Internal problem ID [1347]
Internal file name [OUTPUT/1348_Sunday_June_05_2022_02_12_15_AM_28508719/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. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
Problem number: 67.
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]]
\[ \boxed {3 x^{2} \left (x +1\right )^{2} y^{\prime \prime }-x \left (-11 x^{2}-10 x +1\right ) y^{\prime }+\left (5 x^{2}+1\right ) y=0} \] 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. \[ \left (3 x^{4}+6 x^{3}+3 x^{2}\right ) y^{\prime \prime }+\left (11 x^{3}+10 x^{2}-x \right ) y^{\prime }+\left (5 x^{2}+1\right ) y = 0 \] 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 {11 x -1}{3 x \left (x +1\right )}\\ q(x) &= \frac {5 x^{2}+1}{3 x^{2} \left (x +1\right )^{2}}\\ \end {align*}
| \(p(x)=\frac {11 x -1}{3 x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {5 x^{2}+1}{3 x^{2} \left (x +1\right )^{2}}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 3 x^{2} \left (x^{2}+2 x +1\right ) y^{\prime \prime }+\left (11 x^{3}+10 x^{2}-x \right ) y^{\prime }+\left (5 x^{2}+1\right ) y = 0 \] 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} 3 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 (11 x^{3}+10 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (5 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 }}3 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+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} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 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 }}5 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 }}3 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}11 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}5 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 }}3 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}11 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}10 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 }}5 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 \[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (3 x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (3 r^{2}-4 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}-4 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= {\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (3 r^{2}-4 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {2}{3}}\) 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^{1+n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{3}} \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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -2 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 3 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+6 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 a_{n} \left (n +r \right ) \left (n +r -1\right )+11 a_{n -2} \left (n +r -2\right )+10 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+5 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 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 (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -2 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 3 b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+6 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 b_{n} \left (n +r \right ) \left (n +r -1\right )+11 b_{n -2} \left (n +r -2\right )+10 b_{n -1} \left (n +r -1\right )-b_{n} \left (n +r \right )+5 b_{n -2}+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -b_{n -2}-2 b_{n -1}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{n} = -b_{n -2}-2 b_{n -1}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-2\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=3 \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{2}=3 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-2\) | \(-2\) |
| \(b_{2}\) | \(3\) | \(3\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-4 \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{3}=-4 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-2\) | \(-2\) |
| \(b_{2}\) | \(3\) | \(3\) |
| \(b_{3}\) | \(-4\) | \(-4\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=5 \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{4}=5 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-2\) | \(-2\) |
| \(b_{2}\) | \(3\) | \(3\) |
| \(b_{3}\) | \(-4\) | \(-4\) |
| \(b_{4}\) | \(5\) | \(5\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-6 \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{5}=-6 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-2\) | \(-2\) |
| \(b_{2}\) | \(3\) | \(3\) |
| \(b_{3}\) | \(-4\) | \(-4\) |
| \(b_{4}\) | \(5\) | \(5\) |
| \(b_{5}\) | \(-6\) | \(-6\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 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 ) \\ &= x^{\frac {1}{3}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+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 (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{3}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (x^{2}+2 x +1\right ) y^{\prime \prime }+\left (11 x^{3}+10 x^{2}-x \right ) y^{\prime }+\left (5 x^{2}+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 (5 x^{2}+1\right ) y}{3 x^{2} \left (x^{2}+2 x +1\right )}-\frac {y^{\prime } \left (11 x -1\right )}{3 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} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime } \left (11 x -1\right )}{3 x \left (x +1\right )}+\frac {\left (5 x^{2}+1\right ) y}{3 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 {11 x -1}{3 x \left (x +1\right )}, P_{3}\left (x \right )=\frac {5 x^{2}+1}{3 x^{2} \left (x^{2}+2 x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=4 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=2 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (x +1\right ) \left (x^{2}+2 x +1\right ) y^{\prime \prime }+\left (11 x -1\right ) x \left (x^{2}+2 x +1\right ) y^{\prime }+\left (5 x^{2}+1\right ) \left (x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (3 u^{5}-6 u^{4}+3 u^{3}\right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (11 u^{4}-23 u^{3}+12 u^{2}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (5 u^{3}-10 u^{2}+6 u \right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =3..5 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 3 a_{0} \left (2+r \right ) \left (1+r \right ) u^{1+r}+\left (3 a_{1} \left (3+r \right ) \left (2+r \right )-a_{0} \left (2+r \right ) \left (5+6 r \right )\right ) u^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (3 a_{k -1} \left (k +r +1\right ) \left (k +r \right )-a_{k -2} \left (k +r \right ) \left (6 k -7+6 r \right )+a_{k -3} \left (3 k -4+3 r \right ) \left (k -2+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 \left (2+r \right ) \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, -1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 3 a_{1} \left (3+r \right ) \left (2+r \right )-a_{0} \left (2+r \right ) \left (5+6 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {a_{0} \left (5+6 r \right )}{3 \left (3+r \right )} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 a_{k -1} \left (k +r +1\right ) \left (k +r \right )-6 a_{k -2} \left (k -\frac {7}{6}+r \right ) \left (k +r \right )+a_{k -3} \left (3 k -4+3 r \right ) \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & 3 a_{k +2} \left (k +4+r \right ) \left (k +3+r \right )-6 a_{k +1} \left (k +\frac {11}{6}+r \right ) \left (k +3+r \right )+a_{k} \left (3 k +3 r +5\right ) \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}+6 k r a_{k}-12 k r a_{k +1}+3 r^{2} a_{k}-6 r^{2} a_{k +1}+8 k a_{k}-29 k a_{k +1}+8 r a_{k}-29 r a_{k +1}+5 a_{k}-33 a_{k +1}}{3 \left (k +4+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}-4 k a_{k}-5 k a_{k +1}+a_{k}+a_{k +1}}{3 \left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -2}, a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}-4 k a_{k}-5 k a_{k +1}+a_{k}+a_{k +1}}{3 \left (k +2\right ) \left (k +1\right )}, a_{1}=-\frac {7 a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -2}, a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}-4 k a_{k}-5 k a_{k +1}+a_{k}+a_{k +1}}{3 \left (k +2\right ) \left (k +1\right )}, a_{1}=-\frac {7 a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}+2 k a_{k}-17 k a_{k +1}-10 a_{k +1}}{3 \left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -1}, a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}+2 k a_{k}-17 k a_{k +1}-10 a_{k +1}}{3 \left (k +3\right ) \left (k +2\right )}, a_{1}=-\frac {a_{0}}{6}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -1}, a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{k +1}+2 k a_{k}-17 k a_{k +1}-10 a_{k +1}}{3 \left (k +3\right ) \left (k +2\right )}, a_{1}=-\frac {a_{0}}{6}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k -1}\right ), a_{k +2}=-\frac {3 k^{2} a_{k}-6 k^{2} a_{1+k}-4 k a_{k}-5 k a_{1+k}+a_{k}+a_{1+k}}{3 \left (k +2\right ) \left (1+k \right )}, a_{1}=-\frac {7 a_{0}}{3}, b_{k +2}=-\frac {3 k^{2} b_{k}-6 k^{2} b_{1+k}+2 k b_{k}-17 k b_{1+k}-10 b_{1+k}}{3 \left (k +3\right ) \left (k +2\right )}, b_{1}=-\frac {b_{0}}{6}\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 45
Order:=6; dsolve(3*x^2*(1+x)^2*diff(y(x),x$2)-x*(1-10*x-11*x^2)*diff(y(x),x)+(1+5*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (-6 x^{5}+5 x^{4}-4 x^{3}+3 x^{2}-2 x +1\right ) \left (c_{1} x^{\frac {1}{3}}+c_{2} x \right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 66
AsymptoticDSolveValue[3*x^2*(1+x)^2*y''[x]-x*(1-10*x-11*x^2)*y'[x]+(1+5*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (-6 x^5+5 x^4-4 x^3+3 x^2-2 x+1\right )+c_2 \sqrt [3]{x} \left (-6 x^5+5 x^4-4 x^3+3 x^2-2 x+1\right ) \]