Internal problem ID [1436]
Internal file name [OUTPUT/1437_Sunday_June_05_2022_02_17_20_AM_96646602/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: 20.
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]]
\[ \boxed {x^{2} \left (x +1\right ) y^{\prime \prime }-x \left (6+11 x \right ) y^{\prime }+\left (6+32 x \right ) y=0} \] 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. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (-11 x^{2}-6 x \right ) y^{\prime }+\left (6+32 x \right ) y = 0 \] 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 {6+11 x}{x \left (x +1\right )}\\ q(x) &= \frac {6+32 x}{x^{2} \left (x +1\right )}\\ \end {align*}
| \(p(x)=-\frac {6+11 x}{x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {6+32 x}{x^{2} \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x +1\right ) y^{\prime \prime }+\left (-11 x^{2}-6 x \right ) y^{\prime }+\left (6+32 x \right ) y = 0 \] 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 (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 (-11 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+32 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 }}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 (-11 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 }}32 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 }}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 }}\left (-11 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-11 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}32 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}32 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 }}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 (-11 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 }}32 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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-11 a_{n -1} \left (n +r -1\right )-6 a_{n} \left (n +r \right )+6 a_{n}+32 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-14 n -14 r +45\right )}{n^{2}+2 n r +r^{2}-7 n -7 r +6}\tag {4} \] Which for the root \(r = 6\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n^{2}-2 n -3\right )}{n \left (n +5\right )}\tag {5} \] 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 \[ a_{1}=\frac {-r^{2}+12 r -32}{r \left (r -5\right )} \] Which for the root \(r = 6\) becomes \[ a_{1}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {2}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )} \] Which for the root \(r = 6\) becomes \[ a_{2}={\frac {1}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {1}{7}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )} \] Which for the root \(r = 6\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {1}{7}\) |
| \(a_{3}\) | \(\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{4}-22 r^{3}+167 r^{2}-482 r +336}{\left (r +3\right ) r \left (r +1\right ) \left (r +2\right )} \] Which for the root \(r = 6\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {1}{7}\) |
| \(a_{3}\) | \(\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {r^{4}-22 r^{3}+167 r^{2}-482 r +336}{\left (r +3\right ) r \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r^{4}+25 r^{3}-230 r^{2}+920 r -1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )} \] Which for the root \(r = 6\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {1}{7}\) |
| \(a_{3}\) | \(\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {r^{4}-22 r^{3}+167 r^{2}-482 r +336}{\left (r +3\right ) r \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
| \(a_{5}\) | \(\frac {-r^{4}+25 r^{3}-230 r^{2}+920 r -1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
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 {2 x}{3}+\frac {x^{2}}{7}+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 {-r^{4}+25 r^{3}-230 r^{2}+920 r -1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-r^{4}+25 r^{3}-230 r^{2}+920 r -1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}&= \lim _{r\rightarrow 1}\frac {-r^{4}+25 r^{3}-230 r^{2}+920 r -1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\\ &= -{\frac {21}{4}} \end {align*}
The limit is \(-{\frac {21}{4}}\). 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} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )-11 b_{n -1} \left (n +r -1\right )-6 b_{n} \left (n +r \right )+6 b_{n}+32 b_{n -1} = 0 \end{equation} Which for for the root \(r = 1\) becomes \begin{equation} \tag{4A} b_{n -1} n \left (n -1\right )+b_{n} \left (1+n \right ) n -11 b_{n -1} n -6 b_{n} \left (1+n \right )+6 b_{n}+32 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}+2 n r +r^{2}-14 n -14 r +45\right )}{n^{2}+2 n r +r^{2}-7 n -7 r +6}\tag {5} \] Which for the root \(r = 1\) becomes \[ b_{n} = -\frac {b_{n -1} \left (n^{2}-12 n +32\right )}{n^{2}-5 n}\tag {6} \] 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 \[ b_{1}=-\frac {r^{2}-12 r +32}{r \left (r -5\right )} \] Which for the root \(r = 1\) becomes \[ b_{1}={\frac {21}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {21}{4}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )} \] Which for the root \(r = 1\) becomes \[ b_{2}={\frac {21}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {21}{4}\) |
| \(b_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {21}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {r^{4}-23 r^{3}+188 r^{2}-628 r +672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )} \] Which for the root \(r = 1\) becomes \[ b_{3}={\frac {35}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {21}{4}\) |
| \(b_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {21}{2}\) |
| \(b_{3}\) | \(\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )}\) | \(\frac {35}{4}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{4}-22 r^{3}+167 r^{2}-482 r +336}{\left (r +3\right ) r \left (r +1\right ) \left (r +2\right )} \] Which for the root \(r = 1\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {21}{4}\) |
| \(b_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {21}{2}\) |
| \(b_{3}\) | \(\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )}\) | \(\frac {35}{4}\) |
| \(b_{4}\) | \(\frac {r^{4}-22 r^{3}+167 r^{2}-482 r +336}{\left (r +3\right ) r \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {r^{4}-25 r^{3}+230 r^{2}-920 r +1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )} \] Which for the root \(r = 1\) becomes \[ b_{5}=-{\frac {21}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}+12 r -32}{r \left (r -5\right )}\) | \(\frac {21}{4}\) |
| \(b_{2}\) | \(\frac {r^{3}-18 r^{2}+101 r -168}{\left (r +1\right ) r \left (r -5\right )}\) | \(\frac {21}{2}\) |
| \(b_{3}\) | \(\frac {-r^{4}+23 r^{3}-188 r^{2}+628 r -672}{\left (r +2\right ) \left (r +1\right ) r \left (r -5\right )}\) | \(\frac {35}{4}\) |
| \(b_{4}\) | \(\frac {r^{4}-22 r^{3}+167 r^{2}-482 r +336}{\left (r +3\right ) r \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
| \(b_{5}\) | \(\frac {-r^{4}+25 r^{3}-230 r^{2}+920 r -1344}{\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\) | \(-{\frac {21}{4}}\) |
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+\frac {21 x}{4}+\frac {21 x^{2}}{2}+\frac {35 x^{3}}{4}-\frac {21 x^{5}}{4}+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 {2 x}{3}+\frac {x^{2}}{7}+O\left (x^{6}\right )\right ) + c_{2} x \left (1+\frac {21 x}{4}+\frac {21 x^{2}}{2}+\frac {35 x^{3}}{4}-\frac {21 x^{5}}{4}+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 {2 x}{3}+\frac {x^{2}}{7}+O\left (x^{6}\right )\right )+c_{2} x \left (1+\frac {21 x}{4}+\frac {21 x^{2}}{2}+\frac {35 x^{3}}{4}-\frac {21 x^{5}}{4}+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 {2 x}{3}+\frac {x^{2}}{7}+O\left (x^{6}\right )\right )+c_{2} x \left (1+\frac {21 x}{4}+\frac {21 x^{2}}{2}+\frac {35 x^{3}}{4}-\frac {21 x^{5}}{4}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{6} \left (1+\frac {2 x}{3}+\frac {x^{2}}{7}+O\left (x^{6}\right )\right )+c_{2} x \left (1+\frac {21 x}{4}+\frac {21 x^{2}}{2}+\frac {35 x^{3}}{4}-\frac {21 x^{5}}{4}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) y^{\prime \prime }+\left (-11 x^{2}-6 x \right ) y^{\prime }+\left (6+32 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+16 x \right ) y}{x^{2} \left (x +1\right )}+\frac {\left (6+11 x \right ) y^{\prime }}{x \left (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 (6+11 x \right ) y^{\prime }}{x \left (x +1\right )}+\frac {2 \left (3+16 x \right ) y}{x^{2} \left (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 {6+11 x}{x \left (x +1\right )}, P_{3}\left (x \right )=\frac {2 \left (3+16 x \right )}{x^{2} \left (x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-5 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) y^{\prime \prime }-x \left (6+11 x \right ) y^{\prime }+\left (6+32 x \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-2 u^{2}+u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-11 u^{2}+16 u -5\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-26+32 u \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-6+r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-5+r \right )-2 a_{0} \left (r^{2}-9 r +13\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k -5+r \right )-2 a_{k} \left (k^{2}+2 k r +r^{2}-9 k -9 r +13\right )+a_{k -1} \left (k -5+r \right ) \left (k -9+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-6+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 6\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-5+r \right )-2 a_{0} \left (r^{2}-9 r +13\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+2 \left (\left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r +9 a_{k}-7 a_{k -1}-2 a_{k +1}\right ) k +\left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+2 \left (9 a_{k}-7 a_{k -1}-2 a_{k +1}\right ) r -26 a_{k}+45 a_{k -1}-5 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) \left (k +1\right )^{2}+2 \left (\left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r +9 a_{k +1}-7 a_{k}-2 a_{k +2}\right ) \left (k +1\right )+\left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r^{2}+2 \left (9 a_{k +1}-7 a_{k}-2 a_{k +2}\right ) r -26 a_{k +1}+45 a_{k}-5 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k r a_{k}-4 k r a_{k +1}+r^{2} a_{k}-2 r^{2} a_{k +1}-12 k a_{k}+14 k a_{k +1}-12 r a_{k}+14 r a_{k +1}+32 a_{k}-10 a_{k +1}}{k^{2}+2 k r +r^{2}-2 k -2 r -8} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-12 k a_{k}+14 k a_{k +1}+32 a_{k}-10 a_{k +1}}{k^{2}-2 k -8} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =4 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-12 k a_{k}+14 k a_{k +1}+32 a_{k}-10 a_{k +1}}{k^{2}-2 k -8} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =6 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-10 k a_{k +1}-4 a_{k}+2 a_{k +1}}{k^{2}+10 k +16} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =6 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +6}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-10 k a_{k +1}-4 a_{k}+2 a_{k +1}}{k^{2}+10 k +16}, 7 a_{1}+10 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +6}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-10 k a_{k +1}-4 a_{k}+2 a_{k +1}}{k^{2}+10 k +16}, 7 a_{1}+10 a_{0}=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 37
Order:=6; dsolve(x^2*(1+x)*diff(y(x),x$2)-x*(6+11*x)*diff(y(x),x)+(6+32*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{6} \left (1+\frac {2}{3} x +\frac {1}{7} x^{2}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x \left (2880+15120 x +30240 x^{2}+25200 x^{3}-15120 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.05 (sec). Leaf size: 51
AsymptoticDSolveValue[x^2*(1+x)*y''[x]-x*(6+11*x)*y'[x]+(6+32*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {x^8}{7}+\frac {2 x^7}{3}+x^6\right )+c_1 \left (\frac {35 x^4}{4}+\frac {21 x^3}{2}+\frac {21 x^2}{4}+x\right ) \]