Internal problem ID [2929]
Internal file name [OUTPUT/2421_Sunday_June_05_2022_03_08_58_AM_82109773/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.4. page
758
Problem number: 14.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {3 x^{2} y^{\prime \prime }-x \left (x +8\right ) y^{\prime }+6 y=0} \] 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. \[ 3 x^{2} y^{\prime \prime }+\left (-x^{2}-8 x \right ) y^{\prime }+6 y = 0 \] 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 +8}{3 x}\\ q(x) &= \frac {2}{x^{2}}\\ \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 \[ 3 x^{2} y^{\prime \prime }+\left (-x^{2}-8 x \right ) y^{\prime }+6 y = 0 \] 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} 3 x^{2} \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^{2}-8 x \right ) \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 }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 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 }}\left (-x^{1+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}\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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} 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}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-8 x^{n +r} a_{n} \left (n +r \right )+6 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{r} a_{0} r \left (-1+r \right )-8 x^{r} a_{0} r +6 a_{0} x^{r} = 0 \] Or \[ \left (3 x^{r} r \left (-1+r \right )-8 x^{r} r +6 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (3 r^{2}-11 r +6\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}-11 r +6 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= {\frac {2}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (3 r^{2}-11 r +6\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [3, {\frac {2}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {7}{3}}\) is not 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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +3}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {2}{3}} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 3 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )-8 a_{n} \left (n +r \right )+6 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r -1\right )}{3 n^{2}+6 n r +3 r^{2}-11 n -11 r +6}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = \frac {a_{n -1} \left (n +2\right )}{n \left (3 n +7\right )}\tag {5} \] 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 \[ a_{1}=\frac {r}{3 r^{2}-5 r -2} \] Which for the root \(r = 3\) becomes \[ a_{1}={\frac {3}{10}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(\frac {3}{10}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {3}{65}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(\frac {3}{10}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {3}{65}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56} \] Which for the root \(r = 3\) becomes \[ a_{3}={\frac {1}{208}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(\frac {3}{10}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {3}{65}\) |
| \(a_{3}\) | \(\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56}\) | \(\frac {1}{208}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (2+r \right ) \left (3+r \right )}{81 r^{6}+351 r^{5}-189 r^{4}-1851 r^{3}-620 r^{2}+1668 r +560} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {3}{7904}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(\frac {3}{10}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {3}{65}\) |
| \(a_{3}\) | \(\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56}\) | \(\frac {1}{208}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (3+r \right )}{81 r^{6}+351 r^{5}-189 r^{4}-1851 r^{3}-620 r^{2}+1668 r +560}\) | \(\frac {3}{7904}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (4+r \right ) \left (3+r \right )}{243 r^{7}+2106 r^{6}+3996 r^{5}-8010 r^{4}-25923 r^{3}-3056 r^{2}+23364 r +7280} \] Which for the root \(r = 3\) becomes \[ a_{5}={\frac {21}{869440}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(\frac {3}{10}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {3}{65}\) |
| \(a_{3}\) | \(\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56}\) | \(\frac {1}{208}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (3+r \right )}{81 r^{6}+351 r^{5}-189 r^{4}-1851 r^{3}-620 r^{2}+1668 r +560}\) | \(\frac {3}{7904}\) |
| \(a_{5}\) | \(\frac {\left (4+r \right ) \left (3+r \right )}{243 r^{7}+2106 r^{6}+3996 r^{5}-8010 r^{4}-25923 r^{3}-3056 r^{2}+23364 r +7280}\) | \(\frac {21}{869440}\) |
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}\dots \right ) \\ &= x^{3} \left (1+\frac {3 x}{10}+\frac {3 x^{2}}{65}+\frac {x^{3}}{208}+\frac {3 x^{4}}{7904}+\frac {21 x^{5}}{869440}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now 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{3} 3 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -1} \left (n +r -1\right )-8 b_{n} \left (n +r \right )+6 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (n +r -1\right )}{3 n^{2}+6 n r +3 r^{2}-11 n -11 r +6}\tag {4} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ b_{n} = \frac {3 n b_{n -1}-b_{n -1}}{9 n^{2}-21 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {2}{3}}\) 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 \[ b_{1}=\frac {r}{3 r^{2}-5 r -2} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ b_{1}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(-{\frac {1}{6}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ b_{2}={\frac {5}{36}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(-{\frac {1}{6}}\) |
| \(b_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {5}{36}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ b_{3}={\frac {5}{81}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(-{\frac {1}{6}}\) |
| \(b_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {5}{36}\) |
| \(b_{3}\) | \(\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56}\) | \(\frac {5}{81}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (2+r \right ) \left (3+r \right )}{81 r^{6}+351 r^{5}-189 r^{4}-1851 r^{3}-620 r^{2}+1668 r +560} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ b_{4}={\frac {11}{972}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(-{\frac {1}{6}}\) |
| \(b_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {5}{36}\) |
| \(b_{3}\) | \(\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56}\) | \(\frac {5}{81}\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (3+r \right )}{81 r^{6}+351 r^{5}-189 r^{4}-1851 r^{3}-620 r^{2}+1668 r +560}\) | \(\frac {11}{972}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {\left (4+r \right ) \left (3+r \right )}{243 r^{7}+2106 r^{6}+3996 r^{5}-8010 r^{4}-25923 r^{3}-3056 r^{2}+23364 r +7280} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ b_{5}={\frac {77}{58320}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{3 r^{2}-5 r -2}\) | \(-{\frac {1}{6}}\) |
| \(b_{2}\) | \(\frac {r \left (1+r \right )}{9 r^{4}-12 r^{3}-23 r^{2}+18 r +8}\) | \(\frac {5}{36}\) |
| \(b_{3}\) | \(\frac {\left (1+r \right ) \left (2+r \right )}{27 r^{5}+27 r^{4}-153 r^{3}-107 r^{2}+150 r +56}\) | \(\frac {5}{81}\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (3+r \right )}{81 r^{6}+351 r^{5}-189 r^{4}-1851 r^{3}-620 r^{2}+1668 r +560}\) | \(\frac {11}{972}\) |
| \(b_{5}\) | \(\frac {\left (4+r \right ) \left (3+r \right )}{243 r^{7}+2106 r^{6}+3996 r^{5}-8010 r^{4}-25923 r^{3}-3056 r^{2}+23364 r +7280}\) | \(\frac {77}{58320}\) |
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}\dots \right ) \\ &= x^{\frac {2}{3}} \left (1-\frac {x}{6}+\frac {5 x^{2}}{36}+\frac {5 x^{3}}{81}+\frac {11 x^{4}}{972}+\frac {77 x^{5}}{58320}+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^{3} \left (1+\frac {3 x}{10}+\frac {3 x^{2}}{65}+\frac {x^{3}}{208}+\frac {3 x^{4}}{7904}+\frac {21 x^{5}}{869440}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {2}{3}} \left (1-\frac {x}{6}+\frac {5 x^{2}}{36}+\frac {5 x^{3}}{81}+\frac {11 x^{4}}{972}+\frac {77 x^{5}}{58320}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1+\frac {3 x}{10}+\frac {3 x^{2}}{65}+\frac {x^{3}}{208}+\frac {3 x^{4}}{7904}+\frac {21 x^{5}}{869440}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {2}{3}} \left (1-\frac {x}{6}+\frac {5 x^{2}}{36}+\frac {5 x^{3}}{81}+\frac {11 x^{4}}{972}+\frac {77 x^{5}}{58320}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1+\frac {3 x}{10}+\frac {3 x^{2}}{65}+\frac {x^{3}}{208}+\frac {3 x^{4}}{7904}+\frac {21 x^{5}}{869440}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {2}{3}} \left (1-\frac {x}{6}+\frac {5 x^{2}}{36}+\frac {5 x^{3}}{81}+\frac {11 x^{4}}{972}+\frac {77 x^{5}}{58320}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1+\frac {3 x}{10}+\frac {3 x^{2}}{65}+\frac {x^{3}}{208}+\frac {3 x^{4}}{7904}+\frac {21 x^{5}}{869440}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {2}{3}} \left (1-\frac {x}{6}+\frac {5 x^{2}}{36}+\frac {5 x^{3}}{81}+\frac {11 x^{4}}{972}+\frac {77 x^{5}}{58320}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-x^{2}-8 x \right ) y^{\prime }+6 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {2 y}{x^{2}}+\frac {\left (x +8\right ) y^{\prime }}{3 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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (x +8\right ) y^{\prime }}{3 x}+\frac {2 y}{x^{2}}=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 +8}{3 x}, P_{3}\left (x \right )=\frac {2}{x^{2}}\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}}}=-\frac {8}{3} \\ {} & \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}}}=2 \\ {} & \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} \\ {} & {} & 3 x^{2} \left (\frac {d}{d x}y^{\prime }\right )-x \left (x +8\right ) y^{\prime }+6 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 =1..2 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-2+3 r \right ) \left (-3+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (3 k +3 r -2\right ) \left (k +r -3\right )-a_{k -1} \left (k +r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-2+3 r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{3, \frac {2}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +r -\frac {2}{3}\right ) \left (k +r -3\right ) a_{k}-a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 \left (k +\frac {1}{3}+r \right ) \left (k -2+r \right ) a_{k +1}-a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r \right )}{\left (3 k +1+3 r \right ) \left (k -2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +3\right )}{\left (3 k +10\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 +3\right )}{\left (3 k +10\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +\frac {2}{3}\right )}{\left (3 k +3\right ) \left (k -\frac {4}{3}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {2}{3}}, a_{k +1}=\frac {a_{k} \left (k +\frac {2}{3}\right )}{\left (3 k +3\right ) \left (k -\frac {4}{3}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +3}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {2}{3}}\right ), a_{k +1}=\frac {a_{k} \left (k +3\right )}{\left (3 k +10\right ) \left (k +1\right )}, b_{k +1}=\frac {b_{k} \left (k +\frac {2}{3}\right )}{\left (3 k +3\right ) \left (k -\frac {4}{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) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Kummer successful <- special function solution successful Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 47
Order:=6; dsolve(3*x^2*diff(y(x),x$2)-x*(x+8)*diff(y(x),x)+6*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {2}{3}} \left (1-\frac {1}{6} x +\frac {5}{36} x^{2}+\frac {5}{81} x^{3}+\frac {11}{972} x^{4}+\frac {77}{58320} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{3} \left (1+\frac {3}{10} x +\frac {3}{65} x^{2}+\frac {1}{208} x^{3}+\frac {3}{7904} x^{4}+\frac {21}{869440} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 88
AsymptoticDSolveValue[3*x^2*y''[x]-x*(x+8)*y'[x]+6*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {21 x^5}{869440}+\frac {3 x^4}{7904}+\frac {x^3}{208}+\frac {3 x^2}{65}+\frac {3 x}{10}+1\right ) x^3+c_2 \left (\frac {77 x^5}{58320}+\frac {11 x^4}{972}+\frac {5 x^3}{81}+\frac {5 x^2}{36}-\frac {x}{6}+1\right ) x^{2/3} \]