Internal problem ID [1433]
Internal file name [OUTPUT/1434_Sunday_June_05_2022_02_17_10_AM_51484014/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. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
Problem number: 17.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (x +1\right ) y^{\prime \prime }+x \left (1-10 x \right ) y^{\prime }-\left (9-10 x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (-10 x^{2}+x \right ) y^{\prime }+\left (10 x -9\right ) y = 0 \] 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 {10 x -1}{x \left (x +1\right )}\\ q(x) &= \frac {10 x -9}{x^{2} \left (x +1\right )}\\ \end {align*}
| \(p(x)=-\frac {10 x -1}{x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {10 x -9}{x^{2} \left (x +1\right )}\) | |
| 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 : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x +1\right ) y^{\prime \prime }+\left (-10 x^{2}+x \right ) y^{\prime }+\left (10 x -9\right ) y = 0 \] 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 (-10 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (10 x -9\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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-10 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 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 }}\left (-10 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-10 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}10 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 )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-10 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-9 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -9 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -9 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-9\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-9 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= -3 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-9\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, -3]\).
Since \(r_1 - r_2 = 6\) 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 )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{3}} \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 -3}\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 )-10 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+10 a_{n -1}-9 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-13 n -13 r +22\right )}{n^{2}+2 n r +r^{2}-9}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n^{2}-7 n -8\right )}{n \left (n +6\right )}\tag {5} \] 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 \[ a_{1}=\frac {-r^{2}+11 r -10}{r^{2}+2 r -8} \] Which for the root \(r = 3\) becomes \[ a_{1}=2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {9}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(2\) |
| \(a_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {9}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{r^{4}+13 r^{3}+44 r^{2}-28 r -240} \] Which for the root \(r = 3\) becomes \[ a_{3}={\frac {5}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(2\) |
| \(a_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {9}{4}\) |
| \(a_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{r^{4}+13 r^{3}+44 r^{2}-28 r -240}\) | \(\frac {5}{3}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{r^{5}+20 r^{4}+135 r^{3}+280 r^{2}-436 r -1680} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {5}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(2\) |
| \(a_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {9}{4}\) |
| \(a_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{r^{4}+13 r^{3}+44 r^{2}-28 r -240}\) | \(\frac {5}{3}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{r^{5}+20 r^{4}+135 r^{3}+280 r^{2}-436 r -1680}\) | \(\frac {5}{6}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (r +3\right ) \left (r -6\right ) \left (r -9\right ) \left (r -10\right ) \left (r -8\right ) \left (r -7\right )}{\left (r +8\right ) \left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )} \] Which for the root \(r = 3\) becomes \[ a_{5}={\frac {3}{11}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(2\) |
| \(a_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {9}{4}\) |
| \(a_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{r^{4}+13 r^{3}+44 r^{2}-28 r -240}\) | \(\frac {5}{3}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{r^{5}+20 r^{4}+135 r^{3}+280 r^{2}-436 r -1680}\) | \(\frac {5}{6}\) |
| \(a_{5}\) | \(-\frac {\left (r +3\right ) \left (r -6\right ) \left (r -9\right ) \left (r -10\right ) \left (r -8\right ) \left (r -7\right )}{\left (r +8\right ) \left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(\frac {3}{11}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )} \] Which for the root \(r = 3\) becomes \[ a_{6}={\frac {7}{132}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(2\) |
| \(a_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {9}{4}\) |
| \(a_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{r^{4}+13 r^{3}+44 r^{2}-28 r -240}\) | \(\frac {5}{3}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{r^{5}+20 r^{4}+135 r^{3}+280 r^{2}-436 r -1680}\) | \(\frac {5}{6}\) |
| \(a_{5}\) | \(-\frac {\left (r +3\right ) \left (r -6\right ) \left (r -9\right ) \left (r -10\right ) \left (r -8\right ) \left (r -7\right )}{\left (r +8\right ) \left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(\frac {3}{11}\) |
| \(a_{6}\) | \(\frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}\) | \(\frac {7}{132}\) |
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}\dots \right ) \\ &= x^{3} \left (1+2 x +\frac {9 x^{2}}{4}+\frac {5 x^{3}}{3}+\frac {5 x^{4}}{6}+\frac {3 x^{5}}{11}+\frac {7 x^{6}}{132}+O\left (x^{7}\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 \(N=6\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{6}\left (r \right )\). 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_{6} \\ &= \frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}&= \lim _{r\rightarrow -3}\frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}\\ &= -{\frac {1716}{5}} \end {align*}
The limit is \(-{\frac {1716}{5}}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -3} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )-10 b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )+10 b_{n -1}-9 b_{n} = 0 \end{equation} Which for for the root \(r = -3\) becomes \begin{equation} \tag{4A} b_{n -1} \left (n -4\right ) \left (n -5\right )+b_{n} \left (n -3\right ) \left (n -4\right )-10 b_{n -1} \left (n -4\right )+b_{n} \left (n -3\right )+10 b_{n -1}-9 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}+2 n r +r^{2}-13 n -13 r +22\right )}{n^{2}+2 n r +r^{2}-9}\tag {5} \] Which for the root \(r = -3\) becomes \[ b_{n} = -\frac {b_{n -1} \left (n^{2}-19 n +70\right )}{n^{2}-6 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -3\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {r^{2}-11 r +10}{r^{2}+2 r -8} \] Which for the root \(r = -3\) becomes \[ b_{1}={\frac {52}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(\frac {52}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\left (r -9\right ) r \left (r -10\right )}{\left (r +5\right ) \left (r^{2}+2 r -8\right )} \] Which for the root \(r = -3\) becomes \[ b_{2}={\frac {234}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(\frac {52}{5}\) |
| \(b_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {234}{5}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (r -9\right ) \left (r -10\right ) \left (r^{2}-7 r -8\right )}{\left (r +5\right ) \left (r^{2}+2 r -8\right ) \left (r +6\right )} \] Which for the root \(r = -3\) becomes \[ b_{3}={\frac {572}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(\frac {52}{5}\) |
| \(b_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {234}{5}\) |
| \(b_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{\left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(\frac {572}{5}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (r^{2}-5 r -14\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{\left (r +7\right ) \left (r +5\right ) \left (r^{2}+2 r -8\right ) \left (r +6\right )} \] Which for the root \(r = -3\) becomes \[ b_{4}=143 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(\frac {52}{5}\) |
| \(b_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {234}{5}\) |
| \(b_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{\left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(\frac {572}{5}\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{\left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(143\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (r^{2}-3 r -18\right ) \left (r -9\right ) \left (r -10\right ) \left (r -8\right ) \left (r -7\right )}{\left (r +8\right ) \left (r +7\right ) \left (r +5\right ) \left (r^{2}+2 r -8\right ) \left (r +6\right )} \] Which for the root \(r = -3\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(\frac {52}{5}\) |
| \(b_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {234}{5}\) |
| \(b_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{\left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(\frac {572}{5}\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{\left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(143\) |
| \(b_{5}\) | \(-\frac {\left (r +3\right ) \left (r -6\right ) \left (r -9\right ) \left (r -10\right ) \left (r -8\right ) \left (r -7\right )}{\left (r +8\right ) \left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )} \] Which for the root \(r = -3\) becomes \[ b_{6}=-{\frac {1716}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+11 r -10}{r^{2}+2 r -8}\) | \(\frac {52}{5}\) |
| \(b_{2}\) | \(\frac {\left (r -9\right ) r \left (r -10\right )}{r^{3}+7 r^{2}+2 r -40}\) | \(\frac {234}{5}\) |
| \(b_{3}\) | \(-\frac {\left (r -9\right ) \left (r -10\right ) \left (r +1\right ) \left (r -8\right )}{\left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(\frac {572}{5}\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right )}{\left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(143\) |
| \(b_{5}\) | \(-\frac {\left (r +3\right ) \left (r -6\right ) \left (r -9\right ) \left (r -10\right ) \left (r -8\right ) \left (r -7\right )}{\left (r +8\right ) \left (r +7\right ) \left (r +5\right ) \left (r +4\right ) \left (-2+r \right ) \left (r +6\right )}\) | \(0\) |
| \(b_{6}\) | \(\frac {\left (r -7\right ) \left (r -8\right ) \left (r -10\right ) \left (r -9\right ) \left (r -6\right ) \left (r -5\right )}{\left (r +9\right ) \left (r +6\right ) \left (-2+r \right ) \left (r +5\right ) \left (r +7\right ) \left (r +8\right )}\) | \(-{\frac {1716}{5}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{3} \left (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}+b_{7} x^{7}\dots \right ) \\ &= \frac {1+\frac {52 x}{5}+\frac {234 x^{2}}{5}+\frac {572 x^{3}}{5}+143 x^{4}-\frac {1716 x^{6}}{5}+O\left (x^{7}\right )}{x^{3}} \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^{3} \left (1+2 x +\frac {9 x^{2}}{4}+\frac {5 x^{3}}{3}+\frac {5 x^{4}}{6}+\frac {3 x^{5}}{11}+\frac {7 x^{6}}{132}+O\left (x^{7}\right )\right ) + \frac {c_{2} \left (1+\frac {52 x}{5}+\frac {234 x^{2}}{5}+\frac {572 x^{3}}{5}+143 x^{4}-\frac {1716 x^{6}}{5}+O\left (x^{7}\right )\right )}{x^{3}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1+2 x +\frac {9 x^{2}}{4}+\frac {5 x^{3}}{3}+\frac {5 x^{4}}{6}+\frac {3 x^{5}}{11}+\frac {7 x^{6}}{132}+O\left (x^{7}\right )\right )+\frac {c_{2} \left (1+\frac {52 x}{5}+\frac {234 x^{2}}{5}+\frac {572 x^{3}}{5}+143 x^{4}-\frac {1716 x^{6}}{5}+O\left (x^{7}\right )\right )}{x^{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1+2 x +\frac {9 x^{2}}{4}+\frac {5 x^{3}}{3}+\frac {5 x^{4}}{6}+\frac {3 x^{5}}{11}+\frac {7 x^{6}}{132}+O\left (x^{7}\right )\right )+\frac {c_{2} \left (1+\frac {52 x}{5}+\frac {234 x^{2}}{5}+\frac {572 x^{3}}{5}+143 x^{4}-\frac {1716 x^{6}}{5}+O\left (x^{7}\right )\right )}{x^{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1+2 x +\frac {9 x^{2}}{4}+\frac {5 x^{3}}{3}+\frac {5 x^{4}}{6}+\frac {3 x^{5}}{11}+\frac {7 x^{6}}{132}+O\left (x^{7}\right )\right )+\frac {c_{2} \left (1+\frac {52 x}{5}+\frac {234 x^{2}}{5}+\frac {572 x^{3}}{5}+143 x^{4}-\frac {1716 x^{6}}{5}+O\left (x^{7}\right )\right )}{x^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) y^{\prime \prime }+\left (-10 x^{2}+x \right ) y^{\prime }+\left (10 x -9\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 (10 x -9\right ) y}{x^{2} \left (x +1\right )}+\frac {\left (10 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 (10 x -1\right ) y^{\prime }}{x \left (x +1\right )}+\frac {\left (10 x -9\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 {10 x -1}{x \left (x +1\right )}, P_{3}\left (x \right )=\frac {10 x -9}{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}}}=-11 \\ {} & \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 (10 x -1\right ) y^{\prime }+\left (10 x -9\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 (-10 u^{2}+21 u -11\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (10 u -19\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..1 \\ {} & {} & 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..2 \\ {} & {} & 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 (-12+r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-11+r \right )-a_{0} \left (2 r^{2}-23 r +19\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k -11+r \right )-a_{k} \left (2 k^{2}+4 k r +2 r^{2}-23 k -23 r +19\right )+a_{k -1} \left (k -2+r \right ) \left (k -11+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-12+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 12\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-11+r \right )-a_{0} \left (2 r^{2}-23 r +19\right )=0 \\ \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 +23 a_{k}-13 a_{k -1}-10 a_{k +1}\right ) k +\left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+\left (23 a_{k}-13 a_{k -1}-10 a_{k +1}\right ) r -19 a_{k}+22 a_{k -1}-11 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) \left (k +1\right )^{2}+\left (\left (-4 a_{k +1}+2 a_{k}+2 a_{k +2}\right ) r +23 a_{k +1}-13 a_{k}-10 a_{k +2}\right ) \left (k +1\right )+\left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r^{2}+\left (23 a_{k +1}-13 a_{k}-10 a_{k +2}\right ) r -19 a_{k +1}+22 a_{k}-11 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k r a_{k}-4 k r a_{k +1}+r^{2} a_{k}-2 r^{2} a_{k +1}-11 k a_{k}+19 k a_{k +1}-11 r a_{k}+19 r a_{k +1}+10 a_{k}+2 a_{k +1}}{k^{2}+2 k r +r^{2}-8 k -8 r -20} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-11 k a_{k}+19 k a_{k +1}+10 a_{k}+2 a_{k +1}}{k^{2}-8 k -20} \\ \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 =10 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-11 k a_{k}+19 k a_{k +1}+10 a_{k}+2 a_{k +1}}{k^{2}-8 k -20} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =12 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+13 k a_{k}-29 k a_{k +1}+22 a_{k}-58 a_{k +1}}{k^{2}+16 k +28} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =12 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +12}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+13 k a_{k}-29 k a_{k +1}+22 a_{k}-58 a_{k +1}}{k^{2}+16 k +28}, 13 a_{1}-31 a_{0}=0\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 +12}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+13 k a_{k}-29 k a_{k +1}+22 a_{k}-58 a_{k +1}}{k^{2}+16 k +28}, 13 a_{1}-31 a_{0}=0\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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 45
Order:=6; dsolve(x^2*(1+x)*diff(y(x),x$2)+x*(1-10*x)*diff(y(x),x)-(9-10*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{3} \left (1+2 x +\frac {9}{4} x^{2}+\frac {5}{3} x^{3}+\frac {5}{6} x^{4}+\frac {3}{11} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (-86400-898560 x -4043520 x^{2}-9884160 x^{3}-12355200 x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.046 (sec). Leaf size: 64
AsymptoticDSolveValue[x^2*(1+x)*y''[x]+x*(1-10*x)*y'[x]-(9-10*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{x^3}+\frac {52}{5 x^2}+143 x+\frac {234}{5 x}+\frac {572}{5}\right )+c_2 \left (\frac {5 x^7}{6}+\frac {5 x^6}{3}+\frac {9 x^5}{4}+2 x^4+x^3\right ) \]