Internal problem ID [2940]
Internal file name [OUTPUT/2432_Sunday_June_05_2022_03_09_46_AM_34906402/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: (a).
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]]
\[ \boxed {x^{2} y^{\prime \prime }-\left (-x^{2}+x \right ) y^{\prime }+\left (x^{3}+1\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x^{2} y^{\prime \prime }+\left (x^{2}-x \right ) y^{\prime }+\left (x^{3}+1\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {x -1}{x}\\ q(x) &= \frac {x^{3}+1}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (x^{2}-x \right ) y^{\prime }+\left (x^{3}+1\right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x^{3}+1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +3} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} x^{n +r} \\ \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 )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-1+r \right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (-1+r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-1+r \right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 1]\).
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, \(r = 1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{1+n}\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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {1}{r} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {1}{\left (1+r \right ) r} \] For \(3\le n\) 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 )+a_{n -3}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -1}+r a_{n -1}+a_{n -3}-a_{n -1}}{n^{2}+2 n r +r^{2}-2 n -2 r +1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-n a_{n -1}-a_{n -3}}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{2}-2 r -2}{\left (1+r \right ) r \left (2+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {5}{18}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {-r^{2}-2 r -2}{\left (1+r \right ) r \left (2+r \right )^{2}}\) | \(-{\frac {5}{18}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {2 r^{3}+10 r^{2}+16 r +10}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {19}{144}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {-r^{2}-2 r -2}{\left (1+r \right ) r \left (2+r \right )^{2}}\) | \(-{\frac {5}{18}}\) |
| \(a_{4}\) | \(\frac {2 r^{3}+10 r^{2}+16 r +10}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {19}{144}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-3 r^{4}-28 r^{3}-93 r^{2}-134 r -76}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {167}{3600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {-r^{2}-2 r -2}{\left (1+r \right ) r \left (2+r \right )^{2}}\) | \(-{\frac {5}{18}}\) |
| \(a_{4}\) | \(\frac {2 r^{3}+10 r^{2}+16 r +10}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {19}{144}\) |
| \(a_{5}\) | \(\frac {-3 r^{4}-28 r^{3}-93 r^{2}-134 r -76}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {167}{3600}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= 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 ) \\ &= x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) \\ \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 \(r = 1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(-\frac {1}{r}\) | \(-1\) | \(\frac {1}{r^{2}}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{\left (1+r \right ) r}\) | \(\frac {1}{2}\) | \(\frac {-2 r -1}{\left (1+r \right )^{2} r^{2}}\) | \(-{\frac {3}{4}}\) |
| \(b_{3}\) | \(\frac {-r^{2}-2 r -2}{\left (1+r \right ) r \left (2+r \right )^{2}}\) | \(-{\frac {5}{18}}\) | \(\frac {2 r^{4}+7 r^{3}+14 r^{2}+14 r +4}{\left (1+r \right )^{2} r^{2} \left (2+r \right )^{3}}\) | \(\frac {41}{108}\) |
| \(b_{4}\) | \(\frac {2 r^{3}+10 r^{2}+16 r +10}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {19}{144}\) | \(\frac {-6 r^{6}-54 r^{5}-198 r^{4}-390 r^{3}-446 r^{2}-270 r -60}{\left (1+r \right )^{2} r^{2} \left (2+r \right )^{3} \left (r +3\right )^{3}}\) | \(-{\frac {89}{432}}\) |
| \(b_{5}\) | \(\frac {-3 r^{4}-28 r^{3}-93 r^{2}-134 r -76}{\left (1+r \right ) r \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {167}{3600}}\) | \(\frac {12 r^{8}+203 r^{7}+1453 r^{6}+5761 r^{5}+13901 r^{4}+20986 r^{3}+19276 r^{2}+9576 r +1824}{\left (1+r \right )^{2} r^{2} \left (2+r \right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(\frac {2281}{27000}\) |
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 \\ &= x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x -\frac {3 x^{2}}{4}+\frac {41 x^{3}}{108}-\frac {89 x^{4}}{432}+\frac {2281 x^{5}}{27000}+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 \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) + c_{2} \left (x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x -\frac {3 x^{2}}{4}+\frac {41 x^{3}}{108}-\frac {89 x^{4}}{432}+\frac {2281 x^{5}}{27000}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x -\frac {3 x^{2}}{4}+\frac {41 x^{3}}{108}-\frac {89 x^{4}}{432}+\frac {2281 x^{5}}{27000}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x -\frac {3 x^{2}}{4}+\frac {41 x^{3}}{108}-\frac {89 x^{4}}{432}+\frac {2281 x^{5}}{27000}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1-x +\frac {x^{2}}{2}-\frac {5 x^{3}}{18}+\frac {19 x^{4}}{144}-\frac {167 x^{5}}{3600}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x -\frac {3 x^{2}}{4}+\frac {41 x^{3}}{108}-\frac {89 x^{4}}{432}+\frac {2281 x^{5}}{27000}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+\left (x^{2}-x \right ) y^{\prime }+\left (x^{3}+1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (x^{3}+1\right ) 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 {\left (x^{3}+1\right ) 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 {x^{3}+1}{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}}}=1 \\ {} & \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 }+\left (x^{3}+1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{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 (-1+r \right )^{2} x^{r}+\left (a_{1} r^{2}+a_{0} r \right ) x^{1+r}+\left (a_{2} \left (1+r \right )^{2}+a_{1} \left (1+r \right )\right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k} \left (k +r -1\right )^{2}+a_{k -1} \left (k +r -1\right )+a_{k -3}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =1 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} r^{2}+a_{0} r =0, a_{2} \left (1+r \right )^{2}+a_{1} \left (1+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0}}{r}, a_{2}=\frac {a_{0}}{r \left (1+r \right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r -1\right )^{2}+a_{k -1} \left (k +r -1\right )+a_{k -3}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +3} \left (k +2+r \right )^{2}+a_{k +2} \left (k +2+r \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {k a_{k +2}+r a_{k +2}+a_{k}+2 a_{k +2}}{\left (k +2+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=-\frac {k a_{k +2}+a_{k}+3 a_{k +2}}{\left (k +3\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +3}=-\frac {k a_{k +2}+a_{k}+3 a_{k +2}}{\left (k +3\right )^{2}}, a_{1}=-a_{0}, a_{2}=\frac {a_{0}}{2}\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 -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) -> 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 -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 63
Order:=6; dsolve(x^2*diff(y(x),x$2)-(x-x^2)*diff(y(x),x)+(1+x^3)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\left (x -\frac {3}{4} x^{2}+\frac {41}{108} x^{3}-\frac {89}{432} x^{4}+\frac {2281}{27000} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} +\left (1-x +\frac {1}{2} x^{2}-\frac {5}{18} x^{3}+\frac {19}{144} x^{4}-\frac {167}{3600} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \left (c_{2} \ln \left (x \right )+c_{1} \right )\right ) x \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 114
AsymptoticDSolveValue[x^2*y''[x]-(x-x^2)*y'[x]+(1+x^3)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (-\frac {167 x^5}{3600}+\frac {19 x^4}{144}-\frac {5 x^3}{18}+\frac {x^2}{2}-x+1\right )+c_2 \left (x \left (\frac {2281 x^5}{27000}-\frac {89 x^4}{432}+\frac {41 x^3}{108}-\frac {3 x^2}{4}+x\right )+x \left (-\frac {167 x^5}{3600}+\frac {19 x^4}{144}-\frac {5 x^3}{18}+\frac {x^2}{2}-x+1\right ) \log (x)\right ) \]