Internal problem ID [1203]
Internal file name [OUTPUT/1204_Sunday_June_05_2022_02_05_07_AM_19121030/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. 7.1 Exercises. Page
318
Problem number: 25.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (2 x^{2}+1\right ) y^{\prime \prime }+x \left (2 x^{2}+4\right ) y^{\prime }+2 \left (-x^{2}+1\right ) y=0} \] 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. \[ \left (2 x^{4}+x^{2}\right ) y^{\prime \prime }+\left (2 x^{3}+4 x \right ) y^{\prime }+\left (-2 x^{2}+2\right ) y = 0 \] 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 {2 x^{2}+4}{x \left (2 x^{2}+1\right )}\\ q(x) &= -\frac {2 \left (x^{2}-1\right )}{x^{2} \left (2 x^{2}+1\right )}\\ \end {align*}
| \(p(x)=\frac {2 x^{2}+4}{x \left (2 x^{2}+1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {2}}{2}\) | \(\text {``regular''}\) |
| \(x = \frac {i \sqrt {2}}{2}\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {2 \left (x^{2}-1\right )}{x^{2} \left (2 x^{2}+1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {i \sqrt {2}}{2}\) | \(\text {``regular''}\) |
| \(x = \frac {i \sqrt {2}}{2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -\frac {i \sqrt {2}}{2}, \frac {i \sqrt {2}}{2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (2 x^{2}+1\right ) y^{\prime \prime }+\left (2 x^{3}+4 x \right ) y^{\prime }+\left (-2 x^{2}+2\right ) y = 0 \] 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 (2 x^{2}+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 (2 x^{3}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-2 x^{2}+2\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 }}2 x^{n +r +2} 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 }}2 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}2 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} 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 =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \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 }}2 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+4 x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+3 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+3 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -1\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+3 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-1, -2]\).
Since \(r_1 - r_2 = 1\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= \frac {\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}}{x}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{2}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -1}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -2} \left (n +r -2\right )+4 a_{n} \left (n +r \right )-2 a_{n -2}+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 a_{n -2} \left (n^{2}+2 n r +r^{2}-4 n -4 r +3\right )}{n^{2}+2 n r +r^{2}+3 n +3 r +2}\tag {4} \] Which for the root \(r = -1\) becomes \[ a_{n} = -\frac {2 a_{n -2} \left (n^{2}-6 n +8\right )}{n \left (n +1\right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-2 r^{2}+2}{r^{2}+7 r +12} \] Which for the root \(r = -1\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(0\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {4 \left (-1+r \right ) \left (1+r \right )^{2}}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right )} \] Which for the root \(r = -1\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(0\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {4 \left (-1+r \right ) \left (1+r \right )^{2}}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(0\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {4 \left (-1+r \right ) \left (1+r \right )^{2}}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right )}\) | \(0\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \frac {1}{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 ) \\ &= \frac {1+O\left (x^{6}\right )}{x} \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= 0 \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow -2}0\\ &= 0 \end {align*}
The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2} \end {align*}
Eq (3) derived above is 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(3) gives \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} 2 b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+b_{n} \left (n +r \right ) \left (n +r -1\right )+2 b_{n -2} \left (n +r -2\right )+4 b_{n} \left (n +r \right )-2 b_{n -2}+2 b_{n} = 0 \end{equation} Which for for the root \(r = -2\) becomes \begin{equation} \tag{4A} 2 b_{n -2} \left (n -4\right ) \left (n -5\right )+b_{n} \left (n -2\right ) \left (n -3\right )+2 b_{n -2} \left (n -4\right )+4 b_{n} \left (n -2\right )-2 b_{n -2}+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {2 b_{n -2} \left (n^{2}+2 n r +r^{2}-4 n -4 r +3\right )}{n^{2}+2 n r +r^{2}+3 n +3 r +2}\tag {5} \] Which for the root \(r = -2\) becomes \[ b_{n} = -\frac {2 b_{n -2} \left (n^{2}-8 n +15\right )}{n^{2}-n}\tag {6} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=-\frac {2 \left (r^{2}-1\right )}{r^{2}+7 r +12} \] Which for the root \(r = -2\) becomes \[ b_{2}=-3 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(-3\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(-3\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {4 r^{3}+4 r^{2}-4 r -4}{\left (r^{2}+11 r +30\right ) \left (r +4\right )} \] Which for the root \(r = -2\) becomes \[ b_{4}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(-3\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {4 \left (-1+r \right ) \left (1+r \right )^{2}}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right )}\) | \(-{\frac {1}{2}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {-2 r^{2}+2}{r^{2}+7 r +12}\) | \(-3\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {4 \left (-1+r \right ) \left (1+r \right )^{2}}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right )}\) | \(-{\frac {1}{2}}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \frac {1}{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 {1-3 x^{2}-\frac {x^{4}}{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 ) \\ &= \frac {c_{1} \left (1+O\left (x^{6}\right )\right )}{x} + \frac {c_{2} \left (1-3 x^{2}-\frac {x^{4}}{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1+O\left (x^{6}\right )\right )}{x}+\frac {c_{2} \left (1-3 x^{2}-\frac {x^{4}}{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1+O\left (x^{6}\right )\right )}{x}+\frac {c_{2} \left (1-3 x^{2}-\frac {x^{4}}{2}+O\left (x^{6}\right )\right )}{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1+O\left (x^{6}\right )\right )}{x}+\frac {c_{2} \left (1-3 x^{2}-\frac {x^{4}}{2}+O\left (x^{6}\right )\right )}{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (2 x^{2}+1\right ) y^{\prime \prime }+\left (2 x^{3}+4 x \right ) y^{\prime }+\left (-2 x^{2}+2\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 {2 \left (x^{2}-1\right ) y}{x^{2} \left (2 x^{2}+1\right )}-\frac {2 \left (x^{2}+2\right ) y^{\prime }}{x \left (2 x^{2}+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 {2 \left (x^{2}+2\right ) y^{\prime }}{x \left (2 x^{2}+1\right )}-\frac {2 \left (x^{2}-1\right ) y}{x^{2} \left (2 x^{2}+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 {2 \left (x^{2}+2\right )}{x \left (2 x^{2}+1\right )}, P_{3}\left (x \right )=-\frac {2 \left (x^{2}-1\right )}{x^{2} \left (2 x^{2}+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}}}=4 \\ {} & \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}}}=2 \\ {} & \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} \\ {} & {} & x^{2} \left (2 x^{2}+1\right ) y^{\prime \prime }+2 \left (x^{2}+2\right ) x y^{\prime }+\left (-2 x^{2}+2\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..2 \\ {} & {} & 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..3 \\ {} & {} & 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 y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & 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+r \right ) x^{r}+a_{1} \left (3+r \right ) \left (2+r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +2\right ) \left (k +r +1\right )+2 a_{k -2} \left (k +r -1\right ) \left (k -3+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+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} \\ {} & {} & a_{1} \left (3+r \right ) \left (2+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +2\right ) \left (k +r +1\right )+2 a_{k -2} \left (k +r -1\right ) \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +4+r \right ) \left (k +3+r \right )+2 a_{k} \left (k +r +1\right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 a_{k} \left (k +r +1\right ) \left (k +r -1\right )}{\left (k +4+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {2 a_{k} \left (k -1\right ) \left (k -3\right )}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +2}=-\frac {2 a_{k} \left (k -1\right ) \left (k -3\right )}{\left (k +2\right ) \left (k +1\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=-\frac {2 a_{k} k \left (k -2\right )}{\left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=-\frac {2 a_{k} k \left (k -2\right )}{\left (k +3\right ) \left (k +2\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -1}\right ), a_{k +2}=-\frac {2 a_{k} \left (k -1\right ) \left (k -3\right )}{\left (k +2\right ) \left (1+k \right )}, a_{1}=0, b_{k +2}=-\frac {2 b_{k} k \left (k -2\right )}{\left (k +3\right ) \left (k +2\right )}, b_{1}=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: 31
Order:=6; dsolve(x^2*(1+2*x^2)*diff(y(x),x$2)+x*(4+2*x^2)*diff(y(x),x)+2*(1-x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1+\operatorname {O}\left (x^{6}\right )\right ) x +c_{2} \left (1-3 x^{2}-\frac {1}{2} x^{4}+\operatorname {O}\left (x^{6}\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 25
AsymptoticDSolveValue[x^2*(1+2*x^2)*y''[x]+x*(4+2*x^2)*y'[x]+2*(1-x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {x^2}{2}+\frac {1}{x^2}-3\right )+\frac {c_2}{x} \]