Internal problem ID [2969]
Internal file name [OUTPUT/2461_Sunday_June_05_2022_03_14_31_AM_35036725/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.5. page
771
Problem number: 27.
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }-\left (4 x +3\right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (-4 x -3\right ) y = 0 \] 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) &= 0\\ q(x) &= -\frac {4 x +3}{4 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 \[ 4 x^{2} y^{\prime \prime }+\left (-4 x -3\right ) y = 0 \] 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} 4 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 (-4 x -3\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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 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 (-4 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-3 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-4 r -3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-4 r -3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {3}{2}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-4 r -3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {3}{2}}, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = 2\) 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 {3}{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{\sqrt {x}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {3}{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 -\frac {1}{2}}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )-4 a_{n -1}-3 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {4 a_{n -1}}{4 n^{2}+8 n r +4 r^{2}-4 n -4 r -3}\tag {4} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{n} = \frac {a_{n -1}}{n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {3}{2}}\) 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 \[ a_{1}=\frac {4}{4 r^{2}+4 r -3} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{1}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{4 r^{2}+4 r -3}\) | \(\frac {1}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{2}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{4 r^{2}+4 r -3}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )}\) | \(\frac {1}{24}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {64}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right )} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{3}={\frac {1}{360}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{4 r^{2}+4 r -3}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {64}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right )}\) | \(\frac {1}{360}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right ) \left (4 r^{2}+28 r +45\right )} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{4}={\frac {1}{8640}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{4 r^{2}+4 r -3}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {64}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right )}\) | \(\frac {1}{360}\) |
| \(a_{4}\) | \(\frac {256}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right ) \left (4 r^{2}+28 r +45\right )}\) | \(\frac {1}{8640}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1024}{\left (2 r +3\right )^{2} \left (2 r +9\right ) \left (2 r +7\right )^{2} \left (11+2 r \right ) \left (2 r +1\right ) \left (2 r -1\right ) \left (2 r +5\right )^{2}} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{5}={\frac {1}{302400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4}{4 r^{2}+4 r -3}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {64}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right )}\) | \(\frac {1}{360}\) |
| \(a_{4}\) | \(\frac {256}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right ) \left (4 r^{2}+20 r +21\right ) \left (4 r^{2}+28 r +45\right )}\) | \(\frac {1}{8640}\) |
| \(a_{5}\) | \(\frac {1024}{\left (2 r +3\right )^{2} \left (2 r +9\right ) \left (2 r +7\right )^{2} \left (11+2 r \right ) \left (2 r +1\right ) \left (2 r -1\right ) \left (2 r +5\right )^{2}}\) | \(\frac {1}{302400}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {3}{2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+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 \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). 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_{2} \\ &= \frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\right )}&= \lim _{r\rightarrow -{\frac {1}{2}}}\frac {16}{\left (4 r^{2}+4 r -3\right ) \left (4 r^{2}+12 r +5\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 \(4 x^{2} y^{\prime \prime }+\left (-4 x -3\right ) y = 0\) gives \[ 4 x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (-4 x -3\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (4 x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-4 x -3\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+4 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )\right ) C +4 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 (-4 x -3\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 \[ 4 x^{2} y_{1}^{\prime \prime }\left (x \right )+\left (-4 x -3\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} 4 x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right ) C +4 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 (-4 x -3\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 (8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +4 \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}+\left (-4 x -3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = {\frac {3}{2}}\) and \(r_{2} = -{\frac {1}{2}}\) then the above becomes \begin{equation} \tag{10} \left (8 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{\frac {1}{2}+n} a_{n} \left (n +\frac {3}{2}\right )\right ) x -4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {3}{2}}\right )\right ) C +4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-\frac {5}{2}+n} b_{n} \left (n -\frac {1}{2}\right ) \left (-\frac {3}{2}+n \right )\right ) x^{2}+\left (-4 x -3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} Expanding \(-4 \sqrt {x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -4 \sqrt {x} &= -4 \sqrt {x} + \dots \\ &= -4 \sqrt {x} \end {align*}
Expanding \(-\frac {3}{\sqrt {x}}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\frac {3}{\sqrt {x}} &= -\frac {3}{\sqrt {x}} + \dots \\ &= -\frac {3}{\sqrt {x}} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +12\right ) C a_{n} x^{n +\frac {3}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C a_{n} x^{n +\frac {3}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{\frac {1}{2}+n} b_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -\frac {1}{2}\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -\frac {1}{2}}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (8 n +12\right ) C a_{n} x^{n +\frac {3}{2}} &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} \left (8 n -4\right ) x^{n -\frac {1}{2}} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 C a_{n} x^{n +\frac {3}{2}}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 C a_{n -2} x^{n -\frac {1}{2}}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{\frac {1}{2}+n} b_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 b_{n -1} x^{n -\frac {1}{2}}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n -\frac {1}{2}\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}C a_{n -2} \left (8 n -4\right ) x^{n -\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 C a_{n -2} x^{n -\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -\frac {1}{2}} b_{n} \left (4 n^{2}-8 n +3\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 b_{n -1} x^{n -\frac {1}{2}}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 b_{n} x^{n -\frac {1}{2}}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ -4 b_{1}-4 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -4 b_{1}-4 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=-1 \] For \(n=N\), where \(N=2\) which is the difference between the two roots, we are free to choose \(b_{2} = 0\). Hence for \(n=2\), Eq (2B) gives \[ 8 C +4 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {1}{2}} \] For \(n=3\), Eq (2B) gives \[ 16 C a_{1}-4 b_{2}+12 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 12 b_{3}-\frac {8}{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {2}{9}} \] For \(n=4\), Eq (2B) gives \[ 24 C a_{2}-4 b_{3}+32 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 32 b_{4}-\frac {25}{18} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {25}{576}} \] For \(n=5\), Eq (2B) gives \[ 32 C a_{3}-4 b_{4}+60 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 60 b_{5}-\frac {157}{720} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {157}{43200}} \] 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 \(C=-{\frac {1}{2}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {1}{2}\eslowast \left (x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-x +\frac {2 x^{3}}{9}+\frac {25 x^{4}}{576}+\frac {157 x^{5}}{43200}+O\left (x^{6}\right )}{\sqrt {x}} \] 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 {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right ) + c_{2} \left (-\frac {1}{2}\eslowast \left (x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+\frac {1-x +\frac {2 x^{3}}{9}+\frac {25 x^{4}}{576}+\frac {157 x^{5}}{43200}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+\frac {1-x +\frac {2 x^{3}}{9}+\frac {25 x^{4}}{576}+\frac {157 x^{5}}{43200}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+\frac {1-x +\frac {2 x^{3}}{9}+\frac {25 x^{4}}{576}+\frac {157 x^{5}}{43200}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right )+c_{2} \left (-\frac {x^{\frac {3}{2}} \left (1+\frac {x}{3}+\frac {x^{2}}{24}+\frac {x^{3}}{360}+\frac {x^{4}}{8640}+\frac {x^{5}}{302400}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{2}+\frac {1-x +\frac {2 x^{3}}{9}+\frac {25 x^{4}}{576}+\frac {157 x^{5}}{43200}+O\left (x^{6}\right )}{\sqrt {x}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }+\left (-4 x -3\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 (4 x +3\right ) y}{4 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 {\left (4 x +3\right ) y}{4 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 )=0, P_{3}\left (x \right )=-\frac {4 x +3}{4 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}}}=0 \\ {} & \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 {3}{4} \\ {} & \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} \\ {} & {} & 4 x^{2} y^{\prime \prime }+\left (-4 x -3\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^{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 (1+2 r \right ) \left (-3+2 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r +1\right ) \left (2 k +2 r -3\right )-4 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 (1+2 r \right ) \left (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (k +r +\frac {1}{2}\right ) \left (k +r -\frac {3}{2}\right ) a_{k}-4 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 \left (k +\frac {3}{2}+r \right ) \left (k -\frac {1}{2}+r \right ) a_{k +1}-4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {4 a_{k}}{\left (2 k +3+2 r \right ) \left (2 k -1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {4 a_{k}}{\left (2 k +2\right ) \left (2 k -2\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-\frac {1}{2}\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {4 a_{k}}{\left (2 k +2\right ) \left (2 k -2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +1}=\frac {4 a_{k}}{\left (2 k +6\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +1}=\frac {4 a_{k}}{\left (2 k +6\right ) \left (2 k +2\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: 0.063 (sec). Leaf size: 65
Order:=6; dsolve(4*x^2*diff(y(x),x$2)-(3+4*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{2} \left (1+\frac {1}{3} x +\frac {1}{24} x^{2}+\frac {1}{360} x^{3}+\frac {1}{8640} x^{4}+\frac {1}{302400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (x^{2}+\frac {1}{3} x^{3}+\frac {1}{24} x^{4}+\frac {1}{360} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (-2+2 x -\frac {4}{9} x^{3}-\frac {25}{288} x^{4}-\frac {157}{21600} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 101
AsymptoticDSolveValue[4*x^2*y''[x]-(3+4*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {x^{11/2}}{8640}+\frac {x^{9/2}}{360}+\frac {x^{7/2}}{24}+\frac {x^{5/2}}{3}+x^{3/2}\right )+c_1 \left (\frac {31 x^4+176 x^3+144 x^2-576 x+576}{576 \sqrt {x}}-\frac {1}{48} x^{3/2} \left (x^2+8 x+24\right ) \log (x)\right ) \]