Internal problem ID [1424]
Internal file name [OUTPUT/1425_Sunday_June_05_2022_02_16_34_AM_38405556/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: 8.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+10 y^{\prime } x +\left (x +14\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+10 y^{\prime } x +\left (x +14\right ) y = 0 \] 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 {10}{x}\\ q(x) &= \frac {x +14}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+10 y^{\prime } x +\left (x +14\right ) y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (x +14\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^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 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} &= \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 =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+10 x^{n +r} a_{n} \left (n +r \right )+14 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+10 x^{r} a_{0} r +14 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+10 x^{r} r +14 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (7+r \right ) \left (2+r \right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (7+r \right ) \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -2\\ r_2 &= -7 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (7+r \right ) \left (2+r \right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-2, -7]\).
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 ) &= \frac {\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}}{x^{2}}\\ 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^{7}} \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 -7}\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} \left (n +r \right ) \left (n +r -1\right )+10 a_{n} \left (n +r \right )+a_{n -1}+14 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1}}{n^{2}+2 n r +r^{2}+9 n +9 r +14}\tag {4} \] Which for the root \(r = -2\) becomes \[ a_{n} = -\frac {a_{n -1}}{n \left (n +5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -2\) 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 \[ a_{1}=-\frac {1}{r^{2}+11 r +24} \] Which for the root \(r = -2\) becomes \[ a_{1}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+11 r +24}\) | \(-{\frac {1}{6}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right )} \] Which for the root \(r = -2\) becomes \[ a_{2}={\frac {1}{84}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+11 r +24}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right )}\) | \(\frac {1}{84}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right )} \] Which for the root \(r = -2\) becomes \[ a_{3}=-{\frac {1}{2016}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+11 r +24}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right )}\) | \(\frac {1}{84}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right )}\) | \(-{\frac {1}{2016}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right )} \] Which for the root \(r = -2\) becomes \[ a_{4}={\frac {1}{72576}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+11 r +24}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right )}\) | \(\frac {1}{84}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right )}\) | \(-{\frac {1}{2016}}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right )}\) | \(\frac {1}{72576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right ) \left (r +12\right ) \left (7+r \right )} \] Which for the root \(r = -2\) becomes \[ a_{5}=-{\frac {1}{3628800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+11 r +24}\) | \(-{\frac {1}{6}}\) |
| \(a_{2}\) | \(\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right )}\) | \(\frac {1}{84}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right )}\) | \(-{\frac {1}{2016}}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right )}\) | \(\frac {1}{72576}\) |
| \(a_{5}\) | \(-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right ) \left (r +12\right ) \left (7+r \right )}\) | \(-{\frac {1}{3628800}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )}{x^{2}} \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 {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right ) \left (r +12\right ) \left (7+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right ) \left (r +12\right ) \left (7+r \right )}&= \lim _{r\rightarrow -7}-\frac {1}{\left (r +8\right ) \left (r +3\right ) \left (r +9\right ) \left (r +4\right ) \left (r +10\right ) \left (5+r \right ) \left (r +11\right ) \left (r +6\right ) \left (r +12\right ) \left (7+r \right )}\\ &= \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 \(x^{2} y^{\prime \prime }+10 y^{\prime } x +\left (x +14\right ) y = 0\) gives \[ x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+10 \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) x +\left (x +14\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x^{2} y_{1}^{\prime \prime }\left (x \right )+10 y_{1}^{\prime }\left (x \right ) x +\left (x +14\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+10 y_{1}\left (x \right )\right ) C +x^{2} \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 )+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\left (x +14\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 \[ x^{2} y_{1}^{\prime \prime }\left (x \right )+10 y_{1}^{\prime }\left (x \right ) x +\left (x +14\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+10 y_{1}\left (x \right )\right ) C +x^{2} \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 )+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\left (x +14\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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x +9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\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 ) x^{2}+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x +\left (x +14\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = -2\) and \(r_{2} = -7\) then the above becomes \begin{equation} \tag{10} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-3+n} a_{n} \left (n -2\right )\right ) x +9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -2}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-9+n} b_{n} \left (n -7\right ) \left (-8+n \right )\right ) x^{2}+10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-8+n} b_{n} \left (n -7\right )\right ) x +\left (x +14\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -7}\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 }}9 C a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -7} b_{n} \left (-8+n \right ) \left (n -7\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n -7} b_{n} \left (n -7\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -6} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 b_{n} x^{n -7}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -7\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -7}\) 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 =5}{\sum }}2 C a_{n -5} \left (n -7\right ) x^{n -7} \\ \moverset {\infty }{\munderset {n =0}{\sum }}9 C a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =5}{\sum }}9 C a_{n -5} x^{n -7} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n -6} b_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -7} \\ \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 \(n -7\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =5}{\sum }}2 C a_{n -5} \left (n -7\right ) x^{n -7}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}9 C a_{n -5} x^{n -7}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -7} b_{n} \left (-8+n \right ) \left (n -7\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n -7} b_{n} \left (n -7\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} x^{n -7}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 b_{n} x^{n -7}\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 \[ -4 b_{1}+b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -4 b_{1}+1 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}={\frac {1}{4}} \] For \(n=2\), Eq (2B) gives \[ -6 b_{2}+b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -6 b_{2}+\frac {1}{4} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {1}{24}} \] For \(n=3\), Eq (2B) gives \[ -6 b_{3}+b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -6 b_{3}+\frac {1}{24} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {1}{144}} \] For \(n=4\), Eq (2B) gives \[ -4 b_{4}+b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -4 b_{4}+\frac {1}{144} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {1}{576}} \] For \(n=N\), where \(N=5\) which is the difference between the two roots, we are free to choose \(b_{5} = 0\). Hence for \(n=5\), Eq (2B) gives \[ 5 C +\frac {1}{576} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {1}{2880}} \] 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 \(C=-{\frac {1}{2880}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {1}{2880}\eslowast \left (\frac {1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )}{x^{2}}\right ) \ln \left (x \right )+\frac {1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{144}+\frac {x^{4}}{576}+O\left (x^{6}\right )}{x^{7}} \] Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \frac {c_{1} \left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right )}{x^{2}} + c_{2} \left (-\frac {1}{2880}\eslowast \left (\frac {1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )}{x^{2}}\right ) \ln \left (x \right )+\frac {1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{144}+\frac {x^{4}}{576}+O\left (x^{6}\right )}{x^{7}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right )}{x^{2}}+c_{2} \left (-\frac {\left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2880 x^{2}}+\frac {1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{144}+\frac {x^{4}}{576}+O\left (x^{6}\right )}{x^{7}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right )}{x^{2}}+c_{2} \left (-\frac {\left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2880 x^{2}}+\frac {1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{144}+\frac {x^{4}}{576}+O\left (x^{6}\right )}{x^{7}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right )}{x^{2}}+c_{2} \left (-\frac {\left (1-\frac {x}{6}+\frac {x^{2}}{84}-\frac {x^{3}}{2016}+\frac {x^{4}}{72576}-\frac {x^{5}}{3628800}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2880 x^{2}}+\frac {1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{144}+\frac {x^{4}}{576}+O\left (x^{6}\right )}{x^{7}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+10 y^{\prime } x +\left (x +14\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 {10 y^{\prime }}{x}-\frac {\left (x +14\right ) y}{x^{2}} \\ \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 {10 y^{\prime }}{x}+\frac {\left (x +14\right ) y}{x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {10}{x}, P_{3}\left (x \right )=\frac {x +14}{x^{2}}\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}}}=10 \\ {} & \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}}}=14 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+10 y^{\prime } x +\left (x +14\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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (7+r \right ) \left (2+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +7\right ) \left (k +r +2\right )+a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (7+r \right ) \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-7, -2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +7\right ) \left (k +r +2\right )+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +8+r \right ) \left (k +3+r \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +8+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-7 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +1\right ) \left (k -4\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-7\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =4 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +1\right ) \left (k -4\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{\left (k +6\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +1}=-\frac {a_{k}}{\left (k +6\right ) \left (k +1\right )}\right ] \end {array} \]
Maple trace
`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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 59
Order:=6; dsolve(x^2*diff(y(x),x$2)+10*x*diff(y(x),x)+(14+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {1}{6} x +\frac {1}{84} x^{2}-\frac {1}{2016} x^{3}+\frac {1}{72576} x^{4}-\frac {1}{3628800} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) x^{5}+c_{2} \left (\ln \left (x \right ) \left (-x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (2880+720 x +120 x^{2}+20 x^{3}+5 x^{4}+\operatorname {O}\left (x^{6}\right )\right )\right )}{x^{7}} \]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 68
AsymptoticDSolveValue[x^2*y''[x]+10*x*y'[x]+(14+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {x^2}{72576}+\frac {1}{x^2}-\frac {x}{2016}-\frac {1}{6 x}+\frac {1}{84}\right )+c_1 \left (\frac {1}{x^7}+\frac {1}{4 x^6}+\frac {1}{24 x^5}+\frac {1}{144 x^4}+\frac {1}{576 x^3}\right ) \]