Internal problem ID [1434]
Internal file name [OUTPUT/1435_Sunday_June_05_2022_02_17_13_AM_47093617/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: 18.
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]]
\[ \boxed {x^{2} \left (1+x \right ) y^{\prime \prime }+3 x^{2} y^{\prime }-\left (6-x \right ) y=0} \] 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. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+3 x^{2} y^{\prime }+\left (x -6\right ) y = 0 \] 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 {3}{1+x}\\ q(x) &= \frac {x -6}{x^{2} \left (1+x \right )}\\ \end {align*}
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 \[ x^{2} \left (1+x \right ) y^{\prime \prime }+3 x^{2} y^{\prime }+\left (x -6\right ) y = 0 \] 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+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 )+3 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x -6\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 )+\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 }}x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 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 }}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 }}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 }}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 =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 )+\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 =1}{\sum }}a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-6 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-6 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-6 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-r -6\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-r -6 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-r -6\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, -2]\).
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^{3} \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 )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{2}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +3}\\ 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^{n -2}\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 )+3 a_{n -1} \left (n +r -1\right )+a_{n -1}-6 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}\right )}{n^{2}+2 n r +r^{2}-n -r -6}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n +3\right )^{2}}{n \left (n +5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 3\) 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 \[ a_{1}=-\frac {\left (r +1\right )^{2}}{r^{2}+r -6} \] Which for the root \(r = 3\) becomes \[ a_{1}=-{\frac {8}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(-{\frac {8}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {100}{21}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(-{\frac {8}{3}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(\frac {100}{21}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{5}+6 r^{4}-5 r^{3}-42 r^{2}+40 r} \] Which for the root \(r = 3\) becomes \[ a_{3}=-{\frac {50}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(-{\frac {8}{3}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(\frac {100}{21}\) |
| \(a_{3}\) | \(-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{5}+6 r^{4}-5 r^{3}-42 r^{2}+40 r}\) | \(-{\frac {50}{7}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r +2\right )^{2} \left (r +1\right ) \left (r +3\right ) \left (r +4\right )}{r^{5}+8 r^{4}-r^{3}-68 r^{2}+60 r} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {175}{18}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(-{\frac {8}{3}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(\frac {100}{21}\) |
| \(a_{3}\) | \(-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{5}+6 r^{4}-5 r^{3}-42 r^{2}+40 r}\) | \(-{\frac {50}{7}}\) |
| \(a_{4}\) | \(\frac {\left (r +2\right )^{2} \left (r +1\right ) \left (r +3\right ) \left (r +4\right )}{r^{5}+8 r^{4}-r^{3}-68 r^{2}+60 r}\) | \(\frac {175}{18}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right ) \left (r +5\right )}{r^{5}+10 r^{4}+5 r^{3}-100 r^{2}+84 r} \] Which for the root \(r = 3\) becomes \[ a_{5}=-{\frac {112}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(-{\frac {8}{3}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(\frac {100}{21}\) |
| \(a_{3}\) | \(-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{5}+6 r^{4}-5 r^{3}-42 r^{2}+40 r}\) | \(-{\frac {50}{7}}\) |
| \(a_{4}\) | \(\frac {\left (r +2\right )^{2} \left (r +1\right ) \left (r +3\right ) \left (r +4\right )}{r^{5}+8 r^{4}-r^{3}-68 r^{2}+60 r}\) | \(\frac {175}{18}\) |
| \(a_{5}\) | \(-\frac {\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right ) \left (r +5\right )}{r^{5}+10 r^{4}+5 r^{3}-100 r^{2}+84 r}\) | \(-{\frac {112}{9}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{3} \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^{3} \left (1-\frac {8 x}{3}+\frac {100 x^{2}}{21}-\frac {50 x^{3}}{7}+\frac {175 x^{4}}{18}-\frac {112 x^{5}}{9}+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 {\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right ) \left (r +5\right )}{r^{5}+10 r^{4}+5 r^{3}-100 r^{2}+84 r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right ) \left (r +5\right )}{r^{5}+10 r^{4}+5 r^{3}-100 r^{2}+84 r}&= \lim _{r\rightarrow -2}-\frac {\left (r +4\right ) \left (r +3\right ) \left (r +1\right ) \left (r +2\right ) \left (r +5\right )}{r^{5}+10 r^{4}+5 r^{3}-100 r^{2}+84 r}\\ &= 0 \end {align*}
The limit is \(0\). 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^{n -2} \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 )+3 b_{n -1} \left (n +r -1\right )+b_{n -1}-6 b_{n} = 0 \end{equation} Which for for the root \(r = -2\) becomes \begin{equation} \tag{4A} b_{n -1} \left (n -3\right ) \left (n -4\right )+b_{n} \left (n -2\right ) \left (n -3\right )+3 b_{n -1} \left (n -3\right )+b_{n -1}-6 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}+2 n r +r^{2}\right )}{n^{2}+2 n r +r^{2}-n -r -6}\tag {5} \] Which for the root \(r = -2\) becomes \[ b_{n} = -\frac {b_{n -1} \left (n^{2}-4 n +4\right )}{n^{2}-5 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -2\) 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 \[ b_{1}=-\frac {r^{2}+2 r +1}{r^{2}+r -6} \] Which for the root \(r = -2\) becomes \[ b_{1}={\frac {1}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(\frac {1}{4}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r^{4}+6 r^{3}+13 r^{2}+12 r +4}{\left (r^{2}+r -6\right ) \left (r^{2}+3 r -4\right )} \] Which for the root \(r = -2\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {r^{5}+9 r^{4}+31 r^{3}+51 r^{2}+40 r +12}{\left (r^{2}+3 r -4\right ) \left (r -2\right ) r \left (r +5\right )} \] Which for the root \(r = -2\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{\left (r +5\right ) r \left (-1+r \right ) \left (r -2\right ) \left (r +4\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{5}+12 r^{4}+55 r^{3}+120 r^{2}+124 r +48}{\left (r +6\right ) \left (r +5\right ) r \left (r -2\right ) \left (-1+r \right )} \] Which for the root \(r = -2\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{\left (r +5\right ) r \left (-1+r \right ) \left (r -2\right ) \left (r +4\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (r +2\right )^{2} \left (r +1\right ) \left (r +3\right ) \left (r +4\right )}{\left (r +6\right ) \left (r +5\right ) \left (-1+r \right ) \left (r -2\right ) r}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {r^{5}+15 r^{4}+85 r^{3}+225 r^{2}+274 r +120}{\left (r +7\right ) \left (-1+r \right ) \left (r -2\right ) r \left (r +6\right )} \] Which for the root \(r = -2\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{r^{2}+r -6}\) | \(\frac {1}{4}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (r +2\right )^{2}}{r^{4}+4 r^{3}-7 r^{2}-22 r +24}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (r +3\right ) \left (r +1\right )^{2} \left (r +2\right )^{2}}{\left (r +5\right ) r \left (-1+r \right ) \left (r -2\right ) \left (r +4\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (r +2\right )^{2} \left (r +1\right ) \left (r +3\right ) \left (r +4\right )}{\left (r +6\right ) \left (r +5\right ) \left (-1+r \right ) \left (r -2\right ) r}\) | \(0\) |
| \(b_{5}\) | \(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{\left (r +7\right ) \left (-1+r \right ) \left (r -2\right ) r \left (r +6\right )}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{3} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= \frac {1+\frac {x}{4}+O\left (x^{6}\right )}{x^{2}} \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^{3} \left (1-\frac {8 x}{3}+\frac {100 x^{2}}{21}-\frac {50 x^{3}}{7}+\frac {175 x^{4}}{18}-\frac {112 x^{5}}{9}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {x}{4}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1-\frac {8 x}{3}+\frac {100 x^{2}}{21}-\frac {50 x^{3}}{7}+\frac {175 x^{4}}{18}-\frac {112 x^{5}}{9}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{4}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1-\frac {8 x}{3}+\frac {100 x^{2}}{21}-\frac {50 x^{3}}{7}+\frac {175 x^{4}}{18}-\frac {112 x^{5}}{9}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{4}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1-\frac {8 x}{3}+\frac {100 x^{2}}{21}-\frac {50 x^{3}}{7}+\frac {175 x^{4}}{18}-\frac {112 x^{5}}{9}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {x}{4}+O\left (x^{6}\right )\right )}{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (1+x \right ) \left (\frac {d}{d x}y^{\prime }\right )+3 x^{2} y^{\prime }+\left (x -6\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 {3 y^{\prime }}{1+x}-\frac {\left (x -6\right ) y}{x^{2} \left (1+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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {3 y^{\prime }}{1+x}+\frac {\left (x -6\right ) y}{x^{2} \left (1+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 {3}{1+x}, P_{3}\left (x \right )=\frac {x -6}{x^{2} \left (1+x \right )}\right ] \\ {} & \circ & \left (1+x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=3 \\ {} & \circ & \left (1+x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \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 (1+x \right ) \left (\frac {d}{d x}y^{\prime }\right )+3 x^{2} y^{\prime }+\left (x -6\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}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (3 u^{2}-6 u +3\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u -7\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}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}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}{d u}\frac {d}{d u}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 (2+r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (3+r \right )-a_{0} \left (2 r^{2}+4 r +7\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right ) \left (k +3+r \right )-a_{k} \left (2 k^{2}+4 k r +2 r^{2}+4 k +4 r +7\right )+a_{k -1} \left (k +r \right )^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (3+r \right )-a_{0} \left (2 r^{2}+4 r +7\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k -1} \left (k +r \right )^{2}+a_{k +1} \left (k +r +1\right ) \left (k +3+r \right )-2 a_{k} \left (k^{2}+\left (2 r +2\right ) k +r^{2}+2 r +\frac {7}{2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k} \left (k +r +1\right )^{2}+a_{k +2} \left (k +r +2\right ) \left (k +4+r \right )-2 a_{k +1} \left (\left (k +1\right )^{2}+\left (2 r +2\right ) \left (k +1\right )+r^{2}+2 r +\frac {7}{2}\right )=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}+2 k a_{k}-8 k a_{k +1}+2 r a_{k}-8 r a_{k +1}+a_{k}-13 a_{k +1}}{\left (k +r +2\right ) \left (k +4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-2 k a_{k}+a_{k}-5 a_{k +1}}{k \left (k +2\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-2 k a_{k}+a_{k}-5 a_{k +1}}{k \left (k +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k a_{k}-8 k a_{k +1}+a_{k}-13 a_{k +1}}{\left (k +2\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k a_{k}-8 k a_{k +1}+a_{k}-13 a_{k +1}}{\left (k +2\right ) \left (k +4\right )}, 3 a_{1}-7 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (1+x \right )^{k}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k a_{k}-8 k a_{k +1}+a_{k}-13 a_{k +1}}{\left (k +2\right ) \left (k +4\right )}, 3 a_{1}-7 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) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 39
Order:=6; dsolve(x^2*(1+x)*diff(y(x),x$2)+3*x^2*diff(y(x),x)-(6-x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{3} \left (1-\frac {8}{3} x +\frac {100}{21} x^{2}-\frac {50}{7} x^{3}+\frac {175}{18} x^{4}-\frac {112}{9} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (2880+720 x +\operatorname {O}\left (x^{6}\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.051 (sec). Leaf size: 53
AsymptoticDSolveValue[x^2*(1+x)*y''[x]+3*x^2*y'[x]-(6-x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{x^2}+\frac {1}{4 x}\right )+c_2 \left (\frac {175 x^7}{18}-\frac {50 x^6}{7}+\frac {100 x^5}{21}-\frac {8 x^4}{3}+x^3\right ) \]