Internal problem ID [1321]
Internal file name [OUTPUT/1322_Sunday_June_05_2022_02_10_15_AM_64338585/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: 32.
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]]
\[ \boxed {x^{2} \left (x +6\right ) y^{\prime \prime }+x \left (11+4 x \right ) y^{\prime }+\left (1+2 x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{3}+6 x^{2}\right ) y^{\prime \prime }+\left (4 x^{2}+11 x \right ) y^{\prime }+\left (1+2 x \right ) y = 0 \] 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 {11+4 x}{x \left (x +6\right )}\\ q(x) &= \frac {1+2 x}{x^{2} \left (x +6\right )}\\ \end {align*}
| \(p(x)=\frac {11+4 x}{x \left (x +6\right )}\) | |
| singularity | type |
| \(x = -6\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1+2 x}{x^{2} \left (x +6\right )}\) | |
| singularity | type |
| \(x = -6\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-6, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x +6\right ) y^{\prime \prime }+\left (4 x^{2}+11 x \right ) y^{\prime }+\left (1+2 x \right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (x +6\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 (4 x^{2}+11 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (1+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 6 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+11 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 6 x^{r} a_{0} r \left (-1+r \right )+11 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (6 x^{r} r \left (-1+r \right )+11 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (6 r^{2}+5 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 6 r^{2}+5 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -{\frac {1}{3}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (6 r^{2}+5 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-{\frac {1}{3}}, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{6}}\) 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}{2}} \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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+6 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -1} \left (n +r -1\right )+11 a_{n} \left (n +r \right )+a_{n}+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}+n +r \right )}{6 n^{2}+12 n r +6 r^{2}+5 n +5 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (9 n^{2}+3 n -2\right )}{54 n^{2}+9 n}\tag {5} \] 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 \[ a_{1}=\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{1}=-{\frac {10}{63}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {10}{63}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{2}={\frac {200}{7371}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {10}{63}}\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {200}{7371}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {17600}{3781323}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {10}{63}}\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {200}{7371}\) |
| \(a_{3}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400}\) | \(-{\frac {17600}{3781323}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{4}={\frac {3872}{4861701}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {10}{63}}\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {200}{7371}\) |
| \(a_{3}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400}\) | \(-{\frac {17600}{3781323}}\) |
| \(a_{4}\) | \(\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right )}\) | \(\frac {3872}{4861701}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2} \left (r +6\right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right ) \left (6 r^{2}+65 r +176\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {921536}{6782072895}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {10}{63}}\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {200}{7371}\) |
| \(a_{3}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400}\) | \(-{\frac {17600}{3781323}}\) |
| \(a_{4}\) | \(\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right )}\) | \(\frac {3872}{4861701}\) |
| \(a_{5}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2} \left (r +6\right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right ) \left (6 r^{2}+65 r +176\right )}\) | \(-{\frac {921536}{6782072895}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1-\frac {10 x}{63}+\frac {200 x^{2}}{7371}-\frac {17600 x^{3}}{3781323}+\frac {3872 x^{4}}{4861701}-\frac {921536 x^{5}}{6782072895}+O\left (x^{6}\right )}{x^{\frac {1}{3}}} \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} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+6 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -1} \left (n +r -1\right )+11 b_{n} \left (n +r \right )+b_{n}+2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}+2 n r +r^{2}+n +r \right )}{6 n^{2}+12 n r +6 r^{2}+5 n +5 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{n} = \frac {-4 n^{2} b_{n -1}+b_{n -1}}{24 n^{2}-4 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{1}=-{\frac {3}{20}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {3}{20}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{2}={\frac {9}{352}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {3}{20}}\) |
| \(b_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {9}{352}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{3}=-{\frac {105}{23936}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {3}{20}}\) |
| \(b_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {9}{352}\) |
| \(b_{3}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400}\) | \(-{\frac {105}{23936}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{4}={\frac {6615}{8808448}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {3}{20}}\) |
| \(b_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {9}{352}\) |
| \(b_{3}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400}\) | \(-{\frac {105}{23936}}\) |
| \(b_{4}\) | \(\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right )}\) | \(\frac {6615}{8808448}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2} \left (r +6\right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right ) \left (6 r^{2}+65 r +176\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{5}=-{\frac {11907}{92889088}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -2}{6 r^{2}+17 r +12}\) | \(-{\frac {3}{20}}\) |
| \(b_{2}\) | \(\frac {\left (r +3\right ) \left (r +1\right ) \left (2+r \right )^{2}}{36 r^{4}+276 r^{3}+775 r^{2}+943 r +420}\) | \(\frac {9}{352}\) |
| \(b_{3}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )}{216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400}\) | \(-{\frac {105}{23936}}\) |
| \(b_{4}\) | \(\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right )}\) | \(\frac {6615}{8808448}\) |
| \(b_{5}\) | \(-\frac {\left (r +3\right )^{2} \left (r +1\right ) \left (2+r \right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2} \left (r +6\right )}{\left (216 r^{6}+3132 r^{5}+18486 r^{4}+56753 r^{3}+95433 r^{2}+83230 r +29400\right ) \left (6 r^{2}+53 r +117\right ) \left (6 r^{2}+65 r +176\right )}\) | \(-{\frac {11907}{92889088}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \frac {1}{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-\frac {3 x}{20}+\frac {9 x^{2}}{352}-\frac {105 x^{3}}{23936}+\frac {6615 x^{4}}{8808448}-\frac {11907 x^{5}}{92889088}+O\left (x^{6}\right )}{\sqrt {x}} \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 ) \\ &= \frac {c_{1} \left (1-\frac {10 x}{63}+\frac {200 x^{2}}{7371}-\frac {17600 x^{3}}{3781323}+\frac {3872 x^{4}}{4861701}-\frac {921536 x^{5}}{6782072895}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} + \frac {c_{2} \left (1-\frac {3 x}{20}+\frac {9 x^{2}}{352}-\frac {105 x^{3}}{23936}+\frac {6615 x^{4}}{8808448}-\frac {11907 x^{5}}{92889088}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-\frac {10 x}{63}+\frac {200 x^{2}}{7371}-\frac {17600 x^{3}}{3781323}+\frac {3872 x^{4}}{4861701}-\frac {921536 x^{5}}{6782072895}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}}+\frac {c_{2} \left (1-\frac {3 x}{20}+\frac {9 x^{2}}{352}-\frac {105 x^{3}}{23936}+\frac {6615 x^{4}}{8808448}-\frac {11907 x^{5}}{92889088}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-\frac {10 x}{63}+\frac {200 x^{2}}{7371}-\frac {17600 x^{3}}{3781323}+\frac {3872 x^{4}}{4861701}-\frac {921536 x^{5}}{6782072895}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}}+\frac {c_{2} \left (1-\frac {3 x}{20}+\frac {9 x^{2}}{352}-\frac {105 x^{3}}{23936}+\frac {6615 x^{4}}{8808448}-\frac {11907 x^{5}}{92889088}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-\frac {10 x}{63}+\frac {200 x^{2}}{7371}-\frac {17600 x^{3}}{3781323}+\frac {3872 x^{4}}{4861701}-\frac {921536 x^{5}}{6782072895}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}}+\frac {c_{2} \left (1-\frac {3 x}{20}+\frac {9 x^{2}}{352}-\frac {105 x^{3}}{23936}+\frac {6615 x^{4}}{8808448}-\frac {11907 x^{5}}{92889088}+O\left (x^{6}\right )\right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +6\right ) y^{\prime \prime }+\left (4 x^{2}+11 x \right ) y^{\prime }+\left (1+2 x \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 (1+2 x \right ) y}{x^{2} \left (x +6\right )}-\frac {\left (11+4 x \right ) y^{\prime }}{x \left (x +6\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 (11+4 x \right ) y^{\prime }}{x \left (x +6\right )}+\frac {\left (1+2 x \right ) y}{x^{2} \left (x +6\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 {11+4 x}{x \left (x +6\right )}, P_{3}\left (x \right )=\frac {1+2 x}{x^{2} \left (x +6\right )}\right ] \\ {} & \circ & \left (x +6\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-6 \\ {} & {} & \left (\left (x +6\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-6}}}=\frac {13}{6} \\ {} & \circ & \left (x +6\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-6 \\ {} & {} & \left (\left (x +6\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-6}}}=0 \\ {} & \circ & x =-6\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}=-6 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x +6\right ) y^{\prime \prime }+x \left (11+4 x \right ) y^{\prime }+\left (1+2 x \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -6\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-12 u^{2}+36 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (4 u^{2}-37 u +78\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-11+2 u \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} \\ {} & {} & 6 a_{0} r \left (7+6 r \right ) u^{-1+r}+\left (6 a_{1} \left (1+r \right ) \left (13+6 r \right )-a_{0} \left (12 r^{2}+25 r +11\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (6 a_{k +1} \left (k +r +1\right ) \left (6 k +13+6 r \right )-a_{k} \left (12 k^{2}+24 k r +12 r^{2}+25 k +25 r +11\right )+a_{k -1} \left (k +r +1\right ) \left (k +r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 6 r \left (7+6 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {7}{6}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 6 a_{1} \left (1+r \right ) \left (13+6 r \right )-a_{0} \left (12 r^{2}+25 r +11\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-12 a_{k}+a_{k -1}+36 a_{k +1}\right ) k^{2}+\left (\left (-24 a_{k}+2 a_{k -1}+72 a_{k +1}\right ) r -25 a_{k}+a_{k -1}+114 a_{k +1}\right ) k +\left (-12 a_{k}+a_{k -1}+36 a_{k +1}\right ) r^{2}+\left (-25 a_{k}+a_{k -1}+114 a_{k +1}\right ) r -11 a_{k}+78 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-12 a_{k +1}+a_{k}+36 a_{k +2}\right ) \left (k +1\right )^{2}+\left (\left (-24 a_{k +1}+2 a_{k}+72 a_{k +2}\right ) r -25 a_{k +1}+a_{k}+114 a_{k +2}\right ) \left (k +1\right )+\left (-12 a_{k +1}+a_{k}+36 a_{k +2}\right ) r^{2}+\left (-25 a_{k +1}+a_{k}+114 a_{k +2}\right ) r -11 a_{k +1}+78 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{k +1}+2 k r a_{k}-24 k r a_{k +1}+r^{2} a_{k}-12 r^{2} a_{k +1}+3 k a_{k}-49 k a_{k +1}+3 r a_{k}-49 r a_{k +1}+2 a_{k}-48 a_{k +1}}{6 \left (6 k^{2}+12 k r +6 r^{2}+31 k +31 r +38\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{k +1}+3 k a_{k}-49 k a_{k +1}+2 a_{k}-48 a_{k +1}}{6 \left (6 k^{2}+31 k +38\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 {k^{2} a_{k}-12 k^{2} a_{k +1}+3 k a_{k}-49 k a_{k +1}+2 a_{k}-48 a_{k +1}}{6 \left (6 k^{2}+31 k +38\right )}, 78 a_{1}-11 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +6 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +6\right )^{k}, a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{k +1}+3 k a_{k}-49 k a_{k +1}+2 a_{k}-48 a_{k +1}}{6 \left (6 k^{2}+31 k +38\right )}, 78 a_{1}-11 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {7}{6} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{k +1}+\frac {2}{3} k a_{k}-21 k a_{k +1}-\frac {5}{36} a_{k}-\frac {43}{6} a_{k +1}}{6 \left (6 k^{2}+17 k +10\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {7}{6} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {7}{6}}, a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{k +1}+\frac {2}{3} k a_{k}-21 k a_{k +1}-\frac {5}{36} a_{k}-\frac {43}{6} a_{k +1}}{6 \left (6 k^{2}+17 k +10\right )}, -6 a_{1}+\frac {11 a_{0}}{6}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +6 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +6\right )^{k -\frac {7}{6}}, a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{k +1}+\frac {2}{3} k a_{k}-21 k a_{k +1}-\frac {5}{36} a_{k}-\frac {43}{6} a_{k +1}}{6 \left (6 k^{2}+17 k +10\right )}, -6 a_{1}+\frac {11 a_{0}}{6}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +6\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +6\right )^{k -\frac {7}{6}}\right ), a_{k +2}=-\frac {k^{2} a_{k}-12 k^{2} a_{1+k}+3 k a_{k}-49 k a_{1+k}+2 a_{k}-48 a_{1+k}}{6 \left (6 k^{2}+31 k +38\right )}, 78 a_{1}-11 a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}-12 k^{2} b_{1+k}+\frac {2}{3} k b_{k}-21 k b_{1+k}-\frac {5}{36} b_{k}-\frac {43}{6} b_{1+k}}{6 \left (6 k^{2}+17 k +10\right )}, -6 b_{1}+\frac {11 b_{0}}{6}=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: 0.047 (sec). Leaf size: 47
Order:=6; dsolve(x^2*(6+x)*diff(y(x),x$2)+x*(11+4*x)*diff(y(x),x)+(1+2*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} \left (1-\frac {10}{63} x +\frac {200}{7371} x^{2}-\frac {17600}{3781323} x^{3}+\frac {3872}{4861701} x^{4}-\frac {921536}{6782072895} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) x^{\frac {1}{6}}+c_{1} \left (1-\frac {3}{20} x +\frac {9}{352} x^{2}-\frac {105}{23936} x^{3}+\frac {6615}{8808448} x^{4}-\frac {11907}{92889088} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 90
AsymptoticDSolveValue[x^2*(6+x)*y''[x]+x*(11+4*x)*y'[x]+(1+2*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \frac {c_1 \left (-\frac {921536 x^5}{6782072895}+\frac {3872 x^4}{4861701}-\frac {17600 x^3}{3781323}+\frac {200 x^2}{7371}-\frac {10 x}{63}+1\right )}{\sqrt [3]{x}}+\frac {c_2 \left (-\frac {11907 x^5}{92889088}+\frac {6615 x^4}{8808448}-\frac {105 x^3}{23936}+\frac {9 x^2}{352}-\frac {3 x}{20}+1\right )}{\sqrt {x}} \]