Internal problem ID [1435]
Internal file name [OUTPUT/1436_Sunday_June_05_2022_02_17_17_AM_47446477/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: 19.
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]]
\[ \boxed {x^{2} \left (1+2 x \right ) y^{\prime \prime }-2 x \left (3+14 x \right ) y^{\prime }+\left (6+100 x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (2 x^{3}+x^{2}\right ) y^{\prime \prime }+\left (-28 x^{2}-6 x \right ) y^{\prime }+\left (6+100 x \right ) y = 0 \] 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 {2 \left (3+14 x \right )}{x \left (1+2 x \right )}\\ q(x) &= \frac {6+100 x}{x^{2} \left (1+2 x \right )}\\ \end {align*}
| \(p(x)=-\frac {2 \left (3+14 x \right )}{x \left (1+2 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{2}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {6+100 x}{x^{2} \left (1+2 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{2}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {1}{2}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (1+2 x \right ) y^{\prime \prime }+\left (-28 x^{2}-6 x \right ) y^{\prime }+\left (6+100 x \right ) y = 0 \] 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 (1+2 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 (-28 x^{2}-6 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (6+100 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 }}2 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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-28 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}100 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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-28 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-28 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}100 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}100 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-28 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}100 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-6 x^{n +r} a_{n} \left (n +r \right )+6 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-6 x^{r} a_{0} r +6 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-6 x^{r} r +6 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-7 r +6\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-7 r +6 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 6\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-7 r +6\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([6, 1]\).
Since \(r_1 - r_2 = 5\) 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^{6} \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 \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 +6}\\ 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^{1+n}\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} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-28 a_{n -1} \left (n +r -1\right )-6 a_{n} \left (n +r \right )+6 a_{n}+100 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 \left (n +r -11\right ) a_{n -1}}{n +r -1}\tag {4} \] Which for the root \(r = 6\) becomes \[ a_{n} = -\frac {2 \left (n -5\right ) a_{n -1}}{n +5}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 6\) 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 \[ a_{1}=\frac {20-2 r}{r} \] Which for the root \(r = 6\) becomes \[ a_{1}={\frac {4}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {20-2 r}{r}\) | \(\frac {4}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{2}-76 r +360}{r \left (1+r \right )} \] Which for the root \(r = 6\) becomes \[ a_{2}={\frac {8}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {20-2 r}{r}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(\frac {8}{7}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 6\) becomes \[ a_{3}={\frac {4}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {20-2 r}{r}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(\frac {8}{7}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {4}{7}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 r^{4}-544 r^{3}+6896 r^{2}-38624 r +80640}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 6\) becomes \[ a_{4}={\frac {8}{63}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {20-2 r}{r}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(\frac {8}{7}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {4}{7}\) |
| \(a_{4}\) | \(\frac {16 r^{4}-544 r^{3}+6896 r^{2}-38624 r +80640}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {8}{63}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {32 \left (-10+r \right ) \left (-6+r \right ) \left (-7+r \right ) \left (-8+r \right ) \left (-9+r \right )}{\left (4+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (3+r \right )} \] Which for the root \(r = 6\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {20-2 r}{r}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(\frac {8}{7}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )}\) | \(\frac {4}{7}\) |
| \(a_{4}\) | \(\frac {16 r^{4}-544 r^{3}+6896 r^{2}-38624 r +80640}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {8}{63}\) |
| \(a_{5}\) | \(-\frac {32 \left (-10+r \right ) \left (-6+r \right ) \left (-7+r \right ) \left (-8+r \right ) \left (-9+r \right )}{\left (4+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (3+r \right )}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{6} \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^{6} \left (1+\frac {4 x}{3}+\frac {8 x^{2}}{7}+\frac {4 x^{3}}{7}+\frac {8 x^{4}}{63}+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 \(N=5\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{5}\left (r \right )\). 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_{5} \\ &= -\frac {32 \left (-10+r \right ) \left (-6+r \right ) \left (-7+r \right ) \left (-8+r \right ) \left (-9+r \right )}{\left (4+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (3+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {32 \left (-10+r \right ) \left (-6+r \right ) \left (-7+r \right ) \left (-8+r \right ) \left (-9+r \right )}{\left (4+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (3+r \right )}&= \lim _{r\rightarrow 1}-\frac {32 \left (-10+r \right ) \left (-6+r \right ) \left (-7+r \right ) \left (-8+r \right ) \left (-9+r \right )}{\left (4+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (3+r \right )}\\ &= 4032 \end {align*}
The limit is \(4032\). 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^{1+n} \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} 2 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )-28 b_{n -1} \left (n +r -1\right )-6 b_{n} \left (n +r \right )+6 b_{n}+100 b_{n -1} = 0 \end{equation} Which for for the root \(r = 1\) becomes \begin{equation} \tag{4A} 2 b_{n -1} n \left (n -1\right )+b_{n} \left (1+n \right ) n -28 b_{n -1} n -6 b_{n} \left (1+n \right )+6 b_{n}+100 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {2 \left (n +r -11\right ) b_{n -1}}{n +r -1}\tag {5} \] Which for the root \(r = 1\) becomes \[ b_{n} = -\frac {2 \left (n -10\right ) b_{n -1}}{n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 1\) 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 \[ b_{1}=-\frac {2 \left (-10+r \right )}{r} \] Which for the root \(r = 1\) becomes \[ b_{1}=18 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {20-2 r}{r}\) | \(18\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 r^{2}-76 r +360}{r \left (1+r \right )} \] Which for the root \(r = 1\) becomes \[ b_{2}=144 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {20-2 r}{r}\) | \(18\) |
| \(b_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(144\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {8 \left (r^{3}-27 r^{2}+242 r -720\right )}{r \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ b_{3}=672 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {20-2 r}{r}\) | \(18\) |
| \(b_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(144\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )}\) | \(672\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16 r^{4}-544 r^{3}+6896 r^{2}-38624 r +80640}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ b_{4}=2016 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {20-2 r}{r}\) | \(18\) |
| \(b_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(144\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )}\) | \(672\) |
| \(b_{4}\) | \(\frac {16 r^{4}-544 r^{3}+6896 r^{2}-38624 r +80640}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(2016\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {32 \left (r^{5}-40 r^{4}+635 r^{3}-5000 r^{2}+19524 r -30240\right )}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ b_{5}=4032 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {20-2 r}{r}\) | \(18\) |
| \(b_{2}\) | \(\frac {4 r^{2}-76 r +360}{r \left (1+r \right )}\) | \(144\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+216 r^{2}-1936 r +5760}{r \left (1+r \right ) \left (2+r \right )}\) | \(672\) |
| \(b_{4}\) | \(\frac {16 r^{4}-544 r^{3}+6896 r^{2}-38624 r +80640}{r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(2016\) |
| \(b_{5}\) | \(-\frac {32 \left (-10+r \right ) \left (-6+r \right ) \left (-7+r \right ) \left (-8+r \right ) \left (-9+r \right )}{\left (4+r \right ) \left (2+r \right ) \left (1+r \right ) r \left (3+r \right )}\) | \(4032\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{6} \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 ) \\ &= x \left (1+18 x +144 x^{2}+672 x^{3}+2016 x^{4}+4032 x^{5}+O\left (x^{6}\right )\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^{6} \left (1+\frac {4 x}{3}+\frac {8 x^{2}}{7}+\frac {4 x^{3}}{7}+\frac {8 x^{4}}{63}+O\left (x^{6}\right )\right ) + c_{2} x \left (1+18 x +144 x^{2}+672 x^{3}+2016 x^{4}+4032 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{6} \left (1+\frac {4 x}{3}+\frac {8 x^{2}}{7}+\frac {4 x^{3}}{7}+\frac {8 x^{4}}{63}+O\left (x^{6}\right )\right )+c_{2} x \left (1+18 x +144 x^{2}+672 x^{3}+2016 x^{4}+4032 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{6} \left (1+\frac {4 x}{3}+\frac {8 x^{2}}{7}+\frac {4 x^{3}}{7}+\frac {8 x^{4}}{63}+O\left (x^{6}\right )\right )+c_{2} x \left (1+18 x +144 x^{2}+672 x^{3}+2016 x^{4}+4032 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{6} \left (1+\frac {4 x}{3}+\frac {8 x^{2}}{7}+\frac {4 x^{3}}{7}+\frac {8 x^{4}}{63}+O\left (x^{6}\right )\right )+c_{2} x \left (1+18 x +144 x^{2}+672 x^{3}+2016 x^{4}+4032 x^{5}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (1+2 x \right ) y^{\prime \prime }+\left (-28 x^{2}-6 x \right ) y^{\prime }+\left (6+100 x \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {2 \left (3+50 x \right ) y}{x^{2} \left (1+2 x \right )}+\frac {2 \left (3+14 x \right ) y^{\prime }}{x \left (1+2 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} \\ {} & {} & y^{\prime \prime }-\frac {2 \left (3+14 x \right ) y^{\prime }}{x \left (1+2 x \right )}+\frac {2 \left (3+50 x \right ) y}{x^{2} \left (1+2 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 {2 \left (3+14 x \right )}{x \left (1+2 x \right )}, P_{3}\left (x \right )=\frac {2 \left (3+50 x \right )}{x^{2} \left (1+2 x \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}}}=-6 \\ {} & \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}}}=6 \\ {} & \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^{2} \left (1+2 x \right ) y^{\prime \prime }-2 x \left (3+14 x \right ) y^{\prime }+\left (6+100 x \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..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 =1..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 =2..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} \left (-1+r \right ) \left (-6+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r -1\right ) \left (k +r -6\right )+2 a_{k -1} \left (k +r -6\right ) \left (k -11+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+r \right ) \left (-6+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, 6\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (\left (2 k +2 r -22\right ) a_{k -1}+a_{k} \left (k +r -1\right )\right ) \left (k +r -6\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (2 k +2 r -20\right ) a_{k}+a_{k +1} \left (k +r \right )\right ) \left (k +r -5\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 \left (k +r -10\right ) a_{k}}{k +r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =9 \\ {} & {} & a_{k +1}=-\frac {2 \left (k -9\right ) a_{k}}{k +1} \\ \bullet & {} & \textrm {Recursion relation that defines the terminating series solution of the ODE for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {8}{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=-\frac {2 \left (k -9\right ) a_{k}}{k +1}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =6\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =4 \\ {} & {} & a_{k +1}=-\frac {2 \left (k -4\right ) a_{k}}{k +6} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {4 a_{0}}{3} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {6 a_{1}}{7} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {8 a_{0}}{7} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=\frac {a_{2}}{2} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=\frac {4 a_{0}}{7} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =3 \\ {} & {} & a_{4}=\frac {2 a_{3}}{9} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{4}=\frac {8 a_{0}}{63} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =6\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1+\frac {4}{3} x +\frac {8}{7} x^{2}+\frac {4}{7} x^{3}+\frac {8}{63} x^{4}\right ) \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {8}{\munderset {k =0}{\sum }}a_{k} x^{1+k}\right )+b_{0}\cdot \left (1+\frac {4}{3} x +\frac {8}{7} x^{2}+\frac {4}{7} x^{3}+\frac {8}{63} x^{4}\right ), a_{1+k}=-\frac {2 \left (k -9\right ) a_{k}}{1+k}\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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 43
Order:=6; dsolve(x^2*(1+2*x)*diff(y(x),x$2)-2*x*(3+14*x)*diff(y(x),x)+(6+100*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{6} \left (1+\frac {4}{3} x +\frac {8}{7} x^{2}+\frac {4}{7} x^{3}+\frac {8}{63} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x \left (2880+51840 x +414720 x^{2}+1935360 x^{3}+5806080 x^{4}+11612160 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.049 (sec). Leaf size: 64
AsymptoticDSolveValue[x^2*(1+2*x)*y''[x]-2*x*(3+14*x)*y'[x]+(6+100*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (2016 x^5+672 x^4+144 x^3+18 x^2+x\right )+c_2 \left (\frac {8 x^{10}}{63}+\frac {4 x^9}{7}+\frac {8 x^8}{7}+\frac {4 x^7}{3}+x^6\right ) \]