k +2\right )+\left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) r^{2}+\left (-8 a_{k +2}+a_{k}+7 a_{k +3}\right ) r -4 a_{k +2}-a_{k}+3 a_{k +1}+6 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {k^{2} a_{k +1}-2 k^{2} a_{k +2}+2 k r a_{k +1}-4 k r a_{k +2}+r^{2} a_{k +1}-2 r^{2} a_{k +2}+k a_{k}+4 k a_{k +1}-16 k a_{k +2}+r a_{k}+4 r a_{k +1}-16 r a_{k +2}+a_{k}+7 a_{k +1}-28 a_{k +2}}{k^{2}+2 k r +r^{2}+11 k +11 r +24} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-5 \\ {} & {} & a_{k +3}=-\frac {k^{2} a_{k +1}-2 k^{2} a_{k +2}+k a_{k}-6 k a_{k +1}+4 k a_{k +2}-4 a_{k}+12 a_{k +1}+2 a_{k +2}}{k^{2}+k -6} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-5\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +3}=-\frac {k^{2} a_{k +1}-2 k^{2} a_{k +2}+k a_{k}-6 k a_{k +1}+4 k a_{k +2}-4 a_{k}+12 a_{k +1}+2 a_{k +2}}{k^{2}+k -6} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {k^{2} a_{k +1}-2 k^{2} a_{k +2}+k a_{k}+4 k a_{k +1}-16 k a_{k +2}+a_{k}+7 a_{k +1}-28 a_{k +2}}{k^{2}+11 k +24} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +3}=-\frac {k^{2} a_{k +1}-2 k^{2} a_{k +2}+k a_{k}+4 k a_{k +1}-16 k a_{k +2}+a_{k}+7 a_{k +1}-28 a_{k +2}}{k^{2}+11 k +24}, a_{1}=\frac {2 a_{0}}{3}, a_{2}=\frac {8 a_{0}}{21}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +3}=-\frac {k^{2} a_{k +1}-2 k^{2} a_{k +2}+k a_{k}+4 k a_{k +1}-16 k a_{k +2}+a_{k}+7 a_{k +1}-28 a_{k +2}}{k^{2}+11 k +24}, a_{1}=\frac {2 a_{0}}{3}, a_{2}=\frac {8 a_{0}}{21}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 63
Order:=6; dsolve(x^2*(1+x)*diff(y(x),x$2)-x*(1-6*x-x^2)*diff(y(x),x)+(1+6*x+x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-12 x +\frac {119}{2} x^{2}-\frac {583}{3} x^{3}+\frac {1981}{4} x^{4}-\frac {80287}{75} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (17 x -\frac {471}{4} x^{2}+445 x^{3}-\frac {118285}{96} x^{4}+\frac {702451}{250} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) x \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 114
AsymptoticDSolveValue[x^2*(1+x)*y''[x]-x*(1-6*x-x^2)*y'[x]+(1+6*x+x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (-\frac {80287 x^5}{75}+\frac {1981 x^4}{4}-\frac {583 x^3}{3}+\frac {119 x^2}{2}-12 x+1\right )+c_2 \left (x \left (\frac {702451 x^5}{250}-\frac {118285 x^4}{96}+445 x^3-\frac {471 x^2}{4}+17 x\right )+x \left (-\frac {80287 x^5}{75}+\frac {1981 x^4}{4}-\frac {583 x^3}{3}+\frac {119 x^2}{2}-12 x+1\right ) \log (x)\right ) \]