Internal problem ID [1304]
Internal file name [OUTPUT/1305_Sunday_June_05_2022_02_09_21_AM_89487669/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: 10.
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]]
\[ \boxed {10 x^{2} \left (2 x^{2}+x +1\right ) y^{\prime \prime }+x \left (66 x^{2}+13 x +13\right ) y^{\prime }-\left (10 x^{2}+4 x +1\right ) y=0} \] 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. \[ \left (20 x^{4}+10 x^{3}+10 x^{2}\right ) y^{\prime \prime }+\left (66 x^{3}+13 x^{2}+13 x \right ) y^{\prime }+\left (-10 x^{2}-4 x -1\right ) y = 0 \] 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 {66 x^{2}+13 x +13}{10 x \left (2 x^{2}+x +1\right )}\\ q(x) &= -\frac {10 x^{2}+4 x +1}{10 x^{2} \left (2 x^{2}+x +1\right )}\\ \end {align*}
| \(p(x)=\frac {66 x^{2}+13 x +13}{10 x \left (2 x^{2}+x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {7}}{4}-\frac {1}{4}\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {10 x^{2}+4 x +1}{10 x^{2} \left (2 x^{2}+x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{4}+\frac {i \sqrt {7}}{4}\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {7}}{4}-\frac {1}{4}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -\frac {1}{4}+\frac {i \sqrt {7}}{4}, -\frac {i \sqrt {7}}{4}-\frac {1}{4}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 10 x^{2} \left (2 x^{2}+x +1\right ) y^{\prime \prime }+\left (66 x^{3}+13 x^{2}+13 x \right ) y^{\prime }+\left (-10 x^{2}-4 x -1\right ) y = 0 \] 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} 10 x^{2} \left (2 x^{2}+x +1\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 (66 x^{3}+13 x^{2}+13 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-10 x^{2}-4 x -1\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 }}20 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}66 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-10 x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}20 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}20 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}66 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}66 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}13 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}13 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-10 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-10 a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} 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 =2}{\sum }}20 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}66 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}13 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}13 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-10 a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 10 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+13 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 10 x^{r} a_{0} r \left (-1+r \right )+13 x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (10 x^{r} r \left (-1+r \right )+13 x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (10 r^{2}+3 r -1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 10 r^{2}+3 r -1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{5}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (10 r^{2}+3 r -1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{5}}, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {7}{10}}\) 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}{5}}\\ 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {1-2 r}{2 r +3} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 20 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+10 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+10 a_{n} \left (n +r \right ) \left (n +r -1\right )+66 a_{n -2} \left (n +r -2\right )+13 a_{n -1} \left (n +r -1\right )+13 a_{n} \left (n +r \right )-10 a_{n -2}-4 a_{n -1}-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {20 n^{2} a_{n -2}+10 n^{2} a_{n -1}+40 n r a_{n -2}+20 n r a_{n -1}+20 r^{2} a_{n -2}+10 r^{2} a_{n -1}-34 n a_{n -2}-17 n a_{n -1}-34 r a_{n -2}-17 r a_{n -1}-22 a_{n -2}+3 a_{n -1}}{10 n^{2}+20 n r +10 r^{2}+3 n +3 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{n} = \frac {\left (-20 a_{n -2}-10 a_{n -1}\right ) n^{2}+\left (26 a_{n -2}+13 a_{n -1}\right ) n +28 a_{n -2}}{10 n^{2}+7 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{5}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(\frac {3}{17}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{2}=-{\frac {7}{153}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(\frac {3}{17}\) |
| \(a_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(-{\frac {7}{153}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{3}=-{\frac {547}{5661}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(\frac {3}{17}\) |
| \(a_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(-{\frac {7}{153}}\) |
| \(a_{3}\) | \(\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230}\) | \(-{\frac {547}{5661}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-2000 r^{7}-16800 r^{6}+2360 r^{5}+433536 r^{4}+1687211 r^{3}+2569170 r^{2}+1406065 r -10206}{2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{4}={\frac {26942}{266067}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(\frac {3}{17}\) |
| \(a_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(-{\frac {7}{153}}\) |
| \(a_{3}\) | \(\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230}\) | \(-{\frac {547}{5661}}\) |
| \(a_{4}\) | \(\frac {-2000 r^{7}-16800 r^{6}+2360 r^{5}+433536 r^{4}+1687211 r^{3}+2569170 r^{2}+1406065 r -10206}{2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330}\) | \(\frac {26942}{266067}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-100000 r^{9}-2710000 r^{8}-31054000 r^{7}-196182200 r^{6}-744265370 r^{5}-1721296863 r^{4}-2321242940 r^{3}-1556172021 r^{2}-211154622 r +180891144}{\left (10 r^{2}+103 r +264\right ) \left (2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330\right )} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{5}={\frac {200432}{3991005}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(\frac {3}{17}\) |
| \(a_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(-{\frac {7}{153}}\) |
| \(a_{3}\) | \(\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230}\) | \(-{\frac {547}{5661}}\) |
| \(a_{4}\) | \(\frac {-2000 r^{7}-16800 r^{6}+2360 r^{5}+433536 r^{4}+1687211 r^{3}+2569170 r^{2}+1406065 r -10206}{2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330}\) | \(\frac {26942}{266067}\) |
| \(a_{5}\) | \(\frac {-100000 r^{9}-2710000 r^{8}-31054000 r^{7}-196182200 r^{6}-744265370 r^{5}-1721296863 r^{4}-2321242940 r^{3}-1556172021 r^{2}-211154622 r +180891144}{\left (10 r^{2}+103 r +264\right ) \left (2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330\right )}\) | \(\frac {200432}{3991005}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{5}} \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}{5}} \left (1+\frac {3 x}{17}-\frac {7 x^{2}}{153}-\frac {547 x^{3}}{5661}+\frac {26942 x^{4}}{266067}+\frac {200432 x^{5}}{3991005}+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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = \frac {1-2 r}{2 r +3} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 20 b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+10 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+10 b_{n} \left (n +r \right ) \left (n +r -1\right )+66 b_{n -2} \left (n +r -2\right )+13 b_{n -1} \left (n +r -1\right )+13 b_{n} \left (n +r \right )-10 b_{n -2}-4 b_{n -1}-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {20 n^{2} b_{n -2}+10 n^{2} b_{n -1}+40 n r b_{n -2}+20 n r b_{n -1}+20 r^{2} b_{n -2}+10 r^{2} b_{n -1}-34 n b_{n -2}-17 n b_{n -1}-34 r b_{n -2}-17 r b_{n -1}-22 b_{n -2}+3 b_{n -1}}{10 n^{2}+20 n r +10 r^{2}+3 n +3 r -1}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{n} = \frac {\left (-20 b_{n -2}-10 b_{n -1}\right ) n^{2}+\left (54 b_{n -2}+27 b_{n -1}\right ) n -14 b_{n -1}}{10 n^{2}-7 n}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{2}={\frac {14}{13}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(1\) |
| \(b_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(\frac {14}{13}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{3}=-{\frac {556}{897}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(1\) |
| \(b_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(\frac {14}{13}\) |
| \(b_{3}\) | \(\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230}\) | \(-{\frac {556}{897}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {-2000 r^{7}-16800 r^{6}+2360 r^{5}+433536 r^{4}+1687211 r^{3}+2569170 r^{2}+1406065 r -10206}{2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{4}=-{\frac {5314}{9867}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(1\) |
| \(b_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(\frac {14}{13}\) |
| \(b_{3}\) | \(\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230}\) | \(-{\frac {556}{897}}\) |
| \(b_{4}\) | \(\frac {-2000 r^{7}-16800 r^{6}+2360 r^{5}+433536 r^{4}+1687211 r^{3}+2569170 r^{2}+1406065 r -10206}{2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330}\) | \(-{\frac {5314}{9867}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-100000 r^{9}-2710000 r^{8}-31054000 r^{7}-196182200 r^{6}-744265370 r^{5}-1721296863 r^{4}-2321242940 r^{3}-1556172021 r^{2}-211154622 r +180891144}{\left (10 r^{2}+103 r +264\right ) \left (2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{5}={\frac {2092186}{2121405}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1-2 r}{2 r +3}\) | \(1\) |
| \(b_{2}\) | \(\frac {-20 r^{3}-116 r^{2}-123 r +21}{20 r^{3}+116 r^{2}+219 r +135}\) | \(\frac {14}{13}\) |
| \(b_{3}\) | \(\frac {600 r^{5}+5260 r^{4}+15654 r^{3}+17349 r^{2}+3025 r -3402}{200 r^{5}+2420 r^{4}+11458 r^{3}+26515 r^{2}+29967 r +13230}\) | \(-{\frac {556}{897}}\) |
| \(b_{4}\) | \(\frac {-2000 r^{7}-16800 r^{6}+2360 r^{5}+433536 r^{4}+1687211 r^{3}+2569170 r^{2}+1406065 r -10206}{2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330}\) | \(-{\frac {5314}{9867}}\) |
| \(b_{5}\) | \(\frac {-100000 r^{9}-2710000 r^{8}-31054000 r^{7}-196182200 r^{6}-744265370 r^{5}-1721296863 r^{4}-2321242940 r^{3}-1556172021 r^{2}-211154622 r +180891144}{\left (10 r^{2}+103 r +264\right ) \left (2000 r^{7}+40800 r^{6}+349640 r^{5}+1629984 r^{4}+4459733 r^{3}+7153626 r^{2}+6222447 r +2262330\right )}\) | \(\frac {2092186}{2121405}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{5}} \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 {14 x^{2}}{13}-\frac {556 x^{3}}{897}-\frac {5314 x^{4}}{9867}+\frac {2092186 x^{5}}{2121405}+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 ) \\ &= c_{1} x^{\frac {1}{5}} \left (1+\frac {3 x}{17}-\frac {7 x^{2}}{153}-\frac {547 x^{3}}{5661}+\frac {26942 x^{4}}{266067}+\frac {200432 x^{5}}{3991005}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+x +\frac {14 x^{2}}{13}-\frac {556 x^{3}}{897}-\frac {5314 x^{4}}{9867}+\frac {2092186 x^{5}}{2121405}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{5}} \left (1+\frac {3 x}{17}-\frac {7 x^{2}}{153}-\frac {547 x^{3}}{5661}+\frac {26942 x^{4}}{266067}+\frac {200432 x^{5}}{3991005}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+x +\frac {14 x^{2}}{13}-\frac {556 x^{3}}{897}-\frac {5314 x^{4}}{9867}+\frac {2092186 x^{5}}{2121405}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{5}} \left (1+\frac {3 x}{17}-\frac {7 x^{2}}{153}-\frac {547 x^{3}}{5661}+\frac {26942 x^{4}}{266067}+\frac {200432 x^{5}}{3991005}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+x +\frac {14 x^{2}}{13}-\frac {556 x^{3}}{897}-\frac {5314 x^{4}}{9867}+\frac {2092186 x^{5}}{2121405}+O\left (x^{6}\right )\right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{5}} \left (1+\frac {3 x}{17}-\frac {7 x^{2}}{153}-\frac {547 x^{3}}{5661}+\frac {26942 x^{4}}{266067}+\frac {200432 x^{5}}{3991005}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+x +\frac {14 x^{2}}{13}-\frac {556 x^{3}}{897}-\frac {5314 x^{4}}{9867}+\frac {2092186 x^{5}}{2121405}+O\left (x^{6}\right )\right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 10 x^{2} \left (2 x^{2}+x +1\right ) y^{\prime \prime }+\left (66 x^{3}+13 x^{2}+13 x \right ) y^{\prime }+\left (-10 x^{2}-4 x -1\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 (10 x^{2}+4 x +1\right ) y}{10 x^{2} \left (2 x^{2}+x +1\right )}-\frac {\left (66 x^{2}+13 x +13\right ) y^{\prime }}{10 x \left (2 x^{2}+x +1\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 (66 x^{2}+13 x +13\right ) y^{\prime }}{10 x \left (2 x^{2}+x +1\right )}-\frac {\left (10 x^{2}+4 x +1\right ) y}{10 x^{2} \left (2 x^{2}+x +1\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 {66 x^{2}+13 x +13}{10 x \left (2 x^{2}+x +1\right )}, P_{3}\left (x \right )=-\frac {10 x^{2}+4 x +1}{10 x^{2} \left (2 x^{2}+x +1\right )}\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 {13}{10} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {1}{10} \\ {} & \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} \\ {} & {} & 10 x^{2} \left (2 x^{2}+x +1\right ) y^{\prime \prime }+x \left (66 x^{2}+13 x +13\right ) y^{\prime }+\left (-10 x^{2}-4 x -1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & 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 =2..4 \\ {} & {} & 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} \left (1+2 r \right ) \left (-1+5 r \right ) x^{r}+\left (a_{1} \left (3+2 r \right ) \left (4+5 r \right )+a_{0} \left (4+5 r \right ) \left (-1+2 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (5 k +5 r -1\right )+a_{k -1} \left (5 k +5 r -1\right ) \left (2 k -3+2 r \right )+2 a_{k -2} \left (2 k +2 r +1\right ) \left (5 k -11+5 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+2 r \right ) \left (-1+5 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {1}{5}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+2 r \right ) \left (4+5 r \right )+a_{0} \left (4+5 r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {\left (-1+2 r \right ) a_{0}}{3+2 r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (2 k +2 r +1\right ) \left (5 k +5 r -1\right )+a_{k -1} \left (5 k +5 r -1\right ) \left (2 k -3+2 r \right )+2 a_{k -2} \left (2 k +2 r +1\right ) \left (5 k -11+5 r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (2 k +2 r +5\right ) \left (5 k +9+5 r \right )+a_{k +1} \left (5 k +9+5 r \right ) \left (2 k +2 r +1\right )+2 a_{k} \left (2 k +2 r +5\right ) \left (5 k +5 r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {20 k^{2} a_{k}+10 k^{2} a_{k +1}+40 k r a_{k}+20 k r a_{k +1}+20 r^{2} a_{k}+10 r^{2} a_{k +1}+46 k a_{k}+23 k a_{k +1}+46 r a_{k}+23 r a_{k +1}-10 a_{k}+9 a_{k +1}}{\left (2 k +2 r +5\right ) \left (5 k +9+5 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {20 k^{2} a_{k}+10 k^{2} a_{k +1}+26 k a_{k}+13 k a_{k +1}-28 a_{k}}{\left (2 k +4\right ) \left (5 k +\frac {13}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}}, a_{k +2}=-\frac {20 k^{2} a_{k}+10 k^{2} a_{k +1}+26 k a_{k}+13 k a_{k +1}-28 a_{k}}{\left (2 k +4\right ) \left (5 k +\frac {13}{2}\right )}, a_{1}=a_{0}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{5} \\ {} & {} & a_{k +2}=-\frac {20 k^{2} a_{k}+10 k^{2} a_{k +1}+54 k a_{k}+27 k a_{k +1}+14 a_{k +1}}{\left (2 k +\frac {27}{5}\right ) \left (5 k +10\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{5} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{5}}, a_{k +2}=-\frac {20 k^{2} a_{k}+10 k^{2} a_{k +1}+54 k a_{k}+27 k a_{k +1}+14 a_{k +1}}{\left (2 k +\frac {27}{5}\right ) \left (5 k +10\right )}, a_{1}=\frac {3 a_{0}}{17}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{5}}\right ), a_{k +2}=-\frac {20 k^{2} a_{k}+10 k^{2} a_{1+k}+26 k a_{k}+13 k a_{1+k}-28 a_{k}}{\left (2 k +4\right ) \left (5 k +\frac {13}{2}\right )}, a_{1}=a_{0}, b_{k +2}=-\frac {20 k^{2} b_{k}+10 k^{2} b_{1+k}+54 k b_{k}+27 k b_{1+k}+14 b_{1+k}}{\left (2 k +\frac {27}{5}\right ) \left (5 k +10\right )}, b_{1}=\frac {3 b_{0}}{17}\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: 0.032 (sec). Leaf size: 47
Order:=6; dsolve(10*x^2*(1+x+2*x^2)*diff(y(x),x$2)+x*(13+13*x+66*x^2)*diff(y(x),x)-(1+4*x+10*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1+x +\frac {14}{13} x^{2}-\frac {556}{897} x^{3}-\frac {5314}{9867} x^{4}+\frac {2092186}{2121405} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{\sqrt {x}}+c_{2} x^{\frac {1}{5}} \left (1+\frac {3}{17} x -\frac {7}{153} x^{2}-\frac {547}{5661} x^{3}+\frac {26942}{266067} x^{4}+\frac {200432}{3991005} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 86
AsymptoticDSolveValue[10*x^2*(1+x+2*x^2)*y''[x]+x*(13+13*x+66*x^2)*y'[x]-(1+4*x+10*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [5]{x} \left (\frac {200432 x^5}{3991005}+\frac {26942 x^4}{266067}-\frac {547 x^3}{5661}-\frac {7 x^2}{153}+\frac {3 x}{17}+1\right )+\frac {c_2 \left (\frac {2092186 x^5}{2121405}-\frac {5314 x^4}{9867}-\frac {556 x^3}{897}+\frac {14 x^2}{13}+x+1\right )}{\sqrt {x}} \]