Internal problem ID [14813]
Internal file name [OUTPUT/14493_Monday_April_08_2024_06_26_02_AM_89943141/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: 13.
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]]
\[ \boxed {y^{\prime \prime }+\left (\frac {1}{2 x}-2\right ) y^{\prime }-\frac {35 y}{16 x^{2}}=0} \] 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. \[ y^{\prime \prime }+\left (\frac {1}{2 x}-2\right ) y^{\prime }-\frac {35 y}{16 x^{2}} = 0 \] 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 {4 x -1}{2 x}\\ q(x) &= -\frac {35}{16 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 \[ 16 x^{2} y^{\prime \prime }+\left (-32 x^{2}+8 x \right ) y^{\prime }-35 y = 0 \] 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} 16 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 (-32 x^{2}+8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-35 \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 }}16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-32 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-35 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 }}\left (-32 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-32 a_{n -1} \left (n +r -1\right ) 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 =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-32 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-35 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+8 x^{n +r} a_{n} \left (n +r \right )-35 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 16 x^{r} a_{0} r \left (-1+r \right )+8 x^{r} a_{0} r -35 a_{0} x^{r} = 0 \] Or \[ \left (16 x^{r} r \left (-1+r \right )+8 x^{r} r -35 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (16 r^{2}-8 r -35\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 16 r^{2}-8 r -35 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {7}{4}}\\ r_2 &= -{\frac {5}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (16 r^{2}-8 r -35\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {7}{4}}, -{\frac {5}{4}}\right ]\).
Since \(r_1 - r_2 = 3\) 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^{\frac {7}{4}} \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^{\frac {5}{4}}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {7}{4}}\\ 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 -\frac {5}{4}}\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} 16 a_{n} \left (n +r \right ) \left (n +r -1\right )-32 a_{n -1} \left (n +r -1\right )+8 a_{n} \left (n +r \right )-35 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {32 a_{n -1} \left (n +r -1\right )}{16 n^{2}+32 n r +16 r^{2}-8 n -8 r -35}\tag {4} \] Which for the root \(r = {\frac {7}{4}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (4 n +3\right )}{2 n \left (n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {7}{4}}\) 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 \[ a_{1}=\frac {32 r}{16 r^{2}+24 r -27} \] Which for the root \(r = {\frac {7}{4}}\) becomes \[ a_{1}={\frac {7}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {32 r}{16 r^{2}+24 r -27}\) | \(\frac {7}{8}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1024 r \left (1+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right )} \] Which for the root \(r = {\frac {7}{4}}\) becomes \[ a_{2}={\frac {77}{160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {32 r}{16 r^{2}+24 r -27}\) | \(\frac {7}{8}\) |
| \(a_{2}\) | \(\frac {1024 r \left (1+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right )}\) | \(\frac {77}{160}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right )} \] Which for the root \(r = {\frac {7}{4}}\) becomes \[ a_{3}={\frac {77}{384}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {32 r}{16 r^{2}+24 r -27}\) | \(\frac {7}{8}\) |
| \(a_{2}\) | \(\frac {1024 r \left (1+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right )}\) | \(\frac {77}{160}\) |
| \(a_{3}\) | \(\frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right )}\) | \(\frac {77}{384}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1048576 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right ) \left (16 r^{2}+120 r +189\right )} \] Which for the root \(r = {\frac {7}{4}}\) becomes \[ a_{4}={\frac {209}{3072}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {32 r}{16 r^{2}+24 r -27}\) | \(\frac {7}{8}\) |
| \(a_{2}\) | \(\frac {1024 r \left (1+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right )}\) | \(\frac {77}{160}\) |
| \(a_{3}\) | \(\frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right )}\) | \(\frac {77}{384}\) |
| \(a_{4}\) | \(\frac {1048576 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right ) \left (16 r^{2}+120 r +189\right )}\) | \(\frac {209}{3072}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {33554432 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right ) \left (16 r^{2}+120 r +189\right ) \left (16 r^{2}+152 r +325\right )} \] Which for the root \(r = {\frac {7}{4}}\) becomes \[ a_{5}={\frac {4807}{245760}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {32 r}{16 r^{2}+24 r -27}\) | \(\frac {7}{8}\) |
| \(a_{2}\) | \(\frac {1024 r \left (1+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right )}\) | \(\frac {77}{160}\) |
| \(a_{3}\) | \(\frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right )}\) | \(\frac {77}{384}\) |
| \(a_{4}\) | \(\frac {1048576 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right ) \left (16 r^{2}+120 r +189\right )}\) | \(\frac {209}{3072}\) |
| \(a_{5}\) | \(\frac {33554432 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right ) \left (16 r^{2}+120 r +189\right ) \left (16 r^{2}+152 r +325\right )}\) | \(\frac {4807}{245760}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {7}{4}} \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^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+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 \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). 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_{3} \\ &= \frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\right )}&= \lim _{r\rightarrow -{\frac {5}{4}}}\frac {32768 r \left (1+r \right ) \left (2+r \right )}{\left (16 r^{2}+24 r -27\right ) \left (16 r^{2}+56 r +13\right ) \left (16 r^{2}+88 r +85\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 \(16 x^{2} y^{\prime \prime }+\left (-32 x^{2}+8 x \right ) y^{\prime }-35 y = 0\) gives \[ 16 x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-32 x^{2}+8 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )-35 C y_{1}\left (x \right ) \ln \left (x \right )-35 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (16 x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-32 x^{2}+8 x \right ) y_{1}^{\prime }\left (x \right )-35 y_{1}\left (x \right )\right ) \ln \left (x \right )+16 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-32 x^{2}+8 x \right ) y_{1}\left (x \right )}{x}\right ) C +16 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 (-32 x^{2}+8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )-35 \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 \[ 16 x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-32 x^{2}+8 x \right ) y_{1}^{\prime }\left (x \right )-35 y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (16 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (-32 x^{2}+8 x \right ) y_{1}\left (x \right )}{x}\right ) C +16 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 (-32 x^{2}+8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )-35 \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 (32 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x +8 \left (-4 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +16 \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}+8 \left (-4 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )-35 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {7}{4}}\) and \(r_{2} = -{\frac {5}{4}}\) then the above becomes \begin{equation} \tag{10} \left (32 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +\frac {3}{4}} a_{n} \left (n +\frac {7}{4}\right )\right ) x +8 \left (-4 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {7}{4}}\right )\right ) C +16 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {13}{4}+n} b_{n} \left (n -\frac {5}{4}\right ) \left (-\frac {9}{4}+n \right )\right ) x^{2}+8 \left (-4 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {9}{4}+n} b_{n} \left (n -\frac {5}{4}\right )\right )-35 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {5}{4}}\right ) = 0 \end{equation} Expanding \(-32 C \,x^{\frac {11}{4}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -32 C \,x^{\frac {11}{4}} &= -32 C \,x^{\frac {11}{4}} + \dots \\ &= -32 C \,x^{\frac {11}{4}} \end {align*}
Expanding \(-8 C \,x^{\frac {7}{4}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -8 C \,x^{\frac {7}{4}} &= -8 C \,x^{\frac {7}{4}} + \dots \\ &= -8 C \,x^{\frac {7}{4}} \end {align*}
Expanding \(-\frac {8}{x^{\frac {1}{4}}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {8}{x^{\frac {1}{4}}} &= -\frac {8}{x^{\frac {1}{4}}} + \dots \\ &= -\frac {8}{x^{\frac {1}{4}}} \end {align*}
Expanding \(\frac {2}{x^{\frac {5}{4}}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \frac {2}{x^{\frac {5}{4}}} &= \frac {2}{x^{\frac {5}{4}}} + \dots \\ &= \frac {2}{x^{\frac {5}{4}}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (32 n +56\right ) C a_{n} x^{n +\frac {7}{4}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-32 C \,x^{n +\frac {11}{4}} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,x^{n +\frac {7}{4}} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {5}{4}} b_{n} \left (16 n^{2}-56 n +45\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (-32 n +40\right ) b_{n} x^{n -\frac {1}{4}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n -10\right ) b_{n} x^{n -\frac {5}{4}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-35 b_{n} x^{n -\frac {5}{4}}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -\frac {5}{4}\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -\frac {5}{4}}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (32 n +56\right ) C a_{n} x^{n +\frac {7}{4}} &= \moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} \left (32 n -40\right ) x^{n -\frac {5}{4}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-32 C \,x^{n +\frac {11}{4}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-32 C a_{n -4} x^{n -\frac {5}{4}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 C \,x^{n +\frac {7}{4}} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-8 C a_{n -3} x^{n -\frac {5}{4}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-32 n +40\right ) b_{n} x^{n -\frac {1}{4}} &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (-32 n +72\right ) x^{n -\frac {5}{4}} \\ \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 \(n -\frac {5}{4}\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =3}{\sum }}C a_{n -3} \left (32 n -40\right ) x^{n -\frac {5}{4}}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-32 C a_{n -4} x^{n -\frac {5}{4}}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-8 C a_{n -3} x^{n -\frac {5}{4}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {5}{4}} b_{n} \left (16 n^{2}-56 n +45\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} \left (-32 n +72\right ) x^{n -\frac {5}{4}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n -10\right ) b_{n} x^{n -\frac {5}{4}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-35 b_{n} x^{n -\frac {5}{4}}\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 \[ -32 b_{1}+40 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -32 b_{1}+40 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}={\frac {5}{4}} \] For \(n=2\), Eq (2B) gives \[ -32 b_{2}+8 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -32 b_{2}+10 = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {5}{16}} \] For \(n=N\), where \(N=3\) which is the difference between the two roots, we are free to choose \(b_{3} = 0\). Hence for \(n=3\), Eq (2B) gives \[ 48 C -\frac {15}{2} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C={\frac {5}{32}} \] For \(n=4\), Eq (2B) gives \[ \left (-32 a_{0}+80 a_{1}\right ) C -56 b_{3}+64 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {95}{16}+64 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {95}{1024}} \] For \(n=5\), Eq (2B) gives \[ \left (-32 a_{1}+112 a_{2}\right ) C -88 b_{4}+160 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {1563}{128}+160 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {1563}{20480}} \] 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 \(C={\frac {5}{32}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \frac {5}{32}\eslowast \left (x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1+\frac {5 x}{4}+\frac {5 x^{2}}{16}-\frac {95 x^{4}}{1024}-\frac {1563 x^{5}}{20480}+O\left (x^{6}\right )}{x^{\frac {5}{4}}} \] 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^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right ) + c_{2} \left (\frac {5}{32}\eslowast \left (x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1+\frac {5 x}{4}+\frac {5 x^{2}}{16}-\frac {95 x^{4}}{1024}-\frac {1563 x^{5}}{20480}+O\left (x^{6}\right )}{x^{\frac {5}{4}}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {5 x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{32}+\frac {1+\frac {5 x}{4}+\frac {5 x^{2}}{16}-\frac {95 x^{4}}{1024}-\frac {1563 x^{5}}{20480}+O\left (x^{6}\right )}{x^{\frac {5}{4}}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {5 x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{32}+\frac {1+\frac {5 x}{4}+\frac {5 x^{2}}{16}-\frac {95 x^{4}}{1024}-\frac {1563 x^{5}}{20480}+O\left (x^{6}\right )}{x^{\frac {5}{4}}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right )+c_{2} \left (\frac {5 x^{\frac {7}{4}} \left (1+\frac {7 x}{8}+\frac {77 x^{2}}{160}+\frac {77 x^{3}}{384}+\frac {209 x^{4}}{3072}+\frac {4807 x^{5}}{245760}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{32}+\frac {1+\frac {5 x}{4}+\frac {5 x^{2}}{16}-\frac {95 x^{4}}{1024}-\frac {1563 x^{5}}{20480}+O\left (x^{6}\right )}{x^{\frac {5}{4}}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 16 x^{2} y^{\prime \prime }+\left (-32 x^{2}+8 x \right ) y^{\prime }-35 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 {35 y}{16 x^{2}}+\frac {\left (4 x -1\right ) y^{\prime }}{2 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 {\left (4 x -1\right ) y^{\prime }}{2 x}-\frac {35 y}{16 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 {4 x -1}{2 x}, P_{3}\left (x \right )=-\frac {35}{16 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}}}=\frac {1}{2} \\ {} & \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}}}=-\frac {35}{16} \\ {} & \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} \\ {} & {} & 16 x^{2} y^{\prime \prime }-8 \left (4 x -1\right ) y^{\prime } x -35 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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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 (5+4 r \right ) \left (-7+4 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (4 k +4 r +5\right ) \left (4 k +4 r -7\right )-32 a_{k -1} \left (k +r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (5+4 r \right ) \left (-7+4 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {5}{4}, \frac {7}{4}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 16 \left (k +r +\frac {5}{4}\right ) \left (k +r -\frac {7}{4}\right ) a_{k}-32 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 16 \left (k +\frac {9}{4}+r \right ) \left (k -\frac {3}{4}+r \right ) a_{k +1}-32 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {32 a_{k} \left (k +r \right )}{\left (4 k +9+4 r \right ) \left (4 k -3+4 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {5}{4} \\ {} & {} & a_{k +1}=\frac {32 a_{k} \left (k -\frac {5}{4}\right )}{\left (4 k +4\right ) \left (4 k -8\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-\frac {5}{4}\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {32 a_{k} \left (k -\frac {5}{4}\right )}{\left (4 k +4\right ) \left (4 k -8\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{4} \\ {} & {} & a_{k +1}=\frac {32 a_{k} \left (k +\frac {7}{4}\right )}{\left (4 k +16\right ) \left (4 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {7}{4}}, a_{k +1}=\frac {32 a_{k} \left (k +\frac {7}{4}\right )}{\left (4 k +16\right ) \left (4 k +4\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 -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 65
Order:=6; dsolve(diff(y(x),x$2)+(1/2*1/x-2)*diff(y(x),x)-35/16*1/x^2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{3} \left (1+\frac {7}{8} x +\frac {77}{160} x^{2}+\frac {77}{384} x^{3}+\frac {209}{3072} x^{4}+\frac {4807}{245760} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (\frac {15}{8} x^{3}+\frac {105}{64} x^{4}+\frac {231}{256} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (12+15 x +\frac {15}{4} x^{2}-\frac {13}{2} x^{3}-\frac {1741}{256} x^{4}-\frac {4141}{1024} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{x^{\frac {5}{4}}} \]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 98
AsymptoticDSolveValue[y''[x]+(1/2*1/x-2)*y'[x]-35/16*1/x^2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {209 x^{23/4}}{3072}+\frac {77 x^{19/4}}{384}+\frac {77 x^{15/4}}{160}+\frac {7 x^{11/4}}{8}+x^{7/4}\right )+c_1 \left (\frac {5}{256} x^{7/4} (7 x+8) \log (x)-\frac {627 x^4+608 x^3-320 x^2-1280 x-1024}{1024 x^{5/4}}\right ) \]