Internal problem ID [5578]
Internal file name [OUTPUT/4826_Sunday_June_05_2022_03_07_03_PM_7002621/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
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. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {9 x^{2} y^{\prime \prime }+9 x^{2} y^{\prime }+2 y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ 9 x^{2} y^{\prime \prime }+9 x^{2} y^{\prime }+2 y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= 1\\ q(x) &= \frac {2}{9 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 \[ 9 x^{2} y^{\prime \prime }+9 x^{2} y^{\prime }+2 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} 9 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 )+9 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 }}9 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) 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 =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )+2 a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )+2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (9 r^{2}-9 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 9 r^{2}-9 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {2}{3}}\\ r_2 &= {\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (9 r^{2}-9 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {2}{3}}, {\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{3}}\) 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 +\frac {2}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{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} 9 a_{n} \left (n +r \right ) \left (n +r -1\right )+9 a_{n -1} \left (n +r -1\right )+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {9 a_{n -1} \left (n +r -1\right )}{9 n^{2}+18 n r +9 r^{2}-9 n -9 r +2}\tag {4} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (1-3 n \right )}{3 n^{2}+n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {2}{3}}\) 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 \[ a_{1}=-\frac {9 r}{9 r^{2}+9 r +2} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{2}={\frac {5}{28}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {5}{28}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{3}=-{\frac {1}{21}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {5}{28}\) |
| \(a_{3}\) | \(-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )}\) | \(-{\frac {1}{21}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {6561 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right )} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{4}={\frac {11}{1092}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {5}{28}\) |
| \(a_{3}\) | \(-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )}\) | \(-{\frac {1}{21}}\) |
| \(a_{4}\) | \(\frac {6561 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right )}\) | \(\frac {11}{1092}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {59049 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right ) \left (9 r^{2}+81 r +182\right )} \] Which for the root \(r = {\frac {2}{3}}\) becomes \[ a_{5}=-{\frac {11}{6240}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {5}{28}\) |
| \(a_{3}\) | \(-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )}\) | \(-{\frac {1}{21}}\) |
| \(a_{4}\) | \(\frac {6561 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right )}\) | \(\frac {11}{1092}\) |
| \(a_{5}\) | \(-\frac {59049 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right ) \left (9 r^{2}+81 r +182\right )}\) | \(-{\frac {11}{6240}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {2}{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^{\frac {2}{3}} \left (1-\frac {x}{2}+\frac {5 x^{2}}{28}-\frac {x^{3}}{21}+\frac {11 x^{4}}{1092}-\frac {11 x^{5}}{6240}+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} 9 b_{n} \left (n +r \right ) \left (n +r -1\right )+9 b_{n -1} \left (n +r -1\right )+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {9 b_{n -1} \left (n +r -1\right )}{9 n^{2}+18 n r +9 r^{2}-9 n -9 r +2}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{n} = \frac {b_{n -1} \left (2-3 n \right )}{3 n^{2}-n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) 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 \[ b_{1}=-\frac {9 r}{9 r^{2}+9 r +2} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{2}={\frac {1}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {1}{5}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{3}=-{\frac {7}{120}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {1}{5}\) |
| \(b_{3}\) | \(-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )}\) | \(-{\frac {7}{120}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {6561 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{4}={\frac {7}{528}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {1}{5}\) |
| \(b_{3}\) | \(-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )}\) | \(-{\frac {7}{120}}\) |
| \(b_{4}\) | \(\frac {6561 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right )}\) | \(\frac {7}{528}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {59049 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right ) \left (9 r^{2}+81 r +182\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{5}=-{\frac {13}{5280}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {9 r}{9 r^{2}+9 r +2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {81 r \left (1+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right )}\) | \(\frac {1}{5}\) |
| \(b_{3}\) | \(-\frac {729 r \left (1+r \right ) \left (2+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right )}\) | \(-{\frac {7}{120}}\) |
| \(b_{4}\) | \(\frac {6561 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right )}\) | \(\frac {7}{528}\) |
| \(b_{5}\) | \(-\frac {59049 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (9 r^{2}+9 r +2\right ) \left (9 r^{2}+27 r +20\right ) \left (9 r^{2}+45 r +56\right ) \left (9 r^{2}+63 r +110\right ) \left (9 r^{2}+81 r +182\right )}\) | \(-{\frac {13}{5280}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {2}{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 {1}{3}} \left (1-\frac {x}{2}+\frac {x^{2}}{5}-\frac {7 x^{3}}{120}+\frac {7 x^{4}}{528}-\frac {13 x^{5}}{5280}+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^{\frac {2}{3}} \left (1-\frac {x}{2}+\frac {5 x^{2}}{28}-\frac {x^{3}}{21}+\frac {11 x^{4}}{1092}-\frac {11 x^{5}}{6240}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{3}} \left (1-\frac {x}{2}+\frac {x^{2}}{5}-\frac {7 x^{3}}{120}+\frac {7 x^{4}}{528}-\frac {13 x^{5}}{5280}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {2}{3}} \left (1-\frac {x}{2}+\frac {5 x^{2}}{28}-\frac {x^{3}}{21}+\frac {11 x^{4}}{1092}-\frac {11 x^{5}}{6240}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-\frac {x}{2}+\frac {x^{2}}{5}-\frac {7 x^{3}}{120}+\frac {7 x^{4}}{528}-\frac {13 x^{5}}{5280}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {2}{3}} \left (1-\frac {x}{2}+\frac {5 x^{2}}{28}-\frac {x^{3}}{21}+\frac {11 x^{4}}{1092}-\frac {11 x^{5}}{6240}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-\frac {x}{2}+\frac {x^{2}}{5}-\frac {7 x^{3}}{120}+\frac {7 x^{4}}{528}-\frac {13 x^{5}}{5280}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {2}{3}} \left (1-\frac {x}{2}+\frac {5 x^{2}}{28}-\frac {x^{3}}{21}+\frac {11 x^{4}}{1092}-\frac {11 x^{5}}{6240}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-\frac {x}{2}+\frac {x^{2}}{5}-\frac {7 x^{3}}{120}+\frac {7 x^{4}}{528}-\frac {13 x^{5}}{5280}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 x^{2} y^{\prime \prime }+9 x^{2} y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-y^{\prime }-\frac {2 y}{9 x^{2}} \\ \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 }+y^{\prime }+\frac {2 y}{9 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 )=1, P_{3}\left (x \right )=\frac {2}{9 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}}}=0 \\ {} & \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}}}=\frac {2}{9} \\ {} & \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} \\ {} & {} & 9 x^{2} y^{\prime \prime }+9 x^{2} y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+3 r \right ) \left (-2+3 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (3 k +3 r -1\right ) \left (3 k +3 r -2\right )+9 a_{k -1} \left (k -1+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+3 r \right ) \left (-2+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{3}, \frac {2}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (k +r -\frac {1}{3}\right ) \left (k +r -\frac {2}{3}\right ) a_{k}+9 a_{k -1} \left (k -1+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 9 \left (k +\frac {2}{3}+r \right ) \left (k +\frac {1}{3}+r \right ) a_{k +1}+9 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {9 a_{k} \left (k +r \right )}{\left (3 k +2+3 r \right ) \left (3 k +1+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +1}=-\frac {9 a_{k} \left (k +\frac {1}{3}\right )}{\left (3 k +3\right ) \left (3 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{3}}, a_{k +1}=-\frac {9 a_{k} \left (k +\frac {1}{3}\right )}{\left (3 k +3\right ) \left (3 k +2\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & a_{k +1}=-\frac {9 a_{k} \left (k +\frac {2}{3}\right )}{\left (3 k +4\right ) \left (3 k +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 {9 a_{k} \left (k +\frac {2}{3}\right )}{\left (3 k +4\right ) \left (3 k +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 +\frac {1}{3}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {2}{3}}\right ), a_{k +1}=-\frac {9 a_{k} \left (k +\frac {1}{3}\right )}{\left (3 k +3\right ) \left (3 k +2\right )}, b_{k +1}=-\frac {9 b_{k} \left (k +\frac {2}{3}\right )}{\left (3 k +4\right ) \left (3 k +3\right )}\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 47
Order:=6; dsolve(9*x^2*diff(y(x),x$2)+9*x^2*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{3}} \left (1-\frac {1}{2} x +\frac {1}{5} x^{2}-\frac {7}{120} x^{3}+\frac {7}{528} x^{4}-\frac {13}{5280} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\frac {2}{3}} \left (1-\frac {1}{2} x +\frac {5}{28} x^{2}-\frac {1}{21} x^{3}+\frac {11}{1092} x^{4}-\frac {11}{6240} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 90
AsymptoticDSolveValue[9*x^2*y''[x]+9*x^2*y'[x]+2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \sqrt [3]{x} \left (-\frac {13 x^5}{5280}+\frac {7 x^4}{528}-\frac {7 x^3}{120}+\frac {x^2}{5}-\frac {x}{2}+1\right )+c_1 x^{2/3} \left (-\frac {11 x^5}{6240}+\frac {11 x^4}{1092}-\frac {x^3}{21}+\frac {5 x^2}{28}-\frac {x}{2}+1\right ) \]