Internal problem ID [1200]
Internal file name [OUTPUT/1201_Sunday_June_05_2022_02_04_58_AM_40370122/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. 7.1 Exercises. Page
318
Problem number: 22.
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]]
\[ \boxed {x^{2} \left (1+x \right ) y^{\prime \prime }+x \left (1+2 x \right ) y^{\prime }-\left (4+6 x \right ) y=0} \] 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. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (2 x^{2}+x \right ) y^{\prime }+\left (-6 x -4\right ) y = 0 \] 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 {1+2 x}{x \left (1+x \right )}\\ q(x) &= -\frac {2 \left (2+3 x \right )}{x^{2} \left (1+x \right )}\\ \end {align*}
| \(p(x)=\frac {1+2 x}{x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {2 \left (2+3 x \right )}{x^{2} \left (1+x \right )}\) | |
| 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 \[ x^{2} \left (1+x \right ) y^{\prime \prime }+\left (2 x^{2}+x \right ) y^{\prime }+\left (-6 x -4\right ) y = 0 \] 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 )+\left (2 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-6 x -4\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 }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =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 }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-4 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -4 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -4 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-4\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-4 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-4\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, -2]\).
Since \(r_1 - r_2 = 4\) 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^{2} \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 +2}\\ 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 )+2 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-6 a_{n -1}-4 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (n +r -3\right ) a_{n -1}}{n +r -2}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = -\frac {\left (n -1\right ) a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) 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 \[ a_{1}=\frac {-r +2}{-1+r} \] Which for the root \(r = 2\) becomes \[ a_{1}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r +2}{-1+r}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r -2}{r} \] Which for the root \(r = 2\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r +2}{-1+r}\) | \(0\) |
| \(a_{2}\) | \(\frac {r -2}{r}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r +2}{1+r} \] Which for the root \(r = 2\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r +2}{-1+r}\) | \(0\) |
| \(a_{2}\) | \(\frac {r -2}{r}\) | \(0\) |
| \(a_{3}\) | \(\frac {-r +2}{1+r}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r -2}{2+r} \] Which for the root \(r = 2\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r +2}{-1+r}\) | \(0\) |
| \(a_{2}\) | \(\frac {r -2}{r}\) | \(0\) |
| \(a_{3}\) | \(\frac {-r +2}{1+r}\) | \(0\) |
| \(a_{4}\) | \(\frac {r -2}{2+r}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r +2}{3+r} \] Which for the root \(r = 2\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r +2}{-1+r}\) | \(0\) |
| \(a_{2}\) | \(\frac {r -2}{r}\) | \(0\) |
| \(a_{3}\) | \(\frac {-r +2}{1+r}\) | \(0\) |
| \(a_{4}\) | \(\frac {r -2}{2+r}\) | \(0\) |
| \(a_{5}\) | \(\frac {-r +2}{3+r}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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^{2} \left (1+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 \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). 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_{4} \\ &= \frac {r -2}{2+r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {r -2}{2+r}&= \lim _{r\rightarrow -2}\frac {r -2}{2+r}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(x^{2} \left (1+x \right ) y^{\prime \prime }+\left (2 x^{2}+x \right ) y^{\prime }+\left (-6 x -4\right ) y = 0\) gives \[ x^{2} \left (1+x \right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (2 x^{2}+x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (-6 x -4\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x^{2} \left (1+x \right ) y_{1}^{\prime \prime }\left (x \right )+\left (2 x^{2}+x \right ) y_{1}^{\prime }\left (x \right )+\left (-6 x -4\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+x^{2} \left (1+x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (2 x^{2}+x \right ) y_{1}\left (x \right )}{x}\right ) C +x^{2} \left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (2 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (-6 x -4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ x^{2} \left (1+x \right ) y_{1}^{\prime \prime }\left (x \right )+\left (2 x^{2}+x \right ) y_{1}^{\prime }\left (x \right )+\left (-6 x -4\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x^{2} \left (1+x \right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (2 x^{2}+x \right ) y_{1}\left (x \right )}{x}\right ) C +x^{2} \left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (2 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (-6 x -4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \left (2 x \left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right ) x \right ) C +x^{2} \left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right )+\left (2 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+\left (-6 x -4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = 2\) and \(r_{2} = -2\) then the above becomes \begin{equation} \tag{10} \left (2 x \left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} a_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\right ) x \right ) C +x^{2} \left (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-4+n} b_{n} \left (n -2\right ) \left (-3+n \right )\right )+\left (2 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-3+n} b_{n} \left (n -2\right )\right )+\left (-6 x -4\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +2} a_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +3} a_{n} \left (n +2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n -1} b_{n} \left (n -2\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n -2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n -1} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 b_{n} x^{n -2}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -2\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -2}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +2} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}2 C a_{-4+n} \left (n -2\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +3} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}2 C a_{n -5} \left (-3+n \right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,x^{n +3} a_{n} &= \moverset {\infty }{\munderset {n =5}{\sum }}C a_{n -5} x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (n^{2}-5 n +6\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (\left (n -1\right )^{2}-5 n +11\right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n -1} b_{n} \left (n -2\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 b_{n -1} \left (-3+n \right ) x^{n -2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n -1} b_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 b_{n -1} x^{n -2}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n -2\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =4}{\sum }}2 C a_{-4+n} \left (n -2\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}2 C a_{n -5} \left (-3+n \right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}C a_{n -5} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n^{2}-5 n +6\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (\left (n -1\right )^{2}-5 n +11\right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 b_{n -1} \left (-3+n \right ) x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} \left (n -2\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 b_{n -1} x^{n -2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 b_{n} x^{n -2}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ -3 b_{1}-4 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -3 b_{1}-4 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=-{\frac {4}{3}} \] For \(n=2\), Eq (2B) gives \[ -6 b_{1}-4 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 8-4 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=2 \] For \(n=3\), Eq (2B) gives \[ -3 b_{3}-6 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -3 b_{3}-12 = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=-4 \] For \(n=N\), where \(N=4\) which is the difference between the two roots, we are free to choose \(b_{4} = 0\). Hence for \(n=4\), Eq (2B) gives \[ 4 C +16 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-4 \] For \(n=5\), Eq (2B) gives \[ \left (5 a_{0}+6 a_{1}\right ) C +5 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -20+5 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=4 \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Using the above value found for \(C=-4\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \left (-4\right )\eslowast \left (x^{2} \left (1+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-\frac {4 x}{3}+2 x^{2}-4 x^{3}+4 x^{5}+O\left (x^{6}\right )}{x^{2}} \] 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^{2} \left (1+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-4\right )\eslowast \left (x^{2} \left (1+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-\frac {4 x}{3}+2 x^{2}-4 x^{3}+4 x^{5}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1+O\left (x^{6}\right )\right )+c_{2} \left (-4 x^{2} \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {4 x}{3}+2 x^{2}-4 x^{3}+4 x^{5}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1+O\left (x^{6}\right )\right )+c_{2} \left (-4 x^{2} \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {4 x}{3}+2 x^{2}-4 x^{3}+4 x^{5}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1+O\left (x^{6}\right )\right )+c_{2} \left (-4 x^{2} \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+\frac {1-\frac {4 x}{3}+2 x^{2}-4 x^{3}+4 x^{5}+O\left (x^{6}\right )}{x^{2}}\right ) \] 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 )+\left (2 x^{2}+x \right ) y^{\prime }+\left (-6 x -4\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 {2 \left (2+3 x \right ) y}{x^{2} \left (1+x \right )}-\frac {\left (1+2 x \right ) y^{\prime }}{x \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 {\left (1+2 x \right ) y^{\prime }}{x \left (1+x \right )}-\frac {2 \left (2+3 x \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 {1+2 x}{x \left (1+x \right )}, P_{3}\left (x \right )=-\frac {2 \left (2+3 x \right )}{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}}}=1 \\ {} & \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 )+x \left (1+2 x \right ) y^{\prime }+\left (-6 x -4\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 (2 u^{2}-3 u +1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-6 u +2\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^{2} u^{-1+r}+\left (a_{1} \left (1+r \right )^{2}-a_{0} \left (2 r^{2}+r -2\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right )^{2}-a_{k} \left (2 k^{2}+4 k r +2 r^{2}+k +r -2\right )+a_{k -1} \left (k +2+r \right ) \left (k -3+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right )^{2}-a_{0} \left (2 r^{2}+r -2\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}-k -6\right ) a_{k -1}+\left (-2 k^{2}-k +2\right ) a_{k}+a_{k +1} \left (k +1\right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}-k -7\right ) a_{k}+\left (-2 \left (k +1\right )^{2}-k +1\right ) a_{k +1}+a_{k +2} \left (k +2\right )^{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}+k a_{k}-5 k a_{k +1}-6 a_{k}-a_{k +1}}{\left (k +2\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+k a_{k}-5 k a_{k +1}-6 a_{k}-a_{k +1}}{\left (k +2\right )^{2}} \\ \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}+k a_{k}-5 k a_{k +1}-6 a_{k}-a_{k +1}}{\left (k +2\right )^{2}}, a_{1}+2 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}+k a_{k}-5 k a_{k +1}-6 a_{k}-a_{k +1}}{\left (k +2\right )^{2}}, a_{1}+2 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: 0.015 (sec). Leaf size: 51
Order:=6; dsolve(x^2*(1+x)*diff(y(x),x$2)+x*(1+2*x)*diff(y(x),x)-(4+6*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (\ln \left (x \right ) \left (576 x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\left (-144+192 x -288 x^{2}+576 x^{3}-576 x^{4}-576 x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 48
AsymptoticDSolveValue[x^2*(1+x)*y''[x]+x*(1+2*x)*y'[x]-(4+6*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 x^2+c_1 \left (\frac {3 x^4-12 x^3+6 x^2-4 x+3}{3 x^2}-4 x^2 \log (x)\right ) \]