Internal problem ID [2393]
Internal file name [OUTPUT/2393_Tuesday_February_27_2024_08_36_49_AM_10627226/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: 18.
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]]
\[ \boxed {2 x^{2} \left (1-3 x \right ) y^{\prime \prime }+5 y^{\prime } x -2 y=0} \] 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. \[ \left (-6 x^{3}+2 x^{2}\right ) y^{\prime \prime }+5 y^{\prime } x -2 y = 0 \] 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 {5}{2 x \left (3 x -1\right )}\\ q(x) &= \frac {1}{x^{2} \left (3 x -1\right )}\\ \end {align*}
| \(p(x)=-\frac {5}{2 x \left (3 x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{3}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{x^{2} \left (3 x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{3}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, {\frac {1}{3}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -2 y^{\prime \prime } x^{2} \left (3 x -1\right )+5 y^{\prime } x -2 y = 0 \] 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} -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^{2} \left (3 x -1\right )+5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 (-6 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+5 x^{n +r} a_{n} \left (n +r \right )-2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )+5 x^{r} a_{0} r -2 a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )+5 x^{r} r -2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}+3 r -2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}+3 r -2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}+3 r -2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, -2\right ]\).
Since \(r_1 - r_2 = {\frac {5}{2}}\) 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 {1}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2} \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} -6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n} \left (n +r \right )-2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{2 n^{2}+4 n r +2 r^{2}+3 n +3 r -2}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {3 a_{n -1} \left (4 n^{2}-8 n +3\right )}{4 n^{2}+10 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {3}{14}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-{\frac {3}{14}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}=-{\frac {3}{56}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-{\frac {3}{14}}\) |
| \(a_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(-{\frac {3}{56}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {45}{1232}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-{\frac {3}{14}}\) |
| \(a_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(-{\frac {3}{56}}\) |
| \(a_{3}\) | \(\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900}\) | \(-{\frac {45}{1232}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1296 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{16 r^{7}+368 r^{6}+3448 r^{5}+16904 r^{4}+46201 r^{3}+68903 r^{2}+50010 r +12600} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}=-{\frac {675}{18304}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-{\frac {3}{14}}\) |
| \(a_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(-{\frac {3}{56}}\) |
| \(a_{3}\) | \(\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900}\) | \(-{\frac {45}{1232}}\) |
| \(a_{4}\) | \(\frac {1296 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{16 r^{7}+368 r^{6}+3448 r^{5}+16904 r^{4}+46201 r^{3}+68903 r^{2}+50010 r +12600}\) | \(-{\frac {675}{18304}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {7776 \left (3+r \right ) r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{32 r^{8}+976 r^{7}+12464 r^{6}+86440 r^{5}+352658 r^{4}+854749 r^{3}+1176456 r^{2}+810495 r +198450} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {1701}{36608}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-{\frac {3}{14}}\) |
| \(a_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(-{\frac {3}{56}}\) |
| \(a_{3}\) | \(\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900}\) | \(-{\frac {45}{1232}}\) |
| \(a_{4}\) | \(\frac {1296 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{16 r^{7}+368 r^{6}+3448 r^{5}+16904 r^{4}+46201 r^{3}+68903 r^{2}+50010 r +12600}\) | \(-{\frac {675}{18304}}\) |
| \(a_{5}\) | \(\frac {7776 \left (3+r \right ) r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{32 r^{8}+976 r^{7}+12464 r^{6}+86440 r^{5}+352658 r^{4}+854749 r^{3}+1176456 r^{2}+810495 r +198450}\) | \(-{\frac {1701}{36608}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {x} \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 ) \\ &= \sqrt {x}\, \left (1-\frac {3 x}{14}-\frac {3 x^{2}}{56}-\frac {45 x^{3}}{1232}-\frac {675 x^{4}}{18304}-\frac {1701 x^{5}}{36608}+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} -6 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )+5 b_{n} \left (n +r \right )-2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {6 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{2 n^{2}+4 n r +2 r^{2}+3 n +3 r -2}\tag {4} \] Which for the root \(r = -2\) becomes \[ b_{n} = \frac {6 b_{n -1} \left (n -3\right ) \left (n -4\right )}{n \left (2 n -5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -2\) 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 \[ b_{1}=\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3} \] Which for the root \(r = -2\) becomes \[ b_{1}=-12 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-12\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36} \] Which for the root \(r = -2\) becomes \[ b_{2}=72 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-12\) |
| \(b_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(72\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900} \] Which for the root \(r = -2\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-12\) |
| \(b_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(72\) |
| \(b_{3}\) | \(\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1296 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{16 r^{7}+368 r^{6}+3448 r^{5}+16904 r^{4}+46201 r^{3}+68903 r^{2}+50010 r +12600} \] Which for the root \(r = -2\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-12\) |
| \(b_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(72\) |
| \(b_{3}\) | \(\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900}\) | \(0\) |
| \(b_{4}\) | \(\frac {1296 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{16 r^{7}+368 r^{6}+3448 r^{5}+16904 r^{4}+46201 r^{3}+68903 r^{2}+50010 r +12600}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {7776 \left (3+r \right ) r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{32 r^{8}+976 r^{7}+12464 r^{6}+86440 r^{5}+352658 r^{4}+854749 r^{3}+1176456 r^{2}+810495 r +198450} \] Which for the root \(r = -2\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {6 r \left (-1+r \right )}{2 r^{2}+7 r +3}\) | \(-12\) |
| \(b_{2}\) | \(\frac {36 r^{4}-36 r^{2}}{4 r^{4}+36 r^{3}+107 r^{2}+117 r +36}\) | \(72\) |
| \(b_{3}\) | \(\frac {216 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )}{8 r^{6}+132 r^{5}+854 r^{4}+2739 r^{3}+4502 r^{2}+3465 r +900}\) | \(0\) |
| \(b_{4}\) | \(\frac {1296 r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{16 r^{7}+368 r^{6}+3448 r^{5}+16904 r^{4}+46201 r^{3}+68903 r^{2}+50010 r +12600}\) | \(0\) |
| \(b_{5}\) | \(\frac {7776 \left (3+r \right ) r^{2} \left (-1+r \right ) \left (1+r \right )^{2} \left (2+r \right )^{2}}{32 r^{8}+976 r^{7}+12464 r^{6}+86440 r^{5}+352658 r^{4}+854749 r^{3}+1176456 r^{2}+810495 r +198450}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \sqrt {x} \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 ) \\ &= \frac {1-12 x +72 x^{2}+O\left (x^{6}\right )}{x^{2}} \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} \sqrt {x}\, \left (1-\frac {3 x}{14}-\frac {3 x^{2}}{56}-\frac {45 x^{3}}{1232}-\frac {675 x^{4}}{18304}-\frac {1701 x^{5}}{36608}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-12 x +72 x^{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {3 x}{14}-\frac {3 x^{2}}{56}-\frac {45 x^{3}}{1232}-\frac {675 x^{4}}{18304}-\frac {1701 x^{5}}{36608}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-12 x +72 x^{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {3 x}{14}-\frac {3 x^{2}}{56}-\frac {45 x^{3}}{1232}-\frac {675 x^{4}}{18304}-\frac {1701 x^{5}}{36608}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-12 x +72 x^{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {3 x}{14}-\frac {3 x^{2}}{56}-\frac {45 x^{3}}{1232}-\frac {675 x^{4}}{18304}-\frac {1701 x^{5}}{36608}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-12 x +72 x^{2}+O\left (x^{6}\right )\right )}{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -2 y^{\prime \prime } x^{2} \left (3 x -1\right )+5 y^{\prime } x -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}{x^{2} \left (3 x -1\right )}+\frac {5 y^{\prime }}{2 x \left (3 x -1\right )} \\ \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 {5 y^{\prime }}{2 x \left (3 x -1\right )}+\frac {y}{x^{2} \left (3 x -1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {5}{2 x \left (3 x -1\right )}, P_{3}\left (x \right )=\frac {1}{x^{2} \left (3 x -1\right )}\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 {5}{2} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-1 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime } x^{2} \left (3 x -1\right )-5 y^{\prime } x +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 \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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} \left (2+r \right ) \left (-1+2 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (k +r +2\right ) \left (2 k +2 r -1\right )+6 a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (2+r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 6 a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )-2 \left (k +r +2\right ) \left (k +r -\frac {1}{2}\right ) a_{k}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 6 a_{k} \left (k +r \right ) \left (k +r -1\right )-2 \left (k +3+r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {6 a_{k} \left (k +r \right ) \left (k +r -1\right )}{\left (k +3+r \right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {6 a_{k} \left (k -2\right ) \left (k -3\right )}{\left (k +1\right ) \left (2 k -3\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-12 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-6 a_{1} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=72 a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-2\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1-12 x +72 x^{2}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {6 a_{k} \left (k +\frac {1}{2}\right ) \left (k -\frac {1}{2}\right )}{\left (k +\frac {7}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=\frac {6 a_{k} \left (k +\frac {1}{2}\right ) \left (k -\frac {1}{2}\right )}{\left (k +\frac {7}{2}\right ) \left (2 k +2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1-12 x +72 x^{2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), b_{k +1}=\frac {6 b_{k} \left (k +\frac {1}{2}\right ) \left (k -\frac {1}{2}\right )}{\left (k +\frac {7}{2}\right ) \left (2 k +2\right )}\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 Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 41
Order:=6; dsolve(2*x^2*(1-3*x)*diff(y(x),x$2)+5*x*diff(y(x),x)-2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {5}{2}} \left (1-\frac {3}{14} x -\frac {3}{56} x^{2}-\frac {45}{1232} x^{3}-\frac {675}{18304} x^{4}-\frac {1701}{36608} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1-12 x +72 x^{2}+\operatorname {O}\left (x^{6}\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 63
AsymptoticDSolveValue[2*x^2*(1-3*x)*y''[x]+5*x*y'[x]-2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \frac {c_2 \left (72 x^2-12 x+1\right )}{x^2}+c_1 \sqrt {x} \left (-\frac {1701 x^5}{36608}-\frac {675 x^4}{18304}-\frac {45 x^3}{1232}-\frac {3 x^2}{56}-\frac {3 x}{14}+1\right ) \]