Internal problem ID [1303]
Internal file name [OUTPUT/1304_Sunday_June_05_2022_02_09_17_AM_49582011/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: 9.
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]]
\[ \boxed {x \left (x^{2}+x +3\right ) y^{\prime \prime }+\left (-x^{2}+x +4\right ) y^{\prime }+y x=0} \] 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. \[ \left (x^{3}+x^{2}+3 x \right ) y^{\prime \prime }+\left (-x^{2}+x +4\right ) y^{\prime }+y x = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {x^{2}-x -4}{x \left (x^{2}+x +3\right )}\\ q(x) &= \frac {1}{x^{2}+x +3}\\ \end {align*}
| \(p(x)=-\frac {x^{2}-x -4}{x \left (x^{2}+x +3\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {11}}{2}-\frac {1}{2}\) | \(\text {``regular''}\) |
| \(x = \frac {i \sqrt {11}}{2}-\frac {1}{2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{x^{2}+x +3}\) | |
| singularity | type |
| \(x = -\frac {i \sqrt {11}}{2}-\frac {1}{2}\) | \(\text {``regular''}\) |
| \(x = \frac {i \sqrt {11}}{2}-\frac {1}{2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -\frac {i \sqrt {11}}{2}-\frac {1}{2}, \frac {i \sqrt {11}}{2}-\frac {1}{2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (x^{2}+x +3\right ) y^{\prime \prime }+\left (-x^{2}+x +4\right ) y^{\prime }+y x = 0 \] 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 \left (x^{2}+x +3\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}+x +4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x = 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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r -1} 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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1} \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{-1+r} a_{0} r \left (-1+r \right )+4 r a_{0} x^{-1+r} = 0 \] Or \[ \left (3 x^{-1+r} r \left (-1+r \right )+4 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (3 r^{2}+r \right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}+r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (3 r^{2}+r \right ) x^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{3}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {r^{2}}{3 r^{2}+7 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -2} \left (n +r -2\right )+a_{n -1} \left (n +r -1\right )+4 a_{n} \left (n +r \right )+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -2}+n^{2} a_{n -1}+2 n r a_{n -2}+2 n r a_{n -1}+r^{2} a_{n -2}+r^{2} a_{n -1}-6 n a_{n -2}-2 n a_{n -1}-6 r a_{n -2}-2 r a_{n -1}+9 a_{n -2}+a_{n -1}}{3 n^{2}+6 n r +3 r^{2}+n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-a_{n -1}\right ) n^{2}+\left (6 a_{n -2}+2 a_{n -1}\right ) n -9 a_{n -2}-a_{n -1}}{3 n^{2}+n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56} \] Which for the root \(r = 0\) becomes \[ a_{2}=-{\frac {1}{14}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {1}{14}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{105}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {1}{14}}\) |
| \(a_{3}\) | \(\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )}\) | \(\frac {1}{105}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{7}-6 r^{6}-64 r^{5}-127 r^{4}-55 r^{3}+45 r^{2}+30 r -8}{\left (3 r^{2}+25 r +52\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (3 r +10\right ) \left (r +2\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=-{\frac {1}{3640}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {1}{14}}\) |
| \(a_{3}\) | \(\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )}\) | \(\frac {1}{105}\) |
| \(a_{4}\) | \(\frac {r^{7}-6 r^{6}-64 r^{5}-127 r^{4}-55 r^{3}+45 r^{2}+30 r -8}{\left (3 r^{2}+25 r +52\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (3 r +10\right ) \left (r +2\right )}\) | \(-{\frac {1}{3640}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-16 r^{9}-186 r^{8}-756 r^{7}-1032 r^{6}+1053 r^{5}+4476 r^{4}+3939 r^{3}-246 r^{2}-1952 r -736}{\left (3 r^{2}+31 r +80\right ) \left (r +2\right ) \left (3 r +10\right ) \left (3 r +13\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (r +3\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {23}{54600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {1}{14}}\) |
| \(a_{3}\) | \(\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )}\) | \(\frac {1}{105}\) |
| \(a_{4}\) | \(\frac {r^{7}-6 r^{6}-64 r^{5}-127 r^{4}-55 r^{3}+45 r^{2}+30 r -8}{\left (3 r^{2}+25 r +52\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (3 r +10\right ) \left (r +2\right )}\) | \(-{\frac {1}{3640}}\) |
| \(a_{5}\) | \(\frac {-16 r^{9}-186 r^{8}-756 r^{7}-1032 r^{6}+1053 r^{5}+4476 r^{4}+3939 r^{3}-246 r^{2}-1952 r -736}{\left (3 r^{2}+31 r +80\right ) \left (r +2\right ) \left (3 r +10\right ) \left (3 r +13\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (r +3\right )}\) | \(-{\frac {23}{54600}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1-\frac {x^{2}}{14}+\frac {x^{3}}{105}-\frac {x^{4}}{3640}-\frac {23 x^{5}}{54600}+O\left (x^{6}\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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -\frac {r^{2}}{3 r^{2}+7 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -2} \left (n +r -2\right )+b_{n -1} \left (n +r -1\right )+4 \left (n +r \right ) b_{n}+b_{n -2} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {n^{2} b_{n -2}+n^{2} b_{n -1}+2 n r b_{n -2}+2 n r b_{n -1}+r^{2} b_{n -2}+r^{2} b_{n -1}-6 n b_{n -2}-2 n b_{n -1}-6 r b_{n -2}-2 r b_{n -1}+9 b_{n -2}+b_{n -1}}{3 n^{2}+6 n r +3 r^{2}+n +r}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{n} = \frac {\left (-9 b_{n -2}-9 b_{n -1}\right ) n^{2}+\left (60 b_{n -2}+24 b_{n -1}\right ) n -100 b_{n -2}-16 b_{n -1}}{27 n^{2}-9 n}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(-{\frac {1}{18}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{2}=-{\frac {71}{405}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(-{\frac {1}{18}}\) |
| \(b_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {71}{405}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{3}={\frac {719}{34992}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(-{\frac {1}{18}}\) |
| \(b_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {71}{405}}\) |
| \(b_{3}\) | \(\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )}\) | \(\frac {719}{34992}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{7}-6 r^{6}-64 r^{5}-127 r^{4}-55 r^{3}+45 r^{2}+30 r -8}{\left (3 r^{2}+25 r +52\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (3 r +10\right ) \left (r +2\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{4}=-{\frac {1678}{1082565}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(-{\frac {1}{18}}\) |
| \(b_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {71}{405}}\) |
| \(b_{3}\) | \(\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )}\) | \(\frac {719}{34992}\) |
| \(b_{4}\) | \(\frac {r^{7}-6 r^{6}-64 r^{5}-127 r^{4}-55 r^{3}+45 r^{2}+30 r -8}{\left (3 r^{2}+25 r +52\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (3 r +10\right ) \left (r +2\right )}\) | \(-{\frac {1678}{1082565}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-16 r^{9}-186 r^{8}-756 r^{7}-1032 r^{6}+1053 r^{5}+4476 r^{4}+3939 r^{3}-246 r^{2}-1952 r -736}{\left (3 r^{2}+31 r +80\right ) \left (r +2\right ) \left (3 r +10\right ) \left (3 r +13\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (r +3\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{5}=-{\frac {513547}{992023200}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r^{2}}{3 r^{2}+7 r +4}\) | \(-{\frac {1}{18}}\) |
| \(b_{2}\) | \(\frac {-2 r^{3}+3 r^{2}+5 r -4}{9 r^{3}+51 r^{2}+94 r +56}\) | \(-{\frac {71}{405}}\) |
| \(b_{3}\) | \(\frac {5 r^{5}+10 r^{4}-10 r^{3}-17 r^{2}+2 r +8}{\left (3 r^{2}+19 r +30\right ) \left (9 r^{2}+33 r +28\right ) \left (1+r \right )}\) | \(\frac {719}{34992}\) |
| \(b_{4}\) | \(\frac {r^{7}-6 r^{6}-64 r^{5}-127 r^{4}-55 r^{3}+45 r^{2}+30 r -8}{\left (3 r^{2}+25 r +52\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (3 r +10\right ) \left (r +2\right )}\) | \(-{\frac {1678}{1082565}}\) |
| \(b_{5}\) | \(\frac {-16 r^{9}-186 r^{8}-756 r^{7}-1032 r^{6}+1053 r^{5}+4476 r^{4}+3939 r^{3}-246 r^{2}-1952 r -736}{\left (3 r^{2}+31 r +80\right ) \left (r +2\right ) \left (3 r +10\right ) \left (3 r +13\right ) \left (1+r \right ) \left (9 r^{2}+33 r +28\right ) \left (r +3\right )}\) | \(-{\frac {513547}{992023200}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \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 {x}{18}-\frac {71 x^{2}}{405}+\frac {719 x^{3}}{34992}-\frac {1678 x^{4}}{1082565}-\frac {513547 x^{5}}{992023200}+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} \left (1-\frac {x^{2}}{14}+\frac {x^{3}}{105}-\frac {x^{4}}{3640}-\frac {23 x^{5}}{54600}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {x}{18}-\frac {71 x^{2}}{405}+\frac {719 x^{3}}{34992}-\frac {1678 x^{4}}{1082565}-\frac {513547 x^{5}}{992023200}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {x^{2}}{14}+\frac {x^{3}}{105}-\frac {x^{4}}{3640}-\frac {23 x^{5}}{54600}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{18}-\frac {71 x^{2}}{405}+\frac {719 x^{3}}{34992}-\frac {1678 x^{4}}{1082565}-\frac {513547 x^{5}}{992023200}+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} \left (1-\frac {x^{2}}{14}+\frac {x^{3}}{105}-\frac {x^{4}}{3640}-\frac {23 x^{5}}{54600}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{18}-\frac {71 x^{2}}{405}+\frac {719 x^{3}}{34992}-\frac {1678 x^{4}}{1082565}-\frac {513547 x^{5}}{992023200}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {x^{2}}{14}+\frac {x^{3}}{105}-\frac {x^{4}}{3640}-\frac {23 x^{5}}{54600}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{18}-\frac {71 x^{2}}{405}+\frac {719 x^{3}}{34992}-\frac {1678 x^{4}}{1082565}-\frac {513547 x^{5}}{992023200}+O\left (x^{6}\right )\right )}{x^{\frac {1}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x^{2}+x +3\right ) y^{\prime \prime }+\left (-x^{2}+x +4\right ) y^{\prime }+y x =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 {y}{x^{2}+x +3}+\frac {\left (x^{2}-x -4\right ) y^{\prime }}{x \left (x^{2}+x +3\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 (x^{2}-x -4\right ) y^{\prime }}{x \left (x^{2}+x +3\right )}+\frac {y}{x^{2}+x +3}=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 {x^{2}-x -4}{x \left (x^{2}+x +3\right )}, P_{3}\left (x \right )=\frac {1}{x^{2}+x +3}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {4}{3} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x^{2}+x +3\right ) y^{\prime \prime }+\left (-x^{2}+x +4\right ) y^{\prime }+y x =0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+3 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (4+3 r \right )+a_{0} r^{2}\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (3 k +4+3 r \right )+a_{k} \left (k +r \right )^{2}+a_{k -1} \left (k -2+r \right )^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 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} \\ {} & {} & a_{1} \left (1+r \right ) \left (4+3 r \right )+a_{0} r^{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +r +1\right ) \left (3 k +4+3 r \right )+a_{k} \left (k +r \right )^{2}+a_{k -1} \left (k -2+r \right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (3 k +7+3 r \right )+a_{k +1} \left (k +r +1\right )^{2}+a_{k} \left (k +r -1\right )^{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+k^{2} a_{k +1}+2 k r a_{k}+2 k r a_{k +1}+r^{2} a_{k}+r^{2} a_{k +1}-2 k a_{k}+2 k a_{k +1}-2 r a_{k}+2 r a_{k +1}+a_{k}+a_{k +1}}{\left (k +2+r \right ) \left (3 k +7+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+k^{2} a_{k +1}-2 k a_{k}+2 k a_{k +1}+a_{k}+a_{k +1}}{\left (k +2\right ) \left (3 k +7\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {k^{2} a_{k}+k^{2} a_{k +1}-2 k a_{k}+2 k a_{k +1}+a_{k}+a_{k +1}}{\left (k +2\right ) \left (3 k +7\right )}, 4 a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+k^{2} a_{k +1}-\frac {8}{3} k a_{k}+\frac {4}{3} k a_{k +1}+\frac {16}{9} a_{k}+\frac {4}{9} a_{k +1}}{\left (k +\frac {5}{3}\right ) \left (3 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{3}}, a_{k +2}=-\frac {k^{2} a_{k}+k^{2} a_{k +1}-\frac {8}{3} k a_{k}+\frac {4}{3} k a_{k +1}+\frac {16}{9} a_{k}+\frac {4}{9} a_{k +1}}{\left (k +\frac {5}{3}\right ) \left (3 k +6\right )}, 2 a_{1}+\frac {a_{0}}{9}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {1}{3}}\right ), a_{k +2}=-\frac {k^{2} a_{k}+k^{2} a_{1+k}-2 k a_{k}+2 k a_{1+k}+a_{k}+a_{1+k}}{\left (k +2\right ) \left (3 k +7\right )}, 4 a_{1}=0, b_{k +2}=-\frac {k^{2} b_{k}+k^{2} b_{1+k}-\frac {8}{3} k b_{k}+\frac {4}{3} k b_{1+k}+\frac {16}{9} b_{k}+\frac {4}{9} b_{1+k}}{\left (k +\frac {5}{3}\right ) \left (3 k +6\right )}, 2 b_{1}+\frac {b_{0}}{9}=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 -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 42
Order:=6; dsolve(x*(3+x+x^2)*diff(y(x),x$2)+(4+x-x^2)*diff(y(x),x)+x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {1}{18} x -\frac {71}{405} x^{2}+\frac {719}{34992} x^{3}-\frac {1678}{1082565} x^{4}-\frac {513547}{992023200} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {1}{3}}}+c_{2} \left (1-\frac {1}{14} x^{2}+\frac {1}{105} x^{3}-\frac {1}{3640} x^{4}-\frac {23}{54600} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 80
AsymptoticDSolveValue[x*(3+x+x^2)*y''[x]+(4+x-x^2)*y'[x]+x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {23 x^5}{54600}-\frac {x^4}{3640}+\frac {x^3}{105}-\frac {x^2}{14}+1\right )+\frac {c_2 \left (-\frac {513547 x^5}{992023200}-\frac {1678 x^4}{1082565}+\frac {719 x^3}{34992}-\frac {71 x^2}{405}-\frac {x}{18}+1\right )}{\sqrt [3]{x}} \]