Internal problem ID [6977]
Internal file name [OUTPUT/6220_Friday_August_12_2022_11_05_54_PM_39873401/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.8 Indicial Equation with Difference of Roots a
Positive Integer: Nonlogarithmic Case. Exercises page 380
Problem number: 13.
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]]
\[ \boxed {x \left (x +3\right ) y^{\prime \prime }-9 y^{\prime }-6 y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{2}+3 x \right ) y^{\prime \prime }-6 y-9 y^{\prime } = 0 \] 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 {9}{x \left (x +3\right )}\\ q(x) &= -\frac {6}{x \left (x +3\right )}\\ \end {align*}
| \(p(x)=-\frac {9}{x \left (x +3\right )}\) | |
| singularity | type |
| \(x = -3\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {6}{x \left (x +3\right )}\) | |
| singularity | type |
| \(x = -3\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-3, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (x +3\right ) y^{\prime \prime }-9 y^{\prime }-6 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x \left (x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )-9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-6 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-9 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{-1+r} a_{0} r \left (-1+r \right )-9 r a_{0} x^{-1+r} = 0 \] Or \[ \left (3 x^{-1+r} r \left (-1+r \right )-9 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 3 r \,x^{-1+r} \left (-4+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r \left (-4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 4\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 3 r \,x^{-1+r} \left (-4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([4, 0]\).
Since \(r_1 - r_2 = 4\) 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^{4} \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 +4}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 a_{n} \left (n +r \right ) \left (n +r -1\right )-9 a_{n} \left (n +r \right )-6 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (n +r +1\right ) a_{n -1}}{3 \left (n +r \right )}\tag {4} \] Which for the root \(r = 4\) becomes \[ a_{n} = -\frac {\left (n +5\right ) a_{n -1}}{3 n +12}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 4\) 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 \[ a_{1}=\frac {-2-r}{3 r +3} \] Which for the root \(r = 4\) becomes \[ a_{1}=-{\frac {2}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3+r}{9 r +9} \] Which for the root \(r = 4\) becomes \[ a_{2}={\frac {7}{45}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {7}{45}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-4-r}{27 r +27} \] Which for the root \(r = 4\) becomes \[ a_{3}=-{\frac {8}{135}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {7}{45}\) |
| \(a_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {8}{135}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5+r}{81 r +81} \] Which for the root \(r = 4\) becomes \[ a_{4}={\frac {1}{45}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {7}{45}\) |
| \(a_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {8}{135}}\) |
| \(a_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {1}{45}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-6-r}{243 r +243} \] Which for the root \(r = 4\) becomes \[ a_{5}=-{\frac {2}{243}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {7}{45}\) |
| \(a_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {8}{135}}\) |
| \(a_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {1}{45}\) |
| \(a_{5}\) | \(\frac {-6-r}{243 r +243}\) | \(-{\frac {2}{243}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {7+r}{729 r +729} \] Which for the root \(r = 4\) becomes \[ a_{6}={\frac {11}{3645}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {7}{45}\) |
| \(a_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {8}{135}}\) |
| \(a_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {1}{45}\) |
| \(a_{5}\) | \(\frac {-6-r}{243 r +243}\) | \(-{\frac {2}{243}}\) |
| \(a_{6}\) | \(\frac {7+r}{729 r +729}\) | \(\frac {11}{3645}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-8-r}{2187 r +2187} \] Which for the root \(r = 4\) becomes \[ a_{7}=-{\frac {4}{3645}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{5}}\) |
| \(a_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {7}{45}\) |
| \(a_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {8}{135}}\) |
| \(a_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {1}{45}\) |
| \(a_{5}\) | \(\frac {-6-r}{243 r +243}\) | \(-{\frac {2}{243}}\) |
| \(a_{6}\) | \(\frac {7+r}{729 r +729}\) | \(\frac {11}{3645}\) |
| \(a_{7}\) | \(\frac {-8-r}{2187 r +2187}\) | \(-{\frac {4}{3645}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{4} \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^{4} \left (1-\frac {2 x}{5}+\frac {7 x^{2}}{45}-\frac {8 x^{3}}{135}+\frac {x^{4}}{45}-\frac {2 x^{5}}{243}+\frac {11 x^{6}}{3645}-\frac {4 x^{7}}{3645}+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 \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). 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_{4} \\ &= \frac {5+r}{81 r +81} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {5+r}{81 r +81}&= \lim _{r\rightarrow 0}\frac {5+r}{81 r +81}\\ &= {\frac {5}{81}} \end {align*}
The limit is \(\frac {5}{81}\). 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} \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 )+3 b_{n} \left (n +r \right ) \left (n +r -1\right )-9 \left (n +r \right ) b_{n}-6 b_{n -1} = 0 \end{equation} Which for for the root \(r = 0\) becomes \begin{equation} \tag{4A} b_{n -1} \left (n -1\right ) \left (n -2\right )+3 b_{n} n \left (n -1\right )-9 n b_{n}-6 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {\left (n +r +1\right ) b_{n -1}}{3 \left (n +r \right )}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = -\frac {\left (n +1\right ) b_{n -1}}{3 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) 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 \[ b_{1}=-\frac {2+r}{3 \left (r +1\right )} \] Which for the root \(r = 0\) becomes \[ b_{1}=-{\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {3+r}{9 r +9} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {1}{3}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {4+r}{27 \left (r +1\right )} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {4}{27}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {4}{27}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {5+r}{81 r +81} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {5}{81}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {4}{27}}\) |
| \(b_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {5}{81}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {6+r}{243 \left (r +1\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {2}{81}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {4}{27}}\) |
| \(b_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {5}{81}\) |
| \(b_{5}\) | \(\frac {-6-r}{243 r +243}\) | \(-{\frac {2}{81}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {7+r}{729 r +729} \] Which for the root \(r = 0\) becomes \[ b_{6}={\frac {7}{729}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {4}{27}}\) |
| \(b_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {5}{81}\) |
| \(b_{5}\) | \(\frac {-6-r}{243 r +243}\) | \(-{\frac {2}{81}}\) |
| \(b_{6}\) | \(\frac {7+r}{729 r +729}\) | \(\frac {7}{729}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {8+r}{2187 \left (r +1\right )} \] Which for the root \(r = 0\) becomes \[ b_{7}=-{\frac {8}{2187}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2-r}{3 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {3+r}{9 r +9}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {-4-r}{27 r +27}\) | \(-{\frac {4}{27}}\) |
| \(b_{4}\) | \(\frac {5+r}{81 r +81}\) | \(\frac {5}{81}\) |
| \(b_{5}\) | \(\frac {-6-r}{243 r +243}\) | \(-{\frac {2}{81}}\) |
| \(b_{6}\) | \(\frac {7+r}{729 r +729}\) | \(\frac {7}{729}\) |
| \(b_{7}\) | \(\frac {-8-r}{2187 r +2187}\) | \(-{\frac {8}{2187}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\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}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= 1-\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {4 x^{3}}{27}+\frac {5 x^{4}}{81}-\frac {2 x^{5}}{81}+\frac {7 x^{6}}{729}-\frac {8 x^{7}}{2187}+O\left (x^{8}\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^{4} \left (1-\frac {2 x}{5}+\frac {7 x^{2}}{45}-\frac {8 x^{3}}{135}+\frac {x^{4}}{45}-\frac {2 x^{5}}{243}+\frac {11 x^{6}}{3645}-\frac {4 x^{7}}{3645}+O\left (x^{8}\right )\right ) + c_{2} \left (1-\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {4 x^{3}}{27}+\frac {5 x^{4}}{81}-\frac {2 x^{5}}{81}+\frac {7 x^{6}}{729}-\frac {8 x^{7}}{2187}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{4} \left (1-\frac {2 x}{5}+\frac {7 x^{2}}{45}-\frac {8 x^{3}}{135}+\frac {x^{4}}{45}-\frac {2 x^{5}}{243}+\frac {11 x^{6}}{3645}-\frac {4 x^{7}}{3645}+O\left (x^{8}\right )\right )+c_{2} \left (1-\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {4 x^{3}}{27}+\frac {5 x^{4}}{81}-\frac {2 x^{5}}{81}+\frac {7 x^{6}}{729}-\frac {8 x^{7}}{2187}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{4} \left (1-\frac {2 x}{5}+\frac {7 x^{2}}{45}-\frac {8 x^{3}}{135}+\frac {x^{4}}{45}-\frac {2 x^{5}}{243}+\frac {11 x^{6}}{3645}-\frac {4 x^{7}}{3645}+O\left (x^{8}\right )\right )+c_{2} \left (1-\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {4 x^{3}}{27}+\frac {5 x^{4}}{81}-\frac {2 x^{5}}{81}+\frac {7 x^{6}}{729}-\frac {8 x^{7}}{2187}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{4} \left (1-\frac {2 x}{5}+\frac {7 x^{2}}{45}-\frac {8 x^{3}}{135}+\frac {x^{4}}{45}-\frac {2 x^{5}}{243}+\frac {11 x^{6}}{3645}-\frac {4 x^{7}}{3645}+O\left (x^{8}\right )\right )+c_{2} \left (1-\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {4 x^{3}}{27}+\frac {5 x^{4}}{81}-\frac {2 x^{5}}{81}+\frac {7 x^{6}}{729}-\frac {8 x^{7}}{2187}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x +3\right ) y^{\prime \prime }-9 y^{\prime }-6 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 {6 y}{x \left (x +3\right )}+\frac {9 y^{\prime }}{x \left (x +3\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 {9 y^{\prime }}{x \left (x +3\right )}-\frac {6 y}{x \left (x +3\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 {9}{x \left (x +3\right )}, P_{3}\left (x \right )=-\frac {6}{x \left (x +3\right )}\right ] \\ {} & \circ & \left (x +3\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=3 \\ {} & \circ & \left (x +3\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=0 \\ {} & \circ & x =-3\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}=-3 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x +3\right ) y^{\prime \prime }-9 y^{\prime }-6 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -3\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-3 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )-9 \frac {d}{d u}y \left (u \right )-6 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} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+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..2 \\ {} & {} & 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} \\ {} & {} & -3 a_{0} r \left (2+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-3 a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )+a_{k} \left (k +r +2\right ) \left (k +r -3\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -3 r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -3 a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )+a_{k} \left (k +r +2\right ) \left (k +r -3\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 +r -3\right )}{3 \left (k +1+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =5 \\ {} & {} & a_{k +1}=\frac {a_{k} k \left (k -5\right )}{3 \left (k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {a_{k} k \left (k -5\right )}{3 \left (k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =3 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right ) \left (k -3\right )}{3 \left (k +1\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-\frac {2 a_{0}}{3} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-\frac {a_{1}}{4} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{6} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=-\frac {4 a_{2}}{45} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=-\frac {2 a_{0}}{135} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1-\frac {2}{3} u +\frac {1}{6} u^{2}-\frac {2}{135} u^{3}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y=a_{0} \left (\frac {1}{10}-\frac {1}{15} x +\frac {1}{30} x^{2}-\frac {2}{135} x^{3}\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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 52
Order:=8; dsolve(x*(x+3)*diff(y(x),x$2)-9*diff(y(x),x)-6*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{4} \left (1-\frac {2}{5} x +\frac {7}{45} x^{2}-\frac {8}{135} x^{3}+\frac {1}{45} x^{4}-\frac {2}{243} x^{5}+\frac {11}{3645} x^{6}-\frac {4}{3645} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (-144+96 x -48 x^{2}+\frac {64}{3} x^{3}-\frac {80}{9} x^{4}+\frac {32}{9} x^{5}-\frac {112}{81} x^{6}+\frac {128}{243} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.419 (sec). Leaf size: 98
AsymptoticDSolveValue[x*(x+3)*y''[x]-9*y'[x]-6*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {7 x^6}{729}-\frac {2 x^5}{81}+\frac {5 x^4}{81}-\frac {4 x^3}{27}+\frac {x^2}{3}-\frac {2 x}{3}+1\right )+c_2 \left (\frac {11 x^{10}}{3645}-\frac {2 x^9}{243}+\frac {x^8}{45}-\frac {8 x^7}{135}+\frac {7 x^6}{45}-\frac {2 x^5}{5}+x^4\right ) \]