Internal problem ID [1344]
Internal file name [OUTPUT/1345_Sunday_June_05_2022_02_12_07_AM_78276812/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. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
Problem number: 64.
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]]
\[ \boxed {9 x^{2} \left (x +5\right ) y^{\prime \prime }+9 x \left (5+9 x \right ) y^{\prime }-\left (5-8 x \right ) y=0} \] 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. \[ \left (9 x^{3}+45 x^{2}\right ) y^{\prime \prime }+\left (81 x^{2}+45 x \right ) y^{\prime }+\left (8 x -5\right ) y = 0 \] 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 {5+9 x}{x \left (x +5\right )}\\ q(x) &= \frac {8 x -5}{9 x^{2} \left (x +5\right )}\\ \end {align*}
| \(p(x)=\frac {5+9 x}{x \left (x +5\right )}\) | |
| singularity | type |
| \(x = -5\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {8 x -5}{9 x^{2} \left (x +5\right )}\) | |
| singularity | type |
| \(x = -5\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-5, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 9 x^{2} \left (x +5\right ) y^{\prime \prime }+\left (81 x^{2}+45 x \right ) y^{\prime }+\left (8 x -5\right ) y = 0 \] 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} 9 x^{2} \left (x +5\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 (81 x^{2}+45 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (8 x -5\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 }}9 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}45 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}81 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}45 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 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 ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}81 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}81 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}8 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}8 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 }}9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}45 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}81 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}45 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}8 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 45 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+45 x^{n +r} a_{n} \left (n +r \right )-5 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 45 x^{r} a_{0} r \left (-1+r \right )+45 x^{r} a_{0} r -5 a_{0} x^{r} = 0 \] Or \[ \left (45 x^{r} r \left (-1+r \right )+45 x^{r} r -5 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (45 r^{2}-5\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 45 r^{2}-5 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= -{\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (45 r^{2}-5\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{3}}, -{\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {2}{3}}\) 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 {1}{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 -1} \left (n +r -1\right ) \left (n +r -2\right )+45 a_{n} \left (n +r \right ) \left (n +r -1\right )+81 a_{n -1} \left (n +r -1\right )+45 a_{n} \left (n +r \right )+8 a_{n -1}-5 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (9 n^{2}+18 n r +9 r^{2}+54 n +54 r -55\right )}{5 \left (9 n^{2}+18 n r +9 r^{2}-1\right )}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{n} = -\frac {3 a_{n -1} \left (n^{2}+\frac {20}{3} n -4\right )}{15 n^{2}+10 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) 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 \[ a_{1}=\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{1}=-{\frac {11}{25}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(-{\frac {11}{25}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{2}={\frac {11}{50}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(-{\frac {11}{25}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {11}{50}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {1}{10}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(-{\frac {11}{25}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {11}{50}\) |
| \(a_{3}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(-{\frac {1}{10}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right )}{625 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{4}={\frac {29}{700}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(-{\frac {11}{25}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {11}{50}\) |
| \(a_{3}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{4}\) | \(\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right )}{625 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(\frac {29}{700}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right ) \left (9 r^{2}+144 r +440\right )}{3125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {4727}{297500}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(-{\frac {11}{25}}\) |
| \(a_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(\frac {11}{50}\) |
| \(a_{3}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(-{\frac {1}{10}}\) |
| \(a_{4}\) | \(\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right )}{625 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(\frac {29}{700}\) |
| \(a_{5}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right ) \left (9 r^{2}+144 r +440\right )}{3125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )}\) | \(-{\frac {4727}{297500}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{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 {1}{3}} \left (1-\frac {11 x}{25}+\frac {11 x^{2}}{50}-\frac {x^{3}}{10}+\frac {29 x^{4}}{700}-\frac {4727 x^{5}}{297500}+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 -1} \left (n +r -1\right ) \left (n +r -2\right )+45 b_{n} \left (n +r \right ) \left (n +r -1\right )+81 b_{n -1} \left (n +r -1\right )+45 b_{n} \left (n +r \right )+8 b_{n -1}-5 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (9 n^{2}+18 n r +9 r^{2}+54 n +54 r -55\right )}{5 \left (9 n^{2}+18 n r +9 r^{2}-1\right )}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{n} = -\frac {3 b_{n -1} \left (n^{2}+\frac {16}{3} n -8\right )}{15 n^{2}-10 n}\tag {5} \] 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 \[ b_{1}=\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{1}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{2}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {1}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{3}={\frac {17}{70}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {17}{70}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right )}{625 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{4}=-{\frac {187}{1750}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {17}{70}\) |
| \(b_{4}\) | \(\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right )}{625 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(-{\frac {187}{1750}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right ) \left (9 r^{2}+144 r +440\right )}{3125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{5}={\frac {24497}{568750}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-9 r^{2}-72 r -8}{45 r^{2}+90 r +40}\) | \(1\) |
| \(b_{2}\) | \(\frac {81 r^{4}+1458 r^{3}+7353 r^{2}+7128 r +712}{25 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right )}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right )}{125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right )}\) | \(\frac {17}{70}\) |
| \(b_{4}\) | \(\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right )}{625 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right )}\) | \(-{\frac {187}{1750}}\) |
| \(b_{5}\) | \(-\frac {\left (9 r^{2}+108 r +188\right ) \left (9 r^{2}+90 r +89\right ) \left (9 r^{2}+72 r +8\right ) \left (9 r^{2}+126 r +305\right ) \left (9 r^{2}+144 r +440\right )}{3125 \left (9 r^{2}+18 r +8\right ) \left (9 r^{2}+36 r +35\right ) \left (9 r^{2}+54 r +80\right ) \left (9 r^{2}+72 r +143\right ) \left (9 r^{2}+90 r +224\right )}\) | \(\frac {24497}{568750}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{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 ) \\ &= \frac {1+x -\frac {x^{2}}{2}+\frac {17 x^{3}}{70}-\frac {187 x^{4}}{1750}+\frac {24497 x^{5}}{568750}+O\left (x^{6}\right )}{x^{\frac {1}{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^{\frac {1}{3}} \left (1-\frac {11 x}{25}+\frac {11 x^{2}}{50}-\frac {x^{3}}{10}+\frac {29 x^{4}}{700}-\frac {4727 x^{5}}{297500}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {17 x^{3}}{70}-\frac {187 x^{4}}{1750}+\frac {24497 x^{5}}{568750}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{3}} \left (1-\frac {11 x}{25}+\frac {11 x^{2}}{50}-\frac {x^{3}}{10}+\frac {29 x^{4}}{700}-\frac {4727 x^{5}}{297500}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {17 x^{3}}{70}-\frac {187 x^{4}}{1750}+\frac {24497 x^{5}}{568750}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{3}} \left (1-\frac {11 x}{25}+\frac {11 x^{2}}{50}-\frac {x^{3}}{10}+\frac {29 x^{4}}{700}-\frac {4727 x^{5}}{297500}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {17 x^{3}}{70}-\frac {187 x^{4}}{1750}+\frac {24497 x^{5}}{568750}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{3}} \left (1-\frac {11 x}{25}+\frac {11 x^{2}}{50}-\frac {x^{3}}{10}+\frac {29 x^{4}}{700}-\frac {4727 x^{5}}{297500}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {17 x^{3}}{70}-\frac {187 x^{4}}{1750}+\frac {24497 x^{5}}{568750}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (x +5\right ) y^{\prime \prime }+\left (81 x^{2}+45 x \right ) y^{\prime }+\left (8 x -5\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 (8 x -5\right ) y}{9 x^{2} \left (x +5\right )}-\frac {\left (5+9 x \right ) y^{\prime }}{x \left (x +5\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 (5+9 x \right ) y^{\prime }}{x \left (x +5\right )}+\frac {\left (8 x -5\right ) y}{9 x^{2} \left (x +5\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 {5+9 x}{x \left (x +5\right )}, P_{3}\left (x \right )=\frac {8 x -5}{9 x^{2} \left (x +5\right )}\right ] \\ {} & \circ & \left (x +5\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-5 \\ {} & {} & \left (\left (x +5\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-5}}}=8 \\ {} & \circ & \left (x +5\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-5 \\ {} & {} & \left (\left (x +5\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-5}}}=0 \\ {} & \circ & x =-5\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}=-5 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (x +5\right ) y^{\prime \prime }+9 x \left (5+9 x \right ) y^{\prime }+\left (8 x -5\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -5\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (9 u^{3}-90 u^{2}+225 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (81 u^{2}-765 u +1800\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (8 u -45\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} \\ {} & {} & 225 a_{0} r \left (7+r \right ) u^{-1+r}+\left (225 a_{1} \left (1+r \right ) \left (8+r \right )-45 a_{0} \left (2 r^{2}+15 r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (225 a_{k +1} \left (k +1+r \right ) \left (k +8+r \right )-45 a_{k} \left (2 k^{2}+4 k r +2 r^{2}+15 k +15 r +1\right )+a_{k -1} \left (9 \left (k -1\right )^{2}+18 \left (k -1\right ) r +9 r^{2}+72 k -64+72 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 225 r \left (7+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-7, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 225 a_{1} \left (1+r \right ) \left (8+r \right )-45 a_{0} \left (2 r^{2}+15 r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (-10 a_{k}+a_{k -1}+25 a_{k +1}\right ) k^{2}+9 \left (2 \left (-10 a_{k}+a_{k -1}+25 a_{k +1}\right ) r -75 a_{k}+6 a_{k -1}+225 a_{k +1}\right ) k +9 \left (-10 a_{k}+a_{k -1}+25 a_{k +1}\right ) r^{2}+27 \left (-25 a_{k}+2 a_{k -1}+75 a_{k +1}\right ) r -45 a_{k}-55 a_{k -1}+1800 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 9 \left (-10 a_{k +1}+a_{k}+25 a_{k +2}\right ) \left (k +1\right )^{2}+9 \left (2 \left (-10 a_{k +1}+a_{k}+25 a_{k +2}\right ) r -75 a_{k +1}+6 a_{k}+225 a_{k +2}\right ) \left (k +1\right )+9 \left (-10 a_{k +1}+a_{k}+25 a_{k +2}\right ) r^{2}+27 \left (-25 a_{k +1}+2 a_{k}+75 a_{k +2}\right ) r -45 a_{k +1}-55 a_{k}+1800 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {9 k^{2} a_{k}-90 k^{2} a_{k +1}+18 k r a_{k}-180 k r a_{k +1}+9 r^{2} a_{k}-90 r^{2} a_{k +1}+72 k a_{k}-855 k a_{k +1}+72 r a_{k}-855 r a_{k +1}+8 a_{k}-810 a_{k +1}}{225 \left (k^{2}+2 k r +r^{2}+11 k +11 r +18\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-7 \\ {} & {} & a_{k +2}=-\frac {9 k^{2} a_{k}-90 k^{2} a_{k +1}-54 k a_{k}+405 k a_{k +1}-55 a_{k}+765 a_{k +1}}{225 \left (k^{2}-3 k -10\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-7\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =5 \\ {} & {} & a_{k +2}=-\frac {9 k^{2} a_{k}-90 k^{2} a_{k +1}-54 k a_{k}+405 k a_{k +1}-55 a_{k}+765 a_{k +1}}{225 \left (k^{2}-3 k -10\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {9 k^{2} a_{k}-90 k^{2} a_{k +1}+72 k a_{k}-855 k a_{k +1}+8 a_{k}-810 a_{k +1}}{225 \left (k^{2}+11 k +18\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {9 k^{2} a_{k}-90 k^{2} a_{k +1}+72 k a_{k}-855 k a_{k +1}+8 a_{k}-810 a_{k +1}}{225 \left (k^{2}+11 k +18\right )}, 1800 a_{1}-45 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +5 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +5\right )^{k}, a_{k +2}=-\frac {9 k^{2} a_{k}-90 k^{2} a_{k +1}+72 k a_{k}-855 k a_{k +1}+8 a_{k}-810 a_{k +1}}{225 \left (k^{2}+11 k +18\right )}, 1800 a_{1}-45 a_{0}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 47
Order:=6; dsolve(9*x^2*(5+x)*diff(y(x),x$2)+9*x*(5+9*x)*diff(y(x),x)-(5-8*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {2}{3}} \left (1-\frac {11}{25} x +\frac {11}{50} x^{2}-\frac {1}{10} x^{3}+\frac {29}{700} x^{4}-\frac {4727}{297500} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1+x -\frac {1}{2} x^{2}+\frac {17}{70} x^{3}-\frac {187}{1750} x^{4}+\frac {24497}{568750} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 86
AsymptoticDSolveValue[9*x^2*(5+x)*y''[x]+9*x*(5+9*x)*y'[x]-(5-8*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [3]{x} \left (-\frac {4727 x^5}{297500}+\frac {29 x^4}{700}-\frac {x^3}{10}+\frac {11 x^2}{50}-\frac {11 x}{25}+1\right )+\frac {c_2 \left (\frac {24497 x^5}{568750}-\frac {187 x^4}{1750}+\frac {17 x^3}{70}-\frac {x^2}{2}+x+1\right )}{\sqrt [3]{x}} \]