Internal problem ID [2390]
Internal file name [OUTPUT/2390_Tuesday_February_27_2024_08_36_45_AM_75593104/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 41, page 195
Problem number: 15.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} y^{\prime \prime }-3 \left (x^{2}+x \right ) y^{\prime }+2 y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (-3 x^{2}-3 x \right ) y^{\prime }+2 y = 0 \] 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 {3 \left (x +1\right )}{4 x}\\ q(x) &= \frac {1}{2 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 \[ 4 x^{2} y^{\prime \prime }+\left (-3 x^{2}-3 x \right ) y^{\prime }+2 y = 0 \] 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 (-3 x^{2}-3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \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 (-3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 (-3 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-3 x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-3 x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-7 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-7 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {7}{8}+\frac {\sqrt {17}}{8}\\ r_2 &= \frac {7}{8}-\frac {\sqrt {17}}{8} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-7 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {7}{8}+\frac {\sqrt {17}}{8}, \frac {7}{8}-\frac {\sqrt {17}}{8}\right ]\).
Since \(r_1 - r_2 = \frac {\sqrt {17}}{4}\) is not 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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {7}{8}+\frac {\sqrt {17}}{8}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {7}{8}-\frac {\sqrt {17}}{8}} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 )-3 a_{n -1} \left (n +r -1\right )-3 a_{n} \left (n +r \right )+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {3 a_{n -1} \left (n +r -1\right )}{4 n^{2}+8 n r +4 r^{2}-7 n -7 r +2}\tag {4} \] Which for the root \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) becomes \[ a_{n} = \frac {3 a_{n -1} \left (8 n -1+\sqrt {17}\right )}{8 n \left (\sqrt {17}+4 n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) 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 \[ a_{1}=\frac {3 r}{4 r^{2}+r -1} \] Which for the root \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) becomes \[ a_{1}=\frac {21+3 \sqrt {17}}{32+8 \sqrt {17}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {21+3 \sqrt {17}}{32+8 \sqrt {17}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4} \] Which for the root \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) becomes \[ a_{2}=\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right )}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {21+3 \sqrt {17}}{32+8 \sqrt {17}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right )}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68} \] Which for the root \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) becomes \[ a_{3}=\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right )}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {21+3 \sqrt {17}}{32+8 \sqrt {17}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right )}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}\) |
| \(a_{3}\) | \(\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right )}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584} \] Which for the root \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) becomes \[ a_{4}=\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right )}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {21+3 \sqrt {17}}{32+8 \sqrt {17}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right )}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}\) |
| \(a_{3}\) | \(\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right )}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}\) |
| \(a_{4}\) | \(\frac {81 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584}\) | \(\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right )}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {243 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584\right ) \left (4 r^{2}+33 r +67\right )} \] Which for the root \(r = \frac {7}{8}+\frac {\sqrt {17}}{8}\) becomes \[ a_{5}=\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right )}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {21+3 \sqrt {17}}{32+8 \sqrt {17}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right )}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}\) |
| \(a_{3}\) | \(\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68}\) | \(\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right )}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}\) |
| \(a_{4}\) | \(\frac {81 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584}\) | \(\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right )}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}\) |
| \(a_{5}\) | \(\frac {243 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584\right ) \left (4 r^{2}+33 r +67\right )}\) | \(\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right )}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {7}{8}+\frac {\sqrt {17}}{8}} \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}{8}+\frac {\sqrt {17}}{8}} \left (1+\frac {\left (21+3 \sqrt {17}\right ) x}{32+8 \sqrt {17}}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) x^{2}}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) x^{3}}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}+\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) x^{4}}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}+\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right ) x^{5}}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now 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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )-3 b_{n -1} \left (n +r -1\right )-3 b_{n} \left (n +r \right )+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {3 b_{n -1} \left (n +r -1\right )}{4 n^{2}+8 n r +4 r^{2}-7 n -7 r +2}\tag {4} \] Which for the root \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) becomes \[ b_{n} = \frac {3 b_{n -1} \left (-8 n +1+\sqrt {17}\right )}{8 n \left (\sqrt {17}-4 n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {3 r}{4 r^{2}+r -1} \] Which for the root \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) becomes \[ b_{1}=\frac {-21+3 \sqrt {17}}{-32+8 \sqrt {17}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {-21+3 \sqrt {17}}{-32+8 \sqrt {17}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4} \] Which for the root \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) becomes \[ b_{2}=\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right )}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {-21+3 \sqrt {17}}{-32+8 \sqrt {17}}\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right )}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68} \] Which for the root \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) becomes \[ b_{3}=-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right )}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {-21+3 \sqrt {17}}{-32+8 \sqrt {17}}\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right )}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}\) |
| \(b_{3}\) | \(\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68}\) | \(-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right )}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {81 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584} \] Which for the root \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) becomes \[ b_{4}=\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right )}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {-21+3 \sqrt {17}}{-32+8 \sqrt {17}}\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right )}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}\) |
| \(b_{3}\) | \(\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68}\) | \(-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right )}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}\) |
| \(b_{4}\) | \(\frac {81 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584}\) | \(\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right )}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {243 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584\right ) \left (4 r^{2}+33 r +67\right )} \] Which for the root \(r = \frac {7}{8}-\frac {\sqrt {17}}{8}\) becomes \[ b_{5}=-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right )}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3 r}{4 r^{2}+r -1}\) | \(\frac {-21+3 \sqrt {17}}{-32+8 \sqrt {17}}\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{16 r^{4}+40 r^{3}+21 r^{2}-5 r -4}\) | \(\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right )}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}\) |
| \(b_{3}\) | \(\frac {27 r \left (1+r \right ) \left (2+r \right )}{64 r^{6}+432 r^{5}+1036 r^{4}+1017 r^{3}+256 r^{2}-153 r -68}\) | \(-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right )}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}\) |
| \(b_{4}\) | \(\frac {81 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584}\) | \(\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right )}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}\) |
| \(b_{5}\) | \(\frac {243 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (256 r^{8}+3328 r^{7}+17376 r^{6}+46384 r^{5}+65817 r^{4}+44434 r^{3}+5631 r^{2}-7514 r -2584\right ) \left (4 r^{2}+33 r +67\right )}\) | \(-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right )}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {7}{8}+\frac {\sqrt {17}}{8}} \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 {7}{8}-\frac {\sqrt {17}}{8}} \left (1+\frac {\left (-21+3 \sqrt {17}\right ) x}{-32+8 \sqrt {17}}+\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) x^{2}}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) x^{3}}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}+\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) x^{4}}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right ) x^{5}}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )}+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 {7}{8}+\frac {\sqrt {17}}{8}} \left (1+\frac {\left (21+3 \sqrt {17}\right ) x}{32+8 \sqrt {17}}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) x^{2}}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) x^{3}}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}+\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) x^{4}}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}+\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right ) x^{5}}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {7}{8}-\frac {\sqrt {17}}{8}} \left (1+\frac {\left (-21+3 \sqrt {17}\right ) x}{-32+8 \sqrt {17}}+\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) x^{2}}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) x^{3}}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}+\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) x^{4}}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right ) x^{5}}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {7}{8}+\frac {\sqrt {17}}{8}} \left (1+\frac {\left (21+3 \sqrt {17}\right ) x}{32+8 \sqrt {17}}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) x^{2}}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) x^{3}}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}+\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) x^{4}}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}+\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right ) x^{5}}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {7}{8}-\frac {\sqrt {17}}{8}} \left (1+\frac {\left (-21+3 \sqrt {17}\right ) x}{-32+8 \sqrt {17}}+\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) x^{2}}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) x^{3}}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}+\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) x^{4}}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right ) x^{5}}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {7}{8}+\frac {\sqrt {17}}{8}} \left (1+\frac {\left (21+3 \sqrt {17}\right ) x}{32+8 \sqrt {17}}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) x^{2}}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) x^{3}}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}+\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) x^{4}}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}+\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right ) x^{5}}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {7}{8}-\frac {\sqrt {17}}{8}} \left (1+\frac {\left (-21+3 \sqrt {17}\right ) x}{-32+8 \sqrt {17}}+\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) x^{2}}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) x^{3}}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}+\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) x^{4}}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right ) x^{5}}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {7}{8}+\frac {\sqrt {17}}{8}} \left (1+\frac {\left (21+3 \sqrt {17}\right ) x}{32+8 \sqrt {17}}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) x^{2}}{128 \left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )}+\frac {9 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) x^{3}}{1024 \left (49+12 \sqrt {17}\right ) \left (12+\sqrt {17}\right )}+\frac {27 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) x^{4}}{32768 \left (792+193 \sqrt {17}\right ) \left (16+\sqrt {17}\right )}+\frac {81 \left (7+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (39+\sqrt {17}\right ) x^{5}}{1310720 \left (15953+3880 \sqrt {17}\right ) \left (20+\sqrt {17}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {7}{8}-\frac {\sqrt {17}}{8}} \left (1+\frac {\left (-21+3 \sqrt {17}\right ) x}{-32+8 \sqrt {17}}+\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) x^{2}}{128 \left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )}-\frac {9 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) x^{3}}{1024 \left (-49+12 \sqrt {17}\right ) \left (-12+\sqrt {17}\right )}+\frac {27 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) x^{4}}{32768 \left (-792+193 \sqrt {17}\right ) \left (-16+\sqrt {17}\right )}-\frac {81 \left (-7+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-39+\sqrt {17}\right ) x^{5}}{1310720 \left (-15953+3880 \sqrt {17}\right ) \left (-20+\sqrt {17}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }+\left (-3 x^{2}-3 x \right ) y^{\prime }+2 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 {y}{2 x^{2}}+\frac {3 \left (x +1\right ) y^{\prime }}{4 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 {3 \left (x +1\right ) y^{\prime }}{4 x}+\frac {y}{2 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 {3 \left (x +1\right )}{4 x}, P_{3}\left (x \right )=\frac {1}{2 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 {3}{4} \\ {} & \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 {1}{2} \\ {} & \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 }-3 x \left (x +1\right ) y^{\prime }+2 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 (4 r^{2}-7 r +2\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (4 k^{2}+8 k r +4 r^{2}-7 k -7 r +2\right )-3 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} \\ {} & {} & 4 r^{2}-7 r +2=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {7}{8}-\frac {\sqrt {17}}{8}, \frac {7}{8}+\frac {\sqrt {17}}{8}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (4 k^{2}+\left (8 r -7\right ) k +4 r^{2}-7 r +2\right ) a_{k}-3 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (4 \left (k +1\right )^{2}+\left (8 r -7\right ) \left (k +1\right )+4 r^{2}-7 r +2\right ) a_{k +1}-3 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {3 a_{k} \left (k +r \right )}{4 k^{2}+8 k r +4 r^{2}+k +r -1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{8}-\frac {\sqrt {17}}{8} \\ {} & {} & a_{k +1}=\frac {3 a_{k} \left (k +\frac {7}{8}-\frac {\sqrt {17}}{8}\right )}{4 k^{2}+8 k \left (\frac {7}{8}-\frac {\sqrt {17}}{8}\right )+4 \left (\frac {7}{8}-\frac {\sqrt {17}}{8}\right )^{2}+k -\frac {1}{8}-\frac {\sqrt {17}}{8}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{8}-\frac {\sqrt {17}}{8} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {7}{8}-\frac {\sqrt {17}}{8}}, a_{k +1}=\frac {3 a_{k} \left (k +\frac {7}{8}-\frac {\sqrt {17}}{8}\right )}{4 k^{2}+8 k \left (\frac {7}{8}-\frac {\sqrt {17}}{8}\right )+4 \left (\frac {7}{8}-\frac {\sqrt {17}}{8}\right )^{2}+k -\frac {1}{8}-\frac {\sqrt {17}}{8}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {7}{8}+\frac {\sqrt {17}}{8} \\ {} & {} & a_{k +1}=\frac {3 a_{k} \left (k +\frac {7}{8}+\frac {\sqrt {17}}{8}\right )}{4 k^{2}+8 k \left (\frac {7}{8}+\frac {\sqrt {17}}{8}\right )+4 \left (\frac {7}{8}+\frac {\sqrt {17}}{8}\right )^{2}+k -\frac {1}{8}+\frac {\sqrt {17}}{8}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {7}{8}+\frac {\sqrt {17}}{8} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {7}{8}+\frac {\sqrt {17}}{8}}, a_{k +1}=\frac {3 a_{k} \left (k +\frac {7}{8}+\frac {\sqrt {17}}{8}\right )}{4 k^{2}+8 k \left (\frac {7}{8}+\frac {\sqrt {17}}{8}\right )+4 \left (\frac {7}{8}+\frac {\sqrt {17}}{8}\right )^{2}+k -\frac {1}{8}+\frac {\sqrt {17}}{8}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {7}{8}-\frac {\sqrt {17}}{8}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {7}{8}+\frac {\sqrt {17}}{8}}\right ), a_{k +1}=\frac {3 a_{k} \left (k +\frac {7}{8}-\frac {\sqrt {17}}{8}\right )}{4 k^{2}+8 k \left (\frac {7}{8}-\frac {\sqrt {17}}{8}\right )+4 \left (\frac {7}{8}-\frac {\sqrt {17}}{8}\right )^{2}+k -\frac {1}{8}-\frac {\sqrt {17}}{8}}, b_{k +1}=\frac {3 b_{k} \left (k +\frac {7}{8}+\frac {\sqrt {17}}{8}\right )}{4 k^{2}+8 k \left (\frac {7}{8}+\frac {\sqrt {17}}{8}\right )+4 \left (\frac {7}{8}+\frac {\sqrt {17}}{8}\right )^{2}+k -\frac {1}{8}+\frac {\sqrt {17}}{8}}\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: 0.015 (sec). Leaf size: 879
Order:=6; dsolve(4*x^2*diff(y(x),x$2)-3*(x+x^2)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x^{\frac {7}{8}} \left (c_{2} x^{\frac {\sqrt {17}}{8}} \left (1+\frac {21+3 \sqrt {17}}{8 \sqrt {17}+32} x +\frac {9}{128} \frac {\left (15+\sqrt {17}\right ) \left (7+\sqrt {17}\right )}{\left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right )} x^{2}+\frac {9}{1024} \frac {\left (23+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (7+\sqrt {17}\right )}{\left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right ) \left (12+\sqrt {17}\right )} x^{3}+\frac {27}{32768} \frac {\left (31+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (7+\sqrt {17}\right )}{\left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right ) \left (12+\sqrt {17}\right ) \left (16+\sqrt {17}\right )} x^{4}+\frac {81}{1310720} \frac {\left (39+\sqrt {17}\right ) \left (31+\sqrt {17}\right ) \left (23+\sqrt {17}\right ) \left (15+\sqrt {17}\right ) \left (7+\sqrt {17}\right )}{\left (4+\sqrt {17}\right ) \left (8+\sqrt {17}\right ) \left (12+\sqrt {17}\right ) \left (16+\sqrt {17}\right ) \left (20+\sqrt {17}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} x^{-\frac {\sqrt {17}}{8}} \left (1+\frac {-21+3 \sqrt {17}}{8 \sqrt {17}-32} x +\frac {9}{128} \frac {\left (-15+\sqrt {17}\right ) \left (-7+\sqrt {17}\right )}{\left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right )} x^{2}+\frac {9}{1024} \frac {\left (-23+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-7+\sqrt {17}\right )}{\left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right ) \left (-12+\sqrt {17}\right )} x^{3}+\frac {27}{32768} \frac {\left (-31+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-7+\sqrt {17}\right )}{\left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right ) \left (-12+\sqrt {17}\right ) \left (-16+\sqrt {17}\right )} x^{4}+\frac {81}{1310720} \frac {\left (-39+\sqrt {17}\right ) \left (-31+\sqrt {17}\right ) \left (-23+\sqrt {17}\right ) \left (-15+\sqrt {17}\right ) \left (-7+\sqrt {17}\right )}{\left (-4+\sqrt {17}\right ) \left (-8+\sqrt {17}\right ) \left (-12+\sqrt {17}\right ) \left (-16+\sqrt {17}\right ) \left (-20+\sqrt {17}\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 2028
AsymptoticDSolveValue[4*x^2*y''[x]-3*(x+x^2)*y'[x]+2*y[x]==0,y[x],{x,0,5}]
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