Internal problem ID [14826]
Internal file name [OUTPUT/14506_Monday_April_08_2024_06_26_17_AM_30827800/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.9, page 215
Problem number: 27 (b).
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]]
\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } x +\left (16 x^{2}-25\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (16 x^{2}-25\right ) y = 0 \] 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}{x}\\ q(x) &= \frac {16 x^{2}-25}{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 \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (16 x^{2}-25\right ) y = 0 \] 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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (16 x^{2}-25\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 }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-25 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 }}16 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}16 a_{n -2} 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 }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}16 a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-25 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-25 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -25 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -25 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-25\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-25 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 5\\ r_2 &= -5 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-25\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([5, -5]\).
Since \(r_1 - r_2 = 10\) 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^{5} \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^{5}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +5}\\ 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 -5}\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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+16 a_{n -2}-25 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {16 a_{n -2}}{n^{2}+2 n r +r^{2}-25}\tag {4} \] Which for the root \(r = 5\) becomes \[ a_{n} = -\frac {16 a_{n -2}}{n \left (n +10\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 5\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=-\frac {16}{r^{2}+4 r -21} \] Which for the root \(r = 5\) becomes \[ a_{2}=-{\frac {2}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )} \] Which for the root \(r = 5\) becomes \[ a_{4}={\frac {4}{21}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
| \(a_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=-\frac {4096}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right )} \] Which for the root \(r = 5\) becomes \[ a_{6}=-{\frac {2}{63}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {4096}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right )}\) | \(-{\frac {2}{63}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {4096}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right )}\) | \(-{\frac {2}{63}}\) |
| \(a_{7}\) | \(0\) | \(0\) |
For \(n = 8\), using the above recursive equation gives \[ a_{8}=\frac {65536}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right )} \] Which for the root \(r = 5\) becomes \[ a_{8}={\frac {2}{567}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {4096}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right )}\) | \(-{\frac {2}{63}}\) |
| \(a_{7}\) | \(0\) | \(0\) |
| \(a_{8}\) | \(\frac {65536}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right )}\) | \(\frac {2}{567}\) |
For \(n = 9\), using the above recursive equation gives \[ a_{9}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {4096}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right )}\) | \(-{\frac {2}{63}}\) |
| \(a_{7}\) | \(0\) | \(0\) |
| \(a_{8}\) | \(\frac {65536}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right )}\) | \(\frac {2}{567}\) |
| \(a_{9}\) | \(0\) | \(0\) |
For \(n = 10\), using the above recursive equation gives \[ a_{10}=-\frac {1048576}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right ) \left (r^{2}+20 r +75\right )} \] Which for the root \(r = 5\) becomes \[ a_{10}=-{\frac {4}{14175}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {16}{r^{2}+4 r -21}\) | \(-{\frac {2}{3}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {256}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right )}\) | \(\frac {4}{21}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {4096}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right )}\) | \(-{\frac {2}{63}}\) |
| \(a_{7}\) | \(0\) | \(0\) |
| \(a_{8}\) | \(\frac {65536}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right )}\) | \(\frac {2}{567}\) |
| \(a_{9}\) | \(0\) | \(0\) |
| \(a_{10}\) | \(-\frac {1048576}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right ) \left (r^{2}+20 r +75\right )}\) | \(-{\frac {4}{14175}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{5} \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}+a_{7} x^{7}+a_{8} x^{8}+a_{9} x^{9}+a_{10} x^{10}+a_{11} x^{11}\dots \right ) \\ &= x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\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 \(N=10\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{10}\left (r \right )\). 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_{10} \\ &= -\frac {1048576}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right ) \left (r^{2}+20 r +75\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {1048576}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right ) \left (r^{2}+20 r +75\right )}&= \lim _{r\rightarrow -5}-\frac {1048576}{\left (r^{2}+4 r -21\right ) \left (r^{2}+8 r -9\right ) \left (r^{2}+12 r +11\right ) \left (r^{2}+16 r +39\right ) \left (r^{2}+20 r +75\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 \(x^{2} y^{\prime \prime }+y^{\prime } x +\left (16 x^{2}-25\right ) y = 0\) gives \[ x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right ) x +\left (16 x^{2}-25\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x^{2} y_{1}^{\prime \prime }\left (x \right )+y_{1}^{\prime }\left (x \right ) x +\left (16 x^{2}-25\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 )+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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\left (16 x^{2}-25\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 \[ x^{2} y_{1}^{\prime \prime }\left (x \right )+y_{1}^{\prime }\left (x \right ) x +\left (16 x^{2}-25\right ) y_{1}\left (x \right ) = 0 \] 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 )+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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) x +\left (16 x^{2}-25\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} 2 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (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}+16 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x -25 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = 5\) and \(r_{2} = -5\) then the above becomes \begin{equation} \tag{10} 2 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{4+n} a_{n} \left (n +5\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-7+n} b_{n} \left (n -5\right ) \left (-6+n \right )\right ) x^{2}+16 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -5}\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-6+n} b_{n} \left (n -5\right )\right ) x -25 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -5}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +5} a_{n} \left (n +5\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -5} b_{n} \left (-6+n \right ) \left (n -5\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{-3+n} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -5} b_{n} \left (n -5\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-25 b_{n} x^{n -5}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -5\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -5}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +5} a_{n} \left (n +5\right ) &= \moverset {\infty }{\munderset {n =10}{\sum }}2 C a_{n -10} \left (n -5\right ) x^{n -5} \\ \moverset {\infty }{\munderset {n =0}{\sum }}16 x^{-3+n} b_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}16 b_{n -2} x^{n -5} \\ \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 \(n -5\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =10}{\sum }}2 C a_{n -10} \left (n -5\right ) x^{n -5}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -5} b_{n} \left (-6+n \right ) \left (n -5\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}16 b_{n -2} x^{n -5}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -5} b_{n} \left (n -5\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-25 b_{n} x^{n -5}\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 \[ -9 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -9 b_{1} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=0 \] For \(n=2\), Eq (2B) gives \[ -16 b_{2}+16 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -16 b_{2}+16 = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=1 \] For \(n=3\), Eq (2B) gives \[ -21 b_{3}+16 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -21 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=0 \] For \(n=4\), Eq (2B) gives \[ -24 b_{4}+16 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -24 b_{4}+16 = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {2}{3}} \] For \(n=5\), Eq (2B) gives \[ 16 b_{3}-25 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -25 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=0 \] For \(n=6\), Eq (2B) gives \[ 16 b_{4}-24 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {32}{3}-24 b_{6} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}={\frac {4}{9}} \] For \(n=7\), Eq (2B) gives \[ -21 b_{7}+16 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -21 b_{7} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}=0 \] For \(n=8\), Eq (2B) gives \[ -16 b_{8}+16 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -16 b_{8}+\frac {64}{9} = 0 \] Solving the above for \(b_{8}\) gives \[ b_{8}={\frac {4}{9}} \] For \(n=9\), Eq (2B) gives \[ -9 b_{9}+16 b_{7} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -9 b_{9} = 0 \] Solving the above for \(b_{9}\) gives \[ b_{9}=0 \] For \(n=N\), where \(N=10\) which is the difference between the two roots, we are free to choose \(b_{10} = 0\). Hence for \(n=10\), Eq (2B) gives \[ 10 C +\frac {64}{9} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {32}{45}} \] 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 \(C=-{\frac {32}{45}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {32}{45}\eslowast \left (x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right )\right ) \ln \left (x \right )+\frac {1+x^{2}+\frac {2 x^{4}}{3}+\frac {4 x^{6}}{9}+\frac {4 x^{8}}{9}+O\left (x^{11}\right )}{x^{5}} \] 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^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right ) + c_{2} \left (-\frac {32}{45}\eslowast \left (x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right )\right ) \ln \left (x \right )+\frac {1+x^{2}+\frac {2 x^{4}}{3}+\frac {4 x^{6}}{9}+\frac {4 x^{8}}{9}+O\left (x^{11}\right )}{x^{5}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right )+c_{2} \left (-\frac {32 x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right ) \ln \left (x \right )}{45}+\frac {1+x^{2}+\frac {2 x^{4}}{3}+\frac {4 x^{6}}{9}+\frac {4 x^{8}}{9}+O\left (x^{11}\right )}{x^{5}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right )+c_{2} \left (-\frac {32 x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right ) \ln \left (x \right )}{45}+\frac {1+x^{2}+\frac {2 x^{4}}{3}+\frac {4 x^{6}}{9}+\frac {4 x^{8}}{9}+O\left (x^{11}\right )}{x^{5}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right )+c_{2} \left (-\frac {32 x^{5} \left (1-\frac {2 x^{2}}{3}+\frac {4 x^{4}}{21}-\frac {2 x^{6}}{63}+\frac {2 x^{8}}{567}-\frac {4 x^{10}}{14175}+O\left (x^{11}\right )\right ) \ln \left (x \right )}{45}+\frac {1+x^{2}+\frac {2 x^{4}}{3}+\frac {4 x^{6}}{9}+\frac {4 x^{8}}{9}+O\left (x^{11}\right )}{x^{5}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+y^{\prime } x +\left (16 x^{2}-25\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (16 x^{2}-25\right ) y}{x^{2}}-\frac {y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{x}+\frac {\left (16 x^{2}-25\right ) y}{x^{2}}=0 \\ \bullet & {} & \textrm {Simplify ODE}\hspace {3pt} \\ {} & {} & 16 y x^{2}+x^{2} y^{\prime \prime }+y^{\prime } x -25 y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =4 x \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime } \\ {} & {} & y^{\prime }=4 \frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Compute second derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=16 \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Apply change of variables to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right ) t^{2}+t^{2} \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (\frac {d}{d t}y \left (t \right )\right ) t -25 y \left (t \right )=0 \\ \bullet & {} & \textrm {ODE is now of the Bessel form}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to Bessel ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} \mathit {BesselJ}\left (5, t\right )+c_{2} \mathit {BesselY}\left (5, t\right ) \\ \bullet & {} & \textrm {Make the change from}\hspace {3pt} t \hspace {3pt}\textrm {back to}\hspace {3pt} x \\ {} & {} & y=c_{1} \mathit {BesselJ}\left (5, 4 x \right )+c_{2} \mathit {BesselY}\left (5, 4 x \right ) \end {array} \]
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: 0.047 (sec). Leaf size: 35
Order:=6; dsolve(x^2*diff(y(x),x$2)+x*diff(y(x),x)+(16*x^2-25)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{5} \left (1-\frac {2}{3} x^{2}+\frac {4}{21} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (-1316818944000-1316818944000 x^{2}-877879296000 x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x^{5}} \]
✓ Solution by Mathematica
Time used: 0.009 (sec). Leaf size: 42
AsymptoticDSolveValue[x^2*y''[x]+x*y'[x]+(16*x^2-25)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{x^5}+\frac {1}{x^3}+\frac {2}{3 x}\right )+c_2 \left (\frac {4 x^9}{21}-\frac {2 x^7}{3}+x^5\right ) \]