Internal problem ID [2921]
Internal file name [OUTPUT/2413_Sunday_June_05_2022_03_07_03_AM_688095/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 11.4. page
758
Problem number: 6.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+x \left (1-x \right ) y^{\prime }-7 y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (-x^{2}+x \right ) y^{\prime }-7 y = 0 \] 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 {x -1}{x}\\ q(x) &= -\frac {7}{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 \[ x^{2} y^{\prime \prime }+\left (-x^{2}+x \right ) y^{\prime }-7 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-7 \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 (-x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 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 (-x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-7 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -7 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -7 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-7\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-7 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \sqrt {7}\\ r_2 &= -\sqrt {7} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-7\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\sqrt {7}, -\sqrt {7}\right ]\).
Since \(r_1 - r_2 = 2 \sqrt {7}\) 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 +\sqrt {7}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\sqrt {7}} \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} a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-7 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r -1\right )}{n^{2}+2 n r +r^{2}-7}\tag {4} \] Which for the root \(r = \sqrt {7}\) becomes \[ a_{n} = \frac {a_{n -1} \left (n +\sqrt {7}-1\right )}{n \left (2 \sqrt {7}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \sqrt {7}\) 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 \[ a_{1}=\frac {r}{r^{2}+2 r -6} \] Which for the root \(r = \sqrt {7}\) becomes \[ a_{1}=\frac {\sqrt {7}}{1+2 \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{1+2 \sqrt {7}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )} \] Which for the root \(r = \sqrt {7}\) becomes \[ a_{2}=\frac {\sqrt {7}}{4+8 \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{1+2 \sqrt {7}}\) |
| \(a_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{4+8 \sqrt {7}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )} \] Which for the root \(r = \sqrt {7}\) becomes \[ a_{3}=\frac {\left (2+\sqrt {7}\right ) \sqrt {7}}{372+96 \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{1+2 \sqrt {7}}\) |
| \(a_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{4+8 \sqrt {7}}\) |
| \(a_{3}\) | \(\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )}\) | \(\frac {\left (2+\sqrt {7}\right ) \sqrt {7}}{372+96 \sqrt {7}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right )} \] Which for the root \(r = \sqrt {7}\) becomes \[ a_{4}=\frac {\sqrt {7}\, \left (3+\sqrt {7}\right )}{2976+768 \sqrt {7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{1+2 \sqrt {7}}\) |
| \(a_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{4+8 \sqrt {7}}\) |
| \(a_{3}\) | \(\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )}\) | \(\frac {\left (2+\sqrt {7}\right ) \sqrt {7}}{372+96 \sqrt {7}}\) |
| \(a_{4}\) | \(\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right )}\) | \(\frac {\sqrt {7}\, \left (3+\sqrt {7}\right )}{2976+768 \sqrt {7}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right ) \left (r^{2}+10 r +18\right )} \] Which for the root \(r = \sqrt {7}\) becomes \[ a_{5}=\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right )}{48960 \sqrt {7}+128160} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{1+2 \sqrt {7}}\) |
| \(a_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{4+8 \sqrt {7}}\) |
| \(a_{3}\) | \(\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )}\) | \(\frac {\left (2+\sqrt {7}\right ) \sqrt {7}}{372+96 \sqrt {7}}\) |
| \(a_{4}\) | \(\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right )}\) | \(\frac {\sqrt {7}\, \left (3+\sqrt {7}\right )}{2976+768 \sqrt {7}}\) |
| \(a_{5}\) | \(\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right ) \left (r^{2}+10 r +18\right )}\) | \(\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right )}{48960 \sqrt {7}+128160}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\sqrt {7}} \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^{\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{4+8 \sqrt {7}}+\frac {\left (2+\sqrt {7}\right ) \sqrt {7}\, x^{3}}{372+96 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) x^{4}}{2976+768 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}+128160}+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} b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )-7 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (n +r -1\right )}{n^{2}+2 n r +r^{2}-7}\tag {4} \] Which for the root \(r = -\sqrt {7}\) becomes \[ b_{n} = \frac {b_{n -1} \left (-n +\sqrt {7}+1\right )}{n \left (2 \sqrt {7}-n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -\sqrt {7}\) 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 \[ b_{1}=\frac {r}{r^{2}+2 r -6} \] Which for the root \(r = -\sqrt {7}\) becomes \[ b_{1}=\frac {\sqrt {7}}{-1+2 \sqrt {7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{-1+2 \sqrt {7}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )} \] Which for the root \(r = -\sqrt {7}\) becomes \[ b_{2}=\frac {\sqrt {7}}{-4+8 \sqrt {7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{-1+2 \sqrt {7}}\) |
| \(b_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{-4+8 \sqrt {7}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )} \] Which for the root \(r = -\sqrt {7}\) becomes \[ b_{3}=\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right )}{372-96 \sqrt {7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{-1+2 \sqrt {7}}\) |
| \(b_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{-4+8 \sqrt {7}}\) |
| \(b_{3}\) | \(\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )}\) | \(\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right )}{372-96 \sqrt {7}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right )} \] Which for the root \(r = -\sqrt {7}\) becomes \[ b_{4}=\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right )}{2976-768 \sqrt {7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{-1+2 \sqrt {7}}\) |
| \(b_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{-4+8 \sqrt {7}}\) |
| \(b_{3}\) | \(\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )}\) | \(\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right )}{372-96 \sqrt {7}}\) |
| \(b_{4}\) | \(\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right )}\) | \(\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right )}{2976-768 \sqrt {7}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right ) \left (r^{2}+10 r +18\right )} \] Which for the root \(r = -\sqrt {7}\) becomes \[ b_{5}=\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right )}{48960 \sqrt {7}-128160} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{r^{2}+2 r -6}\) | \(\frac {\sqrt {7}}{-1+2 \sqrt {7}}\) |
| \(b_{2}\) | \(\frac {r \left (r +1\right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right )}\) | \(\frac {\sqrt {7}}{-4+8 \sqrt {7}}\) |
| \(b_{3}\) | \(\frac {r \left (r +1\right ) \left (2+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right )}\) | \(\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right )}{372-96 \sqrt {7}}\) |
| \(b_{4}\) | \(\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right )}\) | \(\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right )}{2976-768 \sqrt {7}}\) |
| \(b_{5}\) | \(\frac {r \left (r +1\right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+2 r -6\right ) \left (r^{2}+4 r -3\right ) \left (r^{2}+6 r +2\right ) \left (r^{2}+8 r +9\right ) \left (r^{2}+10 r +18\right )}\) | \(\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right )}{48960 \sqrt {7}-128160}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\sqrt {7}} \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^{-\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{-1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{-4+8 \sqrt {7}}+\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right ) x^{3}}{372-96 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) x^{4}}{2976-768 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}-128160}+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^{\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{4+8 \sqrt {7}}+\frac {\left (2+\sqrt {7}\right ) \sqrt {7}\, x^{3}}{372+96 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) x^{4}}{2976+768 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}+128160}+O\left (x^{6}\right )\right ) + c_{2} x^{-\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{-1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{-4+8 \sqrt {7}}+\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right ) x^{3}}{372-96 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) x^{4}}{2976-768 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}-128160}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{4+8 \sqrt {7}}+\frac {\left (2+\sqrt {7}\right ) \sqrt {7}\, x^{3}}{372+96 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) x^{4}}{2976+768 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}+128160}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{-1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{-4+8 \sqrt {7}}+\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right ) x^{3}}{372-96 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) x^{4}}{2976-768 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}-128160}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{4+8 \sqrt {7}}+\frac {\left (2+\sqrt {7}\right ) \sqrt {7}\, x^{3}}{372+96 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) x^{4}}{2976+768 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}+128160}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{-1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{-4+8 \sqrt {7}}+\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right ) x^{3}}{372-96 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) x^{4}}{2976-768 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}-128160}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{4+8 \sqrt {7}}+\frac {\left (2+\sqrt {7}\right ) \sqrt {7}\, x^{3}}{372+96 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) x^{4}}{2976+768 \sqrt {7}}+\frac {\sqrt {7}\, \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}+128160}+O\left (x^{6}\right )\right )+c_{2} x^{-\sqrt {7}} \left (1+\frac {\sqrt {7}\, x}{-1+2 \sqrt {7}}+\frac {\sqrt {7}\, x^{2}}{-4+8 \sqrt {7}}+\frac {\sqrt {7}\, \left (-2+\sqrt {7}\right ) x^{3}}{372-96 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) x^{4}}{2976-768 \sqrt {7}}+\frac {\sqrt {7}\, \left (-3+\sqrt {7}\right ) \left (-4+\sqrt {7}\right ) x^{5}}{48960 \sqrt {7}-128160}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+\left (-x^{2}+x \right ) y^{\prime }-7 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 {7 y}{x^{2}}+\frac {\left (x -1\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 (x -1\right ) y^{\prime }}{x}-\frac {7 y}{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 {x -1}{x}, P_{3}\left (x \right )=-\frac {7}{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 \\ {} & \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}}}=-7 \\ {} & \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} \\ {} & {} & x^{2} y^{\prime \prime }-x \left (x -1\right ) y^{\prime }-7 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 (r^{2}-7\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}-7\right )-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} \\ {} & {} & r^{2}-7=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\sqrt {7}, -\sqrt {7}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k^{2}+2 k r +r^{2}-7\right )-a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (\left (k +1\right )^{2}+2 \left (k +1\right ) r +r^{2}-7\right )-a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r \right )}{k^{2}+2 k r +r^{2}+2 k +2 r -6} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\sqrt {7} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +\sqrt {7}\right )}{k^{2}+2 k \sqrt {7}+1+2 k +2 \sqrt {7}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\sqrt {7} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\sqrt {7}}, a_{k +1}=\frac {a_{k} \left (k +\sqrt {7}\right )}{k^{2}+2 k \sqrt {7}+1+2 k +2 \sqrt {7}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\sqrt {7} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -\sqrt {7}\right )}{k^{2}-2 k \sqrt {7}+1+2 k -2 \sqrt {7}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\sqrt {7} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\sqrt {7}}, a_{k +1}=\frac {a_{k} \left (k -\sqrt {7}\right )}{k^{2}-2 k \sqrt {7}+1+2 k -2 \sqrt {7}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\sqrt {7}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\sqrt {7}}\right ), a_{1+k}=\frac {a_{k} \left (k +\sqrt {7}\right )}{k^{2}+2 k \sqrt {7}+1+2 k +2 \sqrt {7}}, b_{1+k}=\frac {b_{k} \left (k -\sqrt {7}\right )}{k^{2}-2 k \sqrt {7}+1+2 k -2 \sqrt {7}}\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 <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 478
Order:=6; dsolve(x^2*diff(y(x),x$2)+x*(1-x)*diff(y(x),x)-7*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-\sqrt {7}} \left (1+\frac {\sqrt {7}}{-1+2 \sqrt {7}} x +\frac {\sqrt {7}}{-4+8 \sqrt {7}} x^{2}+\frac {\sqrt {7}\, \left (\sqrt {7}-2\right )}{372-96 \sqrt {7}} x^{3}+\frac {\sqrt {7}\, \left (\sqrt {7}-3\right )}{2976-768 \sqrt {7}} x^{4}+\frac {\left (\sqrt {7}-4\right ) \left (\sqrt {7}-3\right ) \sqrt {7}}{48960 \sqrt {7}-128160} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{\sqrt {7}} \left (1+\frac {\sqrt {7}}{1+2 \sqrt {7}} x +\frac {\sqrt {7}}{4+8 \sqrt {7}} x^{2}+\frac {\sqrt {7}\, \left (\sqrt {7}+2\right )}{372+96 \sqrt {7}} x^{3}+\frac {\left (\sqrt {7}+3\right ) \sqrt {7}}{2976+768 \sqrt {7}} x^{4}+\frac {\left (\sqrt {7}+4\right ) \left (\sqrt {7}+3\right ) \sqrt {7}}{48960 \sqrt {7}+128160} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 1066
AsymptoticDSolveValue[x^2*y''[x]+x*(1-x)*y'[x]-7*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \left (\frac {\sqrt {7} \left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right ) \left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right ) x^5}{\left (-6+\sqrt {7}+\sqrt {7} \left (1+\sqrt {7}\right )\right ) \left (-5+\sqrt {7}+\left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right )\right ) \left (-4+\sqrt {7}+\left (2+\sqrt {7}\right ) \left (3+\sqrt {7}\right )\right ) \left (-3+\sqrt {7}+\left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right )\right ) \left (-2+\sqrt {7}+\left (4+\sqrt {7}\right ) \left (5+\sqrt {7}\right )\right )}+\frac {\sqrt {7} \left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right ) \left (3+\sqrt {7}\right ) x^4}{\left (-6+\sqrt {7}+\sqrt {7} \left (1+\sqrt {7}\right )\right ) \left (-5+\sqrt {7}+\left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right )\right ) \left (-4+\sqrt {7}+\left (2+\sqrt {7}\right ) \left (3+\sqrt {7}\right )\right ) \left (-3+\sqrt {7}+\left (3+\sqrt {7}\right ) \left (4+\sqrt {7}\right )\right )}+\frac {\sqrt {7} \left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right ) x^3}{\left (-6+\sqrt {7}+\sqrt {7} \left (1+\sqrt {7}\right )\right ) \left (-5+\sqrt {7}+\left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right )\right ) \left (-4+\sqrt {7}+\left (2+\sqrt {7}\right ) \left (3+\sqrt {7}\right )\right )}+\frac {\sqrt {7} \left (1+\sqrt {7}\right ) x^2}{\left (-6+\sqrt {7}+\sqrt {7} \left (1+\sqrt {7}\right )\right ) \left (-5+\sqrt {7}+\left (1+\sqrt {7}\right ) \left (2+\sqrt {7}\right )\right )}+\frac {\sqrt {7} x}{-6+\sqrt {7}+\sqrt {7} \left (1+\sqrt {7}\right )}+1\right ) c_1 x^{\sqrt {7}}+\left (-\frac {\sqrt {7} \left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right ) \left (3-\sqrt {7}\right ) \left (4-\sqrt {7}\right ) x^5}{\left (-6-\sqrt {7}-\sqrt {7} \left (1-\sqrt {7}\right )\right ) \left (-5-\sqrt {7}+\left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right )\right ) \left (-4-\sqrt {7}+\left (2-\sqrt {7}\right ) \left (3-\sqrt {7}\right )\right ) \left (-3-\sqrt {7}+\left (3-\sqrt {7}\right ) \left (4-\sqrt {7}\right )\right ) \left (-2-\sqrt {7}+\left (4-\sqrt {7}\right ) \left (5-\sqrt {7}\right )\right )}-\frac {\sqrt {7} \left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right ) \left (3-\sqrt {7}\right ) x^4}{\left (-6-\sqrt {7}-\sqrt {7} \left (1-\sqrt {7}\right )\right ) \left (-5-\sqrt {7}+\left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right )\right ) \left (-4-\sqrt {7}+\left (2-\sqrt {7}\right ) \left (3-\sqrt {7}\right )\right ) \left (-3-\sqrt {7}+\left (3-\sqrt {7}\right ) \left (4-\sqrt {7}\right )\right )}-\frac {\sqrt {7} \left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right ) x^3}{\left (-6-\sqrt {7}-\sqrt {7} \left (1-\sqrt {7}\right )\right ) \left (-5-\sqrt {7}+\left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right )\right ) \left (-4-\sqrt {7}+\left (2-\sqrt {7}\right ) \left (3-\sqrt {7}\right )\right )}-\frac {\sqrt {7} \left (1-\sqrt {7}\right ) x^2}{\left (-6-\sqrt {7}-\sqrt {7} \left (1-\sqrt {7}\right )\right ) \left (-5-\sqrt {7}+\left (1-\sqrt {7}\right ) \left (2-\sqrt {7}\right )\right )}-\frac {\sqrt {7} x}{-6-\sqrt {7}-\sqrt {7} \left (1-\sqrt {7}\right )}+1\right ) c_2 x^{-\sqrt {7}} \]