Internal problem ID [2964]
Internal file name [OUTPUT/2456_Sunday_June_05_2022_03_14_10_AM_15449460/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.5. page
771
Problem number: 22.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+x \left (5-x \right ) y^{\prime }+4 y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (-x^{2}+5 x \right ) y^{\prime }+4 y = 0 \] 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 -5}{x}\\ q(x) &= \frac {4}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (-x^{2}+5 x \right ) y^{\prime }+4 y = 0 \] 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}+5 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+4 \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 }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 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 }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+5 x^{n +r} a_{n} \left (n +r \right )+4 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+5 x^{r} a_{0} r +4 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+5 x^{r} r +4 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r +2\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (r +2\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -2\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r +2\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-2, -2]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = -2\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -2}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n -2}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). 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 )+5 a_{n} \left (n +r \right )+4 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r -1\right )}{n^{2}+2 n r +r^{2}+4 n +4 r +4}\tag {4} \] Which for the root \(r = -2\) becomes \[ a_{n} = \frac {a_{n -1} \left (n -3\right )}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -2\) 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 \[ a_{1}=\frac {r}{\left (r +3\right )^{2}} \] Which for the root \(r = -2\) becomes \[ a_{1}=-2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (r +3\right )^{2}}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (1+r \right )}{\left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = -2\) becomes \[ a_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (r +3\right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = -2\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (r +3\right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}} \] Which for the root \(r = -2\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (r +3\right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right ) \left (r +5\right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}} \] Which for the root \(r = -2\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{\left (r +3\right )^{2}}\) | \(-2\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}}\) | \(0\) |
| \(a_{5}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right ) \left (r +5\right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}}\) | \(0\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{x^{2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= \frac {-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )}{x^{2}} \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = -2\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-2\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {r}{\left (r +3\right )^{2}}\) | \(-2\) | \(\frac {-r +3}{\left (r +3\right )^{3}}\) | \(5\) |
| \(b_{2}\) | \(\frac {r \left (1+r \right )}{\left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{2}\) | \(\frac {-2 r^{3}-3 r^{2}+17 r +12}{\left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(-{\frac {9}{4}}\) |
| \(b_{3}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(0\) | \(\frac {-3 r^{5}-24 r^{4}-35 r^{3}+108 r^{2}+266 r +120}{\left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(\frac {1}{18}\) |
| \(b_{4}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}}\) | \(0\) | \(\frac {-4 r^{6}-54 r^{5}-228 r^{4}-180 r^{3}+874 r^{2}+1716 r +720}{\left (r +3\right )^{2} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (6+r \right )^{3}}\) | \(\frac {1}{288}\) |
| \(b_{5}\) | \(\frac {r \left (1+r \right ) \left (r +2\right )}{\left (r +3\right ) \left (r +4\right ) \left (r +5\right )^{2} \left (6+r \right )^{2} \left (7+r \right )^{2}}\) | \(0\) | \(\frac {-5 r^{7}-100 r^{6}-730 r^{5}-2190 r^{4}-1089 r^{3}+7258 r^{2}+12552 r +5040}{\left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{3} \left (6+r \right )^{3} \left (7+r \right )^{3}}\) | \(\frac {1}{3600}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+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 \\ &= \frac {\left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{2}}+\frac {5 x -\frac {9 x^{2}}{4}+\frac {x^{3}}{18}+\frac {x^{4}}{288}+\frac {x^{5}}{3600}+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 ) \\ &= \frac {c_{1} \left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right )}{x^{2}} + c_{2} \left (\frac {\left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{2}}+\frac {5 x -\frac {9 x^{2}}{4}+\frac {x^{3}}{18}+\frac {x^{4}}{288}+\frac {x^{5}}{3600}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right )}{x^{2}}+c_{2} \left (\frac {\left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{2}}+\frac {5 x -\frac {9 x^{2}}{4}+\frac {x^{3}}{18}+\frac {x^{4}}{288}+\frac {x^{5}}{3600}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right )}{x^{2}}+c_{2} \left (\frac {\left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{2}}+\frac {5 x -\frac {9 x^{2}}{4}+\frac {x^{3}}{18}+\frac {x^{4}}{288}+\frac {x^{5}}{3600}+O\left (x^{6}\right )}{x^{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right )}{x^{2}}+c_{2} \left (\frac {\left (-2 x +1+\frac {x^{2}}{2}+O\left (x^{6}\right )\right ) \ln \left (x \right )}{x^{2}}+\frac {5 x -\frac {9 x^{2}}{4}+\frac {x^{3}}{18}+\frac {x^{4}}{288}+\frac {x^{5}}{3600}+O\left (x^{6}\right )}{x^{2}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+\left (-x^{2}+5 x \right ) y^{\prime }+4 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 {4 y}{x^{2}}+\frac {\left (x -5\right ) y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {\left (x -5\right ) y^{\prime }}{x}+\frac {4 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 -5}{x}, P_{3}\left (x \right )=\frac {4}{x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=5 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=4 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }-x \left (x -5\right ) y^{\prime }+4 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 (2+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +2\right )^{2}-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} \\ {} & {} & \left (2+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-2 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +2\right )^{2}-a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +3+r \right )^{2}-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 )}{\left (k +3+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -2\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-2 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-\frac {a_{1}}{4} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{2} \\ \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 (-2 x +1+\frac {1}{2} x^{2}\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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 57
Order:=6; dsolve(x^2*diff(y(x),x$2)+x*(5-x)*diff(y(x),x)+4*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-2 x +\frac {1}{2} x^{2}+\operatorname {O}\left (x^{6}\right )\right )+\left (5 x -\frac {9}{4} x^{2}+\frac {1}{18} x^{3}+\frac {1}{288} x^{4}+\frac {1}{3600} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2}}{x^{2}} \]
✓ Solution by Mathematica
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AsymptoticDSolveValue[x^2*y''[x]+x*(5-x)*y'[x]+4*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \frac {c_1 \left (\frac {x^2}{2}-2 x+1\right )}{x^2}+c_2 \left (\frac {\left (\frac {x^2}{2}-2 x+1\right ) \log (x)}{x^2}+\frac {\frac {x^5}{3600}+\frac {x^4}{288}+\frac {x^3}{18}-\frac {9 x^2}{4}+5 x}{x^2}\right ) \]