Internal problem ID [1300]
Internal file name [OUTPUT/1301_Sunday_June_05_2022_02_09_05_AM_93571922/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: 6.
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], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {x^{2} \left (10 x^{2}+x +5\right ) y^{\prime \prime }+x \left (48 x^{2}+3 x +4\right ) y^{\prime }+\left (36 x^{2}+x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The ODE is \[ \left (10 x^{4}+x^{3}+5 x^{2}\right ) y^{\prime \prime }+\left (48 x^{3}+3 x^{2}+4 x \right ) y^{\prime }+\left (36 x^{2}+x \right ) y = 0 \] Or \[ x \left (10 x^{3} y^{\prime \prime }+x^{2} y^{\prime \prime }+48 y^{\prime } x^{2}+5 y^{\prime \prime } x +3 x y^{\prime }+36 y x +4 y^{\prime }+y\right ) = 0 \] For \(x \neq 0\) the above simplifies to \[ \left (10 x^{3}+x^{2}+5 x \right ) y^{\prime \prime }+\left (48 x^{2}+3 x +4\right ) y^{\prime }+\left (36 x +1\right ) y = 0 \] The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (10 x^{4}+x^{3}+5 x^{2}\right ) y^{\prime \prime }+\left (48 x^{3}+3 x^{2}+4 x \right ) y^{\prime }+\left (36 x^{2}+x \right ) y = 0 \] 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 {48 x^{2}+3 x +4}{x \left (10 x^{2}+x +5\right )}\\ q(x) &= \frac {36 x +1}{x \left (10 x^{2}+x +5\right )}\\ \end {align*}
| \(p(x)=\frac {48 x^{2}+3 x +4}{x \left (10 x^{2}+x +5\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {199}}{20}-\frac {1}{20}\) | \(\text {``regular''}\) |
| \(x = \frac {i \sqrt {199}}{20}-\frac {1}{20}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {36 x +1}{x \left (10 x^{2}+x +5\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {199}}{20}-\frac {1}{20}\) | \(\text {``regular''}\) |
| \(x = \frac {i \sqrt {199}}{20}-\frac {1}{20}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -\frac {i \sqrt {199}}{20}-\frac {1}{20}, \frac {i \sqrt {199}}{20}-\frac {1}{20}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (10 x^{2}+x +5\right ) y^{\prime \prime }+\left (48 x^{3}+3 x^{2}+4 x \right ) y^{\prime }+\left (36 x^{2}+x \right ) y = 0 \] 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 (10 x^{2}+x +5\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (48 x^{3}+3 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (36 x^{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 }}10 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\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 }}5 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}48 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}36 x^{n +r +2} a_{n}\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 \(n +r\) 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 }}10 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \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 }}48 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}48 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}36 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}36 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =2}{\sum }}10 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\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 }}5 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}48 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}36 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 5 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ 5 x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r = 0 \] Or \[ \left (5 x^{r} r \left (-1+r \right )+4 x^{r} r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r \left (-1+5 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 5 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{5}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} r \left (-1+5 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{5}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{5}}\) 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} \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 \[ a_{1} = \frac {-1-r}{5 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 10 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 )+5 a_{n} \left (n +r \right ) \left (n +r -1\right )+48 a_{n -2} \left (n +r -2\right )+3 a_{n -1} \left (n +r -1\right )+4 a_{n} \left (n +r \right )+36 a_{n -2}+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {10 n a_{n -2}+n a_{n -1}+10 r a_{n -2}+r a_{n -1}-2 a_{n -2}}{5 n -1+5 r}\tag {4} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{n} = \frac {\left (-50 a_{n -2}-5 a_{n -1}\right ) n -a_{n -1}}{25 n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {6}{25}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{2}=-{\frac {1217}{625}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {6}{25}}\) |
| \(a_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {1217}{625}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{3}={\frac {41972}{46875}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {6}{25}}\) |
| \(a_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {1217}{625}}\) |
| \(a_{3}\) | \(\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504}\) | \(\frac {41972}{46875}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {2351 r^{4}+21570 r^{3}+68329 r^{2}+86470 r +35392}{625 r^{4}+5750 r^{3}+18275 r^{2}+23230 r +9576} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{4}={\frac {1447799}{390625}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {6}{25}}\) |
| \(a_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {1217}{625}}\) |
| \(a_{3}\) | \(\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504}\) | \(\frac {41972}{46875}\) |
| \(a_{4}\) | \(\frac {2351 r^{4}+21570 r^{3}+68329 r^{2}+86470 r +35392}{625 r^{4}+5750 r^{3}+18275 r^{2}+23230 r +9576}\) | \(\frac {1447799}{390625}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-7301 r^{5}-103595 r^{4}-551337 r^{3}-1357653 r^{2}-1517298 r -598304}{3125 r^{5}+43750 r^{4}+229375 r^{3}+554750 r^{2}+605400 r +229824} \] Which for the root \(r = {\frac {1}{5}}\) becomes \[ a_{5}=-{\frac {375253322}{146484375}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {6}{25}}\) |
| \(a_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {1217}{625}}\) |
| \(a_{3}\) | \(\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504}\) | \(\frac {41972}{46875}\) |
| \(a_{4}\) | \(\frac {2351 r^{4}+21570 r^{3}+68329 r^{2}+86470 r +35392}{625 r^{4}+5750 r^{3}+18275 r^{2}+23230 r +9576}\) | \(\frac {1447799}{390625}\) |
| \(a_{5}\) | \(\frac {-7301 r^{5}-103595 r^{4}-551337 r^{3}-1357653 r^{2}-1517298 r -598304}{3125 r^{5}+43750 r^{4}+229375 r^{3}+554750 r^{2}+605400 r +229824}\) | \(-{\frac {375253322}{146484375}}\) |
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 {6 x}{25}-\frac {1217 x^{2}}{625}+\frac {41972 x^{3}}{46875}+\frac {1447799 x^{4}}{390625}-\frac {375253322 x^{5}}{146484375}+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 \[ b_{1} = \frac {-1-r}{5 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 10 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 )+5 b_{n} \left (n +r \right ) \left (n +r -1\right )+48 b_{n -2} \left (n +r -2\right )+3 b_{n -1} \left (n +r -1\right )+4 b_{n} \left (n +r \right )+36 b_{n -2}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {10 n b_{n -2}+n b_{n -1}+10 r b_{n -2}+r b_{n -1}-2 b_{n -2}}{5 n -1+5 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {\left (-10 b_{n -2}-b_{n -1}\right ) n +2 b_{n -2}}{5 n -1}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {1}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36} \] Which for the root \(r = 0\) becomes \[ b_{2}=-{\frac {35}{18}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {35}{18}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {11}{12}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {35}{18}}\) |
| \(b_{3}\) | \(\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504}\) | \(\frac {11}{12}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {2351 r^{4}+21570 r^{3}+68329 r^{2}+86470 r +35392}{625 r^{4}+5750 r^{3}+18275 r^{2}+23230 r +9576} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {632}{171}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {35}{18}}\) |
| \(b_{3}\) | \(\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504}\) | \(\frac {11}{12}\) |
| \(b_{4}\) | \(\frac {2351 r^{4}+21570 r^{3}+68329 r^{2}+86470 r +35392}{625 r^{4}+5750 r^{3}+18275 r^{2}+23230 r +9576}\) | \(\frac {632}{171}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-7301 r^{5}-103595 r^{4}-551337 r^{3}-1357653 r^{2}-1517298 r -598304}{3125 r^{5}+43750 r^{4}+229375 r^{3}+554750 r^{2}+605400 r +229824} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {2671}{1026}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{5 r +4}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {-49 r^{2}-127 r -70}{25 r^{2}+65 r +36}\) | \(-{\frac {35}{18}}\) |
| \(b_{3}\) | \(\frac {99 r^{3}+554 r^{2}+933 r +462}{125 r^{3}+675 r^{2}+1090 r +504}\) | \(\frac {11}{12}\) |
| \(b_{4}\) | \(\frac {2351 r^{4}+21570 r^{3}+68329 r^{2}+86470 r +35392}{625 r^{4}+5750 r^{3}+18275 r^{2}+23230 r +9576}\) | \(\frac {632}{171}\) |
| \(b_{5}\) | \(\frac {-7301 r^{5}-103595 r^{4}-551337 r^{3}-1357653 r^{2}-1517298 r -598304}{3125 r^{5}+43750 r^{4}+229375 r^{3}+554750 r^{2}+605400 r +229824}\) | \(-{\frac {2671}{1026}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 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 \\ &= 1-\frac {x}{4}-\frac {35 x^{2}}{18}+\frac {11 x^{3}}{12}+\frac {632 x^{4}}{171}-\frac {2671 x^{5}}{1026}+O\left (x^{6}\right ) \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 {6 x}{25}-\frac {1217 x^{2}}{625}+\frac {41972 x^{3}}{46875}+\frac {1447799 x^{4}}{390625}-\frac {375253322 x^{5}}{146484375}+O\left (x^{6}\right )\right ) + c_{2} \left (1-\frac {x}{4}-\frac {35 x^{2}}{18}+\frac {11 x^{3}}{12}+\frac {632 x^{4}}{171}-\frac {2671 x^{5}}{1026}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{25}-\frac {1217 x^{2}}{625}+\frac {41972 x^{3}}{46875}+\frac {1447799 x^{4}}{390625}-\frac {375253322 x^{5}}{146484375}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{4}-\frac {35 x^{2}}{18}+\frac {11 x^{3}}{12}+\frac {632 x^{4}}{171}-\frac {2671 x^{5}}{1026}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{25}-\frac {1217 x^{2}}{625}+\frac {41972 x^{3}}{46875}+\frac {1447799 x^{4}}{390625}-\frac {375253322 x^{5}}{146484375}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{4}-\frac {35 x^{2}}{18}+\frac {11 x^{3}}{12}+\frac {632 x^{4}}{171}-\frac {2671 x^{5}}{1026}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{5}} \left (1-\frac {6 x}{25}-\frac {1217 x^{2}}{625}+\frac {41972 x^{3}}{46875}+\frac {1447799 x^{4}}{390625}-\frac {375253322 x^{5}}{146484375}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{4}-\frac {35 x^{2}}{18}+\frac {11 x^{3}}{12}+\frac {632 x^{4}}{171}-\frac {2671 x^{5}}{1026}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (10 x^{2}+x +5\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (48 x^{3}+3 x^{2}+4 x \right ) y^{\prime }+\left (36 x^{2}+x \right ) 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 (36 x +1\right ) y}{x \left (10 x^{2}+x +5\right )}-\frac {\left (48 x^{2}+3 x +4\right ) y^{\prime }}{x \left (10 x^{2}+x +5\right )} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (48 x^{2}+3 x +4\right ) y^{\prime }}{x \left (10 x^{2}+x +5\right )}+\frac {\left (36 x +1\right ) y}{x \left (10 x^{2}+x +5\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {48 x^{2}+3 x +4}{x \left (10 x^{2}+x +5\right )}, P_{3}\left (x \right )=\frac {36 x +1}{x \left (10 x^{2}+x +5\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 {4}{5} \\ {} & \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} \\ {} & {} & \left (36 x +1\right ) y+\left (48 x^{2}+3 x +4\right ) y^{\prime }+x \left (10 x^{2}+x +5\right ) \left (\frac {d}{d x}y^{\prime }\right )=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..1 \\ {} & {} & 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 =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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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+5 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (4+5 r \right )+a_{0} \left (1+r \right )^{2}\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (5 k +4+5 r \right )+a_{k} \left (k +r +1\right )^{2}+2 a_{k -1} \left (k +r +1\right ) \left (5 k +4+5 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+5 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{5}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (4+5 r \right )+a_{0} \left (1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r +1\right ) \left (\left (a_{k}+10 a_{k -1}+5 a_{k +1}\right ) k +\left (a_{k}+10 a_{k -1}+5 a_{k +1}\right ) r +a_{k}+8 a_{k -1}+4 a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r +2\right ) \left (\left (a_{k +1}+10 a_{k}+5 a_{k +2}\right ) \left (k +1\right )+\left (a_{k +1}+10 a_{k}+5 a_{k +2}\right ) r +a_{k +1}+8 a_{k}+4 a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {10 k a_{k}+k a_{k +1}+10 r a_{k}+r a_{k +1}+18 a_{k}+2 a_{k +1}}{5 k +5 r +9} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {10 k a_{k}+k a_{k +1}+18 a_{k}+2 a_{k +1}}{5 k +9} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {10 k a_{k}+k a_{k +1}+18 a_{k}+2 a_{k +1}}{5 k +9}, 4 a_{1}+a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{5} \\ {} & {} & a_{k +2}=-\frac {10 k a_{k}+k a_{k +1}+20 a_{k}+\frac {11}{5} a_{k +1}}{5 k +10} \\ \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 {10 k a_{k}+k a_{k +1}+20 a_{k}+\frac {11}{5} a_{k +1}}{5 k +10}, 6 a_{1}+\frac {36 a_{0}}{25}=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}{5}}\right ), a_{k +2}=-\frac {10 k a_{k}+k a_{k +1}+18 a_{k}+2 a_{k +1}}{5 k +9}, 4 a_{1}+a_{0}=0, b_{k +2}=-\frac {10 k b_{k}+k b_{k +1}+20 b_{k}+\frac {11}{5} b_{k +1}}{5 k +10}, 6 b_{1}+\frac {36 b_{0}}{25}=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) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> 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 -> 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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 44
Order:=6; dsolve(x^2*(5+x+10*x^2)*diff(y(x),x$2)+x*(4+3*x+48*x^2)*diff(y(x),x)+(x+36*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{5}} \left (1-\frac {6}{25} x -\frac {1217}{625} x^{2}+\frac {41972}{46875} x^{3}+\frac {1447799}{390625} x^{4}-\frac {375253322}{146484375} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1-\frac {1}{4} x -\frac {35}{18} x^{2}+\frac {11}{12} x^{3}+\frac {632}{171} x^{4}-\frac {2671}{1026} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 85
AsymptoticDSolveValue[x^2*(5+x+10*x^2)*y''[x]+x*(4+3*x+48*x^2)*y'[x]+(x+36*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [5]{x} \left (-\frac {375253322 x^5}{146484375}+\frac {1447799 x^4}{390625}+\frac {41972 x^3}{46875}-\frac {1217 x^2}{625}-\frac {6 x}{25}+1\right )+c_2 \left (-\frac {2671 x^5}{1026}+\frac {632 x^4}{171}+\frac {11 x^3}{12}-\frac {35 x^2}{18}-\frac {x}{4}+1\right ) \]