Internal problem ID [1411]
Internal file name [OUTPUT/1412_Sunday_June_05_2022_02_15_42_AM_81537289/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS II. Exercises 7.6. Page 374
Problem number: 64.
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]]
\[ \boxed {x^{2} \left (1-x \right )^{2} y^{\prime \prime }-x \left (-3 x^{2}+2 x +1\right ) y^{\prime }+\left (x^{2}+1\right ) y=0} \] 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. \[ \left (x^{4}-2 x^{3}+x^{2}\right ) y^{\prime \prime }+\left (3 x^{3}-2 x^{2}-x \right ) y^{\prime }+\left (x^{2}+1\right ) y = 0 \] 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 {3 x +1}{\left (x -1\right ) x}\\ q(x) &= \frac {x^{2}+1}{x^{2} \left (x -1\right )^{2}}\\ \end {align*}
| \(p(x)=\frac {3 x +1}{\left (x -1\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x^{2}+1}{x^{2} \left (x -1\right )^{2}}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}-2 x +1\right ) y^{\prime \prime }+\left (3 x^{3}-2 x^{2}-x \right ) y^{\prime }+\left (x^{2}+1\right ) y = 0 \] 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 (x^{2}-2 x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (3 x^{3}-2 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x^{2}+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 +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\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 }}3 x^{n +r +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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 +2} 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^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\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 =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 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 =2}{\sum }}a_{n -2} 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 \[ x^{r} \left (-1+r \right )^{2} = 0 \] 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 \[ x^{r} \left (-1+r \right )^{2} = 0 \] 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 \[ a_{1} = 2 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -2} \left (n +r -2\right )-2 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n -2}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -a_{n -2}+2 a_{n -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -a_{n -2}+2 a_{n -1}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(2\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=3 \] Which for the root \(r = 1\) becomes \[ a_{2}=3 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(2\) | \(2\) |
| \(a_{2}\) | \(3\) | \(3\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=4 \] Which for the root \(r = 1\) becomes \[ a_{3}=4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(2\) | \(2\) |
| \(a_{2}\) | \(3\) | \(3\) |
| \(a_{3}\) | \(4\) | \(4\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=5 \] Which for the root \(r = 1\) becomes \[ a_{4}=5 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(2\) | \(2\) |
| \(a_{2}\) | \(3\) | \(3\) |
| \(a_{3}\) | \(4\) | \(4\) |
| \(a_{4}\) | \(5\) | \(5\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=6 \] Which for the root \(r = 1\) becomes \[ a_{5}=6 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(2\) | \(2\) |
| \(a_{2}\) | \(3\) | \(3\) |
| \(a_{3}\) | \(4\) | \(4\) |
| \(a_{4}\) | \(5\) | \(5\) |
| \(a_{5}\) | \(6\) | \(6\) |
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 (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+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 |
| \(b_{1}\) | \(2\) | \(2\) | \(0\) | \(0\) |
| \(b_{2}\) | \(3\) | \(3\) | \(0\) | \(0\) |
| \(b_{3}\) | \(4\) | \(4\) | \(0\) | \(0\) |
| \(b_{4}\) | \(5\) | \(5\) | \(0\) | \(0\) |
| \(b_{5}\) | \(6\) | \(6\) | \(0\) | \(0\) |
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 (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x O\left (x^{6}\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 (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right ) + c_{2} \left (x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right )+c_{2} \left (x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right )+c_{2} \left (x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right )+c_{2} \left (x \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}-2 x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (3 x^{3}-2 x^{2}-x \right ) y^{\prime }+\left (x^{2}+1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (x^{2}+1\right ) y}{x^{2} \left (x^{2}-2 x +1\right )}-\frac {y^{\prime } \left (3 x +1\right )}{x \left (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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {y^{\prime } \left (3 x +1\right )}{x \left (x -1\right )}+\frac {\left (x^{2}+1\right ) y}{x^{2} \left (x^{2}-2 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 {3 x +1}{\left (x -1\right ) x}, P_{3}\left (x \right )=\frac {x^{2}+1}{x^{2} \left (x^{2}-2 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}}}=-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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x -1\right ) x^{2} \left (x^{2}-2 x +1\right )+\left (3 x +1\right ) x \left (x^{2}-2 x +1\right ) y^{\prime }+\left (x^{2}+1\right ) \left (x -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..4 \\ {} & {} & 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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..5 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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 (-1+r \right )^{2} x^{r}+\left (-r^{2} a_{1}+a_{0} \left (3 r^{2}-2 r +1\right )\right ) x^{1+r}+\left (-a_{2} \left (1+r \right )^{2}+a_{1} \left (3 r^{2}+4 r +2\right )-a_{0} \left (3 r^{2}+2 r +1\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 (3 \left (k -1\right )^{2}+6 \left (k -1\right ) r +3 r^{2}-2 k +3-2 r \right )-a_{k -2} \left (3 \left (k -2\right )^{2}+6 \left (k -2\right ) r +3 r^{2}+2 k -3+2 r \right )+a_{k -3} \left (k -2+r \right )^{2}\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 [-r^{2} a_{1}+a_{0} \left (3 r^{2}-2 r +1\right )=0, -a_{2} \left (1+r \right )^{2}+a_{1} \left (3 r^{2}+4 r +2\right )-a_{0} \left (3 r^{2}+2 r +1\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (3 r^{2}-2 r +1\right )}{r^{2}}, a_{2}=\frac {2 a_{0} \left (3 r^{4}+2 r^{3}+1\right )}{r^{2} \left (r^{2}+2 r +1\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 -3} \left (k -2+r \right )^{2}+\left (-3 k^{2}+\left (-6 r +10\right ) k -3 r^{2}+10 r -9\right ) a_{k -2}+3 \left (k^{2}+\left (2 r -\frac {8}{3}\right ) k +r^{2}-\frac {8 r}{3}+2\right ) a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & -a_{k +3} \left (k +2+r \right )^{2}+a_{k} \left (k +r +1\right )^{2}+\left (-3 \left (k +3\right )^{2}+\left (-6 r +10\right ) \left (k +3\right )-3 r^{2}+10 r -9\right ) a_{k +1}+3 \left (\left (k +3\right )^{2}+\left (2 r -\frac {8}{3}\right ) \left (k +3\right )+r^{2}-\frac {8 r}{3}+2\right ) a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-3 k^{2} a_{k +1}+3 k^{2} a_{k +2}+2 k r a_{k}-6 k r a_{k +1}+6 k r a_{k +2}+r^{2} a_{k}-3 r^{2} a_{k +1}+3 r^{2} a_{k +2}+2 k a_{k}-8 k a_{k +1}+10 k a_{k +2}+2 r a_{k}-8 r a_{k +1}+10 r a_{k +2}+a_{k}-6 a_{k +1}+9 a_{k +2}}{\left (k +2+r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-3 k^{2} a_{k +1}+3 k^{2} a_{k +2}+4 k a_{k}-14 k a_{k +1}+16 k a_{k +2}+4 a_{k}-17 a_{k +1}+22 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^{2} a_{k}-3 k^{2} a_{k +1}+3 k^{2} a_{k +2}+4 k a_{k}-14 k a_{k +1}+16 k a_{k +2}+4 a_{k}-17 a_{k +1}+22 a_{k +2}}{\left (k +3\right )^{2}}, a_{1}=2 a_{0}, a_{2}=3 a_{0}\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: 0.016 (sec). Leaf size: 45
Order:=6; dsolve(x^2*(1-x)^2*diff(y(x),x$2)-x*(1+2*x-3*x^2)*diff(y(x),x)+(1+x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (6 x^{5}+5 x^{4}+4 x^{3}+3 x^{2}+2 x +1\right ) x \left (c_{2} \ln \left (x \right )+c_{1} \right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 64
AsymptoticDSolveValue[x^2*(1-x)^2*y''[x]-x*(1+2*x-3*x^2)*y'[x]+(1+x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (6 x^5+5 x^4+4 x^3+3 x^2+2 x+1\right )+c_2 x \left (6 x^5+5 x^4+4 x^3+3 x^2+2 x+1\right ) \log (x) \]