Internal problem ID [1432]
Internal file name [OUTPUT/1433_Sunday_June_05_2022_02_17_06_AM_65559690/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. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
Problem number: 16.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {3 x^{2} \left (3+x \right ) y^{\prime \prime }-x \left (15+x \right ) y^{\prime }-20 y=0} \] 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. \[ \left (3 x^{3}+9 x^{2}\right ) y^{\prime \prime }+\left (-x^{2}-15 x \right ) y^{\prime }-20 y = 0 \] 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 {15+x}{3 x \left (3+x \right )}\\ q(x) &= -\frac {20}{3 x^{2} \left (3+x \right )}\\ \end {align*}
| \(p(x)=-\frac {15+x}{3 x \left (3+x \right )}\) | |
| singularity | type |
| \(x = -3\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {20}{3 x^{2} \left (3+x \right )}\) | |
| singularity | type |
| \(x = -3\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-3, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 3 x^{2} \left (3+x \right ) y^{\prime \prime }+\left (-x^{2}-15 x \right ) y^{\prime }-20 y = 0 \] 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 (3+x \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 (-x^{2}-15 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-20 \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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 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 (-15 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-20 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 }}3 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \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 =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 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 (-15 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-20 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-15 x^{n +r} a_{n} \left (n +r \right )-20 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )-15 x^{r} a_{0} r -20 a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )-15 x^{r} r -20 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (9 r^{2}-24 r -20\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 9 r^{2}-24 r -20 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {10}{3}}\\ r_2 &= -{\frac {2}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (9 r^{2}-24 r -20\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {10}{3}}, -{\frac {2}{3}}\right ]\).
Since \(r_1 - r_2 = 4\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= x^{\frac {10}{3}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{\frac {2}{3}}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {10}{3}}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {2}{3}}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )-15 a_{n} \left (n +r \right )-20 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (3 n^{2}+6 n r +3 r^{2}-10 n -10 r +7\right )}{9 n^{2}+18 n r +9 r^{2}-24 n -24 r -20}\tag {4} \] Which for the root \(r = {\frac {10}{3}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (3 n^{2}+10 n +7\right )}{9 n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {10}{3}}\) 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 \[ a_{1}=\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35} \] Which for the root \(r = {\frac {10}{3}}\) becomes \[ a_{1}=-{\frac {4}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(-{\frac {4}{9}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )} \] Which for the root \(r = {\frac {10}{3}}\) becomes \[ a_{2}={\frac {13}{81}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(-{\frac {4}{9}}\) |
| \(a_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(\frac {13}{81}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080} \] Which for the root \(r = {\frac {10}{3}}\) becomes \[ a_{3}=-{\frac {832}{15309}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(-{\frac {4}{9}}\) |
| \(a_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(\frac {13}{81}\) |
| \(a_{3}\) | \(\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080}\) | \(-{\frac {832}{15309}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624} \] Which for the root \(r = {\frac {10}{3}}\) becomes \[ a_{4}={\frac {2470}{137781}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(-{\frac {4}{9}}\) |
| \(a_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(\frac {13}{81}\) |
| \(a_{3}\) | \(\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080}\) | \(-{\frac {832}{15309}}\) |
| \(a_{4}\) | \(\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624}\) | \(\frac {2470}{137781}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (r +4\right ) \left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{243 r^{5}+3240 r^{4}+12420 r^{3}-90 r^{2}-76503 r -91630} \] Which for the root \(r = {\frac {10}{3}}\) becomes \[ a_{5}=-{\frac {21736}{3720087}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(-{\frac {4}{9}}\) |
| \(a_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(\frac {13}{81}\) |
| \(a_{3}\) | \(\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080}\) | \(-{\frac {832}{15309}}\) |
| \(a_{4}\) | \(\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624}\) | \(\frac {2470}{137781}\) |
| \(a_{5}\) | \(-\frac {\left (r +4\right ) \left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{243 r^{5}+3240 r^{4}+12420 r^{3}-90 r^{2}-76503 r -91630}\) | \(-{\frac {21736}{3720087}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {10}{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 {10}{3}} \left (1-\frac {4 x}{9}+\frac {13 x^{2}}{81}-\frac {832 x^{3}}{15309}+\frac {2470 x^{4}}{137781}-\frac {21736 x^{5}}{3720087}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{4} \\ &= \frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624}&= \lim _{r\rightarrow -{\frac {2}{3}}}\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624}\\ &= {\frac {7}{59049}} \end {align*}
The limit is \(\frac {7}{59049}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {2}{3}} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} 3 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -1} \left (n +r -1\right )-15 b_{n} \left (n +r \right )-20 b_{n} = 0 \end{equation} Which for for the root \(r = -{\frac {2}{3}}\) becomes \begin{equation} \tag{4A} 3 b_{n -1} \left (n -\frac {5}{3}\right ) \left (n -\frac {8}{3}\right )+9 b_{n} \left (n -\frac {2}{3}\right ) \left (n -\frac {5}{3}\right )-b_{n -1} \left (n -\frac {5}{3}\right )-15 b_{n} \left (n -\frac {2}{3}\right )-20 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (3 n^{2}+6 n r +3 r^{2}-10 n -10 r +7\right )}{9 n^{2}+18 n r +9 r^{2}-24 n -24 r -20}\tag {5} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{n} = -\frac {b_{n -1} \left (3 n^{2}-14 n +15\right )}{9 n^{2}-36 n}\tag {6} \] 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 \[ b_{1}=-\frac {r \left (3 r -4\right )}{9 r^{2}-6 r -35} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{1}={\frac {4}{27}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{2}=-{\frac {1}{243}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(b_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(-{\frac {1}{243}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (3 r^{2}+8 r +4\right ) r \left (r +1\right )}{\left (3 r +11\right ) \left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(b_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(-{\frac {1}{243}}\) |
| \(b_{3}\) | \(\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (r +1\right ) r \left (r +2\right ) \left (r +3\right )}{\left (3 r +14\right ) \left (3 r -7\right ) \left (3 r +8\right ) \left (3 r +11\right )} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{4}={\frac {7}{59049}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(b_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(-{\frac {1}{243}}\) |
| \(b_{3}\) | \(\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624}\) | \(\frac {7}{59049}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (r +4\right ) \left (r +1\right ) r \left (r +2\right ) \left (r +3\right )}{\left (3 r +11\right ) \left (3 r -7\right ) \left (3 r +14\right ) \left (9 r^{2}+66 r +85\right )} \] Which for the root \(r = -{\frac {2}{3}}\) becomes \[ b_{5}=-{\frac {28}{531441}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3 r^{2}+4 r}{9 r^{2}-6 r -35}\) | \(\frac {4}{27}\) |
| \(b_{2}\) | \(\frac {r \left (3 r^{2}+2 r -1\right )}{\left (3 r +8\right ) \left (9 r^{2}-6 r -35\right )}\) | \(-{\frac {1}{243}}\) |
| \(b_{3}\) | \(\frac {-3 r^{4}-11 r^{3}-12 r^{2}-4 r}{81 r^{4}+459 r^{3}+135 r^{2}-2523 r -3080}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (r +2\right ) \left (r +1\right ) r \left (r +3\right )}{81 r^{4}+702 r^{3}+1107 r^{2}-3738 r -8624}\) | \(\frac {7}{59049}\) |
| \(b_{5}\) | \(-\frac {\left (r +4\right ) \left (r +1\right ) r \left (r +2\right ) \left (r +3\right )}{\left (3 r +11\right ) \left (3 r -7\right ) \left (3 r +14\right ) \left (9 r^{2}+66 r +85\right )}\) | \(-{\frac {28}{531441}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {10}{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 {4 x}{27}-\frac {x^{2}}{243}+\frac {7 x^{4}}{59049}-\frac {28 x^{5}}{531441}+O\left (x^{6}\right )}{x^{\frac {2}{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 {10}{3}} \left (1-\frac {4 x}{9}+\frac {13 x^{2}}{81}-\frac {832 x^{3}}{15309}+\frac {2470 x^{4}}{137781}-\frac {21736 x^{5}}{3720087}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {4 x}{27}-\frac {x^{2}}{243}+\frac {7 x^{4}}{59049}-\frac {28 x^{5}}{531441}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {10}{3}} \left (1-\frac {4 x}{9}+\frac {13 x^{2}}{81}-\frac {832 x^{3}}{15309}+\frac {2470 x^{4}}{137781}-\frac {21736 x^{5}}{3720087}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {4 x}{27}-\frac {x^{2}}{243}+\frac {7 x^{4}}{59049}-\frac {28 x^{5}}{531441}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {10}{3}} \left (1-\frac {4 x}{9}+\frac {13 x^{2}}{81}-\frac {832 x^{3}}{15309}+\frac {2470 x^{4}}{137781}-\frac {21736 x^{5}}{3720087}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {4 x}{27}-\frac {x^{2}}{243}+\frac {7 x^{4}}{59049}-\frac {28 x^{5}}{531441}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {10}{3}} \left (1-\frac {4 x}{9}+\frac {13 x^{2}}{81}-\frac {832 x^{3}}{15309}+\frac {2470 x^{4}}{137781}-\frac {21736 x^{5}}{3720087}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {4 x}{27}-\frac {x^{2}}{243}+\frac {7 x^{4}}{59049}-\frac {28 x^{5}}{531441}+O\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (3+x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-x^{2}-15 x \right ) y^{\prime }-20 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 {\left (15+x \right ) y^{\prime }}{3 x \left (3+x \right )}+\frac {20 y}{3 x^{2} \left (3+x \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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (15+x \right ) y^{\prime }}{3 x \left (3+x \right )}-\frac {20 y}{3 x^{2} \left (3+x \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 {15+x}{3 x \left (3+x \right )}, P_{3}\left (x \right )=-\frac {20}{3 x^{2} \left (3+x \right )}\right ] \\ {} & \circ & \left (3+x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (3+x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=\frac {4}{3} \\ {} & \circ & \left (3+x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (3+x \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=0 \\ {} & \circ & x =-3\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-3 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (3+x \right ) \left (\frac {d}{d x}y^{\prime }\right )-x \left (15+x \right ) y^{\prime }-20 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -3\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (3 u^{3}-18 u^{2}+27 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-u^{2}-9 u +36\right ) \left (\frac {d}{d u}y \left (u \right )\right )-20 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 \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}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}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}{d u}\frac {d}{d u}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} \\ {} & {} & 9 a_{0} r \left (1+3 r \right ) u^{-1+r}+\left (9 a_{1} \left (1+r \right ) \left (4+3 r \right )-a_{0} \left (18 r^{2}-9 r +20\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (9 a_{k +1} \left (k +1+r \right ) \left (3 k +4+3 r \right )-a_{k} \left (18 k^{2}+36 k r +18 r^{2}-9 k -9 r +20\right )+a_{k -1} \left (k +r -1\right ) \left (3 k -7+3 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 9 r \left (1+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {1}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 9 a_{1} \left (1+r \right ) \left (4+3 r \right )-a_{0} \left (18 r^{2}-9 r +20\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (-6 a_{k}+a_{k -1}+9 a_{k +1}\right ) k^{2}+\left (6 \left (-6 a_{k}+a_{k -1}+9 a_{k +1}\right ) r +9 a_{k}-10 a_{k -1}+63 a_{k +1}\right ) k +3 \left (-6 a_{k}+a_{k -1}+9 a_{k +1}\right ) r^{2}+\left (9 a_{k}-10 a_{k -1}+63 a_{k +1}\right ) r -20 a_{k}+7 a_{k -1}+36 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 \left (-6 a_{k +1}+a_{k}+9 a_{k +2}\right ) \left (k +1\right )^{2}+\left (6 \left (-6 a_{k +1}+a_{k}+9 a_{k +2}\right ) r +9 a_{k +1}-10 a_{k}+63 a_{k +2}\right ) \left (k +1\right )+3 \left (-6 a_{k +1}+a_{k}+9 a_{k +2}\right ) r^{2}+\left (9 a_{k +1}-10 a_{k}+63 a_{k +2}\right ) r -20 a_{k +1}+7 a_{k}+36 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}+6 k r a_{k}-36 k r a_{k +1}+3 r^{2} a_{k}-18 r^{2} a_{k +1}-4 k a_{k}-27 k a_{k +1}-4 r a_{k}-27 r a_{k +1}-29 a_{k +1}}{9 \left (3 k^{2}+6 k r +3 r^{2}+13 k +13 r +14\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}-4 k a_{k}-27 k a_{k +1}-29 a_{k +1}}{9 \left (3 k^{2}+13 k +14\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 {3 k^{2} a_{k}-18 k^{2} a_{k +1}-4 k a_{k}-27 k a_{k +1}-29 a_{k +1}}{9 \left (3 k^{2}+13 k +14\right )}, 36 a_{1}-20 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =3+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (3+x \right )^{k}, a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}-4 k a_{k}-27 k a_{k +1}-29 a_{k +1}}{9 \left (3 k^{2}+13 k +14\right )}, 36 a_{1}-20 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}-6 k a_{k}-15 k a_{k +1}+\frac {5}{3} a_{k}-22 a_{k +1}}{9 \left (3 k^{2}+11 k +10\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {1}{3}}, a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}-6 k a_{k}-15 k a_{k +1}+\frac {5}{3} a_{k}-22 a_{k +1}}{9 \left (3 k^{2}+11 k +10\right )}, 18 a_{1}-25 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =3+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (3+x \right )^{k -\frac {1}{3}}, a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}-6 k a_{k}-15 k a_{k +1}+\frac {5}{3} a_{k}-22 a_{k +1}}{9 \left (3 k^{2}+11 k +10\right )}, 18 a_{1}-25 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (3+x \right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (3+x \right )^{k -\frac {1}{3}}\right ), a_{k +2}=-\frac {3 k^{2} a_{k}-18 k^{2} a_{k +1}-4 k a_{k}-27 k a_{k +1}-29 a_{k +1}}{9 \left (3 k^{2}+13 k +14\right )}, 36 a_{1}-20 a_{0}=0, b_{k +2}=-\frac {3 k^{2} b_{k}-18 k^{2} b_{k +1}-6 k b_{k}-15 k b_{k +1}+\frac {5}{3} b_{k}-22 b_{k +1}}{9 \left (3 k^{2}+11 k +10\right )}, 18 b_{1}-25 b_{0}=0\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 45
Order:=6; dsolve(3*x^2*(3+x)*diff(y(x),x$2)-x*(15+x)*diff(y(x),x)-20*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{4} \left (1-\frac {4}{9} x +\frac {13}{81} x^{2}-\frac {832}{15309} x^{3}+\frac {2470}{137781} x^{4}-\frac {21736}{3720087} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (-144-\frac {64}{3} x +\frac {16}{27} x^{2}-\frac {112}{6561} x^{4}+\frac {448}{59049} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {2}{3}}} \]
✓ Solution by Mathematica
Time used: 0.044 (sec). Leaf size: 85
AsymptoticDSolveValue[3*x^2*(3+x)*y''[x]-x*(15+x)*y'[x]-20*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {7 x^{10/3}}{59049}-\frac {x^{4/3}}{243}+\frac {1}{x^{2/3}}+\frac {4 \sqrt [3]{x}}{27}\right )+c_2 \left (\frac {2470 x^{22/3}}{137781}-\frac {832 x^{19/3}}{15309}+\frac {13 x^{16/3}}{81}-\frac {4 x^{13/3}}{9}+x^{10/3}\right ) \]