Internal problem ID [2941]
Internal file name [OUTPUT/2433_Sunday_June_05_2022_03_09_50_AM_7140631/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: (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]]
\[ \boxed {x^{2} y^{\prime \prime }-\left (-1+2 \sqrt {5}\right ) x y^{\prime }+\left (\frac {19}{4}-3 x^{2}\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+x \left (1-2 \sqrt {5}\right ) y^{\prime }+\left (\frac {19}{4}-3 x^{2}\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {-1+2 \sqrt {5}}{x}\\ q(x) &= -\frac {12 x^{2}-19}{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 \[ x^{2} y^{\prime \prime }+x \left (1-2 \sqrt {5}\right ) y^{\prime }+\left (\frac {19}{4}-3 x^{2}\right ) y = 0 \] 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 )+x \left (1-2 \sqrt {5}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\frac {19}{4}-3 x^{2}\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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (-1+2 \sqrt {5}\right ) x^{n} x^{r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {19 a_{n} x^{n +r}}{4}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r +2} a_{n}\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 (-3 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} 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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (-1+2 \sqrt {5}\right ) x^{n} x^{r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {19 a_{n} x^{n +r}}{4}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right ) \left (1-2 \sqrt {5}\right )+\frac {19 a_{n} x^{n +r}}{4} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r \left (1-2 \sqrt {5}\right )+\frac {19 a_{0} x^{r}}{4} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r \left (1-2 \sqrt {5}\right )+\frac {19 x^{r}}{4}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-2 r \sqrt {5}+\frac {19}{4}\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-2 r \sqrt {5}+\frac {19}{4} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {1}{2}+\sqrt {5}\\ r_2 &= -\frac {1}{2}+\sqrt {5} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-2 r \sqrt {5}+\frac {19}{4}\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {1}{2}+\sqrt {5}, -\frac {1}{2}+\sqrt {5}\right ]\).
Since \(r_1 - r_2 = 1\) 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 {1}{2}+\sqrt {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 )+x^{-\frac {1}{2}+\sqrt {5}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}+\sqrt {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 -\frac {1}{2}+\sqrt {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 (1-2 \sqrt {5}\right ) \left (n +r \right )+\frac {19 a_{n}}{4}-3 a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {12 a_{n -2}}{-19+8 \sqrt {5}\, n +8 r \sqrt {5}-4 n^{2}-8 n r -4 r^{2}}\tag {4} \] Which for the root \(r = \frac {1}{2}+\sqrt {5}\) becomes \[ a_{n} = \frac {3 a_{n -2}}{n \left (n +1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {1}{2}+\sqrt {5}\) 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 \[ a_{2}=\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35} \] Which for the root \(r = \frac {1}{2}+\sqrt {5}\) becomes \[ a_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {144}{\left (-64 r^{3}-576 r^{2}-1968 r -2448\right ) \sqrt {5}+16 r^{4}+192 r^{3}+1304 r^{2}+4368 r +5465} \] Which for the root \(r = \frac {1}{2}+\sqrt {5}\) becomes \[ a_{4}={\frac {3}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {144}{\left (-64 r^{3}-576 r^{2}-1968 r -2448\right ) \sqrt {5}+16 r^{4}+192 r^{3}+1304 r^{2}+4368 r +5465}\) | \(\frac {3}{40}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {144}{\left (-64 r^{3}-576 r^{2}-1968 r -2448\right ) \sqrt {5}+16 r^{4}+192 r^{3}+1304 r^{2}+4368 r +5465}\) | \(\frac {3}{40}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{2}+\sqrt {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}\dots \right ) \\ &= x^{\frac {1}{2}+\sqrt {5}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{40}+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 \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). 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_{1} \\ &= 0 \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow -\frac {1}{2}+\sqrt {5}}0\\ &= 0 \end {align*}
The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}+\sqrt {5}} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (-\frac {1}{2}+\sqrt {5}\right ) b_{n} \left (n +r \right )+\frac {19 b_{n}}{4}-3 b_{n -2} = 0 \end{equation} Which for for the root \(r = -\frac {1}{2}+\sqrt {5}\) becomes \begin{equation} \tag{4A} b_{n} \left (n -\frac {1}{2}+\sqrt {5}\right ) \left (n -\frac {3}{2}+\sqrt {5}\right )-2 \left (-\frac {1}{2}+\sqrt {5}\right ) b_{n} \left (n -\frac {1}{2}+\sqrt {5}\right )+\frac {19 b_{n}}{4}-3 b_{n -2} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {12 b_{n -2}}{-19+8 \sqrt {5}\, n +8 r \sqrt {5}-4 n^{2}-8 n r -4 r^{2}}\tag {5} \] Which for the root \(r = -\frac {1}{2}+\sqrt {5}\) becomes \[ b_{n} = -\frac {12 b_{n -2}}{-19+8 \sqrt {5}\, n +8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 n^{2}-8 n \left (-\frac {1}{2}+\sqrt {5}\right )-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -\frac {1}{2}+\sqrt {5}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=-\frac {12}{-35+16 \sqrt {5}+8 r \sqrt {5}-16 r -4 r^{2}} \] Which for the root \(r = -\frac {1}{2}+\sqrt {5}\) becomes \[ b_{2}=-\frac {12}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(-\frac {12}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(-\frac {12}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=-\frac {144}{-1304 r^{2}+64 \sqrt {5}\, r^{3}-16 r^{4}-4368 r +576 \sqrt {5}\, r^{2}-192 r^{3}-5465+1968 r \sqrt {5}+2448 \sqrt {5}} \] Which for the root \(r = -\frac {1}{2}+\sqrt {5}\) becomes \[ b_{4}=-\frac {144}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(-\frac {12}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {144}{\left (-64 r^{3}-576 r^{2}-1968 r -2448\right ) \sqrt {5}+16 r^{4}+192 r^{3}+1304 r^{2}+4368 r +5465}\) | \(-\frac {144}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {12}{\left (-8 r -16\right ) \sqrt {5}+4 r^{2}+16 r +35}\) | \(-\frac {12}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {144}{\left (-64 r^{3}-576 r^{2}-1968 r -2448\right ) \sqrt {5}+16 r^{4}+192 r^{3}+1304 r^{2}+4368 r +5465}\) | \(-\frac {144}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{2}+\sqrt {5}} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= x^{-\frac {1}{2}+\sqrt {5}} \left (1-\frac {12 x^{2}}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}-\frac {144 x^{4}}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}+O\left (x^{6}\right )\right ) \end {align*}
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 {1}{2}+\sqrt {5}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{40}+O\left (x^{6}\right )\right ) + c_{2} x^{-\frac {1}{2}+\sqrt {5}} \left (1-\frac {12 x^{2}}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}-\frac {144 x^{4}}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{2}+\sqrt {5}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{40}+O\left (x^{6}\right )\right )+c_{2} x^{-\frac {1}{2}+\sqrt {5}} \left (1-\frac {12 x^{2}}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}-\frac {144 x^{4}}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}+\sqrt {5}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{40}+O\left (x^{6}\right )\right )+c_{2} x^{-\frac {1}{2}+\sqrt {5}} \left (1-\frac {12 x^{2}}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}-\frac {144 x^{4}}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{2}+\sqrt {5}} \left (1+\frac {x^{2}}{2}+\frac {3 x^{4}}{40}+O\left (x^{6}\right )\right )+c_{2} x^{-\frac {1}{2}+\sqrt {5}} \left (1-\frac {12 x^{2}}{-27+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}}-\frac {144 x^{4}}{-1304 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}+64 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{3}-16 \left (-\frac {1}{2}+\sqrt {5}\right )^{4}-3281-1920 \sqrt {5}+576 \sqrt {5}\, \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-192 \left (-\frac {1}{2}+\sqrt {5}\right )^{3}+1968 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+x \left (1-2 \sqrt {5}\right ) y^{\prime }+\left (\frac {19}{4}-3 x^{2}\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 (12 x^{2}-19\right ) y}{4 x^{2}}+\frac {\left (-1+2 \sqrt {5}\right ) 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 {\left (-1+2 \sqrt {5}\right ) y^{\prime }}{x}-\frac {\left (12 x^{2}-19\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 )=-\frac {-1+2 \sqrt {5}}{x}, P_{3}\left (x \right )=-\frac {12 x^{2}-19}{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}}}=1-2 \sqrt {5} \\ {} & \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 {19}{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 }-4 \left (-1+2 \sqrt {5}\right ) x y^{\prime }+\left (-12 x^{2}+19\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..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & \left (1+2 \sqrt {5}-2 r \right ) \left (-1+2 \sqrt {5}-2 r \right ) a_{0} x^{r}+\left (-1+2 \sqrt {5}-2 r \right ) \left (-3+2 \sqrt {5}-2 r \right ) a_{1} x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (\left (-2 k +1+2 \sqrt {5}-2 r \right ) \left (-2 k -1+2 \sqrt {5}-2 r \right ) a_{k}-12 a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+2 \sqrt {5}-2 r \right ) \left (-1+2 \sqrt {5}-2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}+\sqrt {5}, \frac {1}{2}+\sqrt {5}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (-1+2 \sqrt {5}-2 r \right ) \left (-3+2 \sqrt {5}-2 r \right ) a_{1}=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -8 a_{k} \left (k +r \right ) \sqrt {5}+\left (4 k^{2}+8 k r +4 r^{2}+19\right ) a_{k}-12 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -8 a_{k +2} \left (k +2+r \right ) \sqrt {5}+\left (4 \left (k +2\right )^{2}+8 \left (k +2\right ) r +4 r^{2}+19\right ) a_{k +2}-12 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {12 a_{k}}{-35+8 k \sqrt {5}+8 \sqrt {5}\, r -4 k^{2}-8 k r -4 r^{2}+16 \sqrt {5}-16 k -16 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}+\sqrt {5} \\ {} & {} & a_{k +2}=-\frac {12 a_{k}}{-27+8 k \sqrt {5}+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 k^{2}-8 k \left (-\frac {1}{2}+\sqrt {5}\right )-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-16 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}+\sqrt {5} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}+\sqrt {5}}, a_{k +2}=-\frac {12 a_{k}}{-27+8 k \sqrt {5}+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 k^{2}-8 k \left (-\frac {1}{2}+\sqrt {5}\right )-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-16 k}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\sqrt {5} \\ {} & {} & a_{k +2}=-\frac {12 a_{k}}{-43+8 k \sqrt {5}+8 \left (\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 k^{2}-8 k \left (\frac {1}{2}+\sqrt {5}\right )-4 \left (\frac {1}{2}+\sqrt {5}\right )^{2}-16 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\sqrt {5} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\sqrt {5}}, a_{k +2}=-\frac {12 a_{k}}{-43+8 k \sqrt {5}+8 \left (\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 k^{2}-8 k \left (\frac {1}{2}+\sqrt {5}\right )-4 \left (\frac {1}{2}+\sqrt {5}\right )^{2}-16 k}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}+\sqrt {5}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}+\sqrt {5}}\right ), a_{2+k}=-\frac {12 a_{k}}{-27+8 k \sqrt {5}+8 \left (-\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 k^{2}-8 k \left (-\frac {1}{2}+\sqrt {5}\right )-4 \left (-\frac {1}{2}+\sqrt {5}\right )^{2}-16 k}, a_{1}=0, b_{2+k}=-\frac {12 b_{k}}{-43+8 k \sqrt {5}+8 \left (\frac {1}{2}+\sqrt {5}\right ) \sqrt {5}-4 k^{2}-8 k \left (\frac {1}{2}+\sqrt {5}\right )-4 \left (\frac {1}{2}+\sqrt {5}\right )^{2}-16 k}, b_{1}=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.046 (sec). Leaf size: 325
Order:=6; dsolve(x^2*diff(y(x),x$2)-(2*sqrt(5)-1)*x*diff(y(x),x)+(19/4-3*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{-\frac {1}{2}+\sqrt {5}} \left (\left (1+\frac {3}{2} x^{2}+\frac {3}{8} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) c_{1} +c_{2} x \left (\left (1+\frac {1}{2} x^{2}+\frac {3}{40} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \ln \left (x \right )+\left (-\frac {5}{12} x^{2}-\frac {77}{800} x^{4}+\operatorname {O}\left (x^{6}\right )\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.054 (sec). Leaf size: 94
AsymptoticDSolveValue[x^2*y''[x]-(2*Sqrt[5]-1)*x*y'[x]+(19/4-3*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {3}{8} x^{\frac {7}{2}+\sqrt {5}}+\frac {3}{2} x^{\frac {3}{2}+\sqrt {5}}+x^{\sqrt {5}-\frac {1}{2}}\right )+c_2 \left (\frac {3}{40} x^{\frac {9}{2}+\sqrt {5}}+\frac {1}{2} x^{\frac {5}{2}+\sqrt {5}}+x^{\frac {1}{2}+\sqrt {5}}\right ) \]