Internal problem ID [1245]
Internal file name [OUTPUT/1246_Sunday_June_05_2022_02_06_21_AM_14952766/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. 7.3 SERIES SOLUTIONS NEAR
AN ORDINARY POINT II. Exercises 7.3. Page 338
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Ordinary point", "second order series method. Taylor series method"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (3 x^{2}+x +1\right ) y^{\prime \prime }+\left (2+15 x \right ) y^{\prime }+12 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
With the expansion point for the power series method at \(x = 0\).
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=\frac {2+15 x}{3 x^{2}+x +1}\\ q(x) &=\frac {12}{3 x^{2}+x +1}\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+\frac {\left (2+15 x \right ) y^{\prime }}{3 x^{2}+x +1}+\frac {12 y}{3 x^{2}+x +1} = 0 \end {align*}
The domain of \(p(x)=\frac {2+15 x}{3 x^{2}+x +1}\) is \[
\{-\infty Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let \[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \] Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change
of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let
initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives\begin {align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \end {align*}
But \begin {align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end {align}
And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as \begin {align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6} \end {align}
Therefore (6) can be used from now on along with \begin {equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7} \end {equation} To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence
\begin {align*} F_0 &= -\frac {15 y^{\prime } x +2 y^{\prime }+12 y}{3 x^{2}+x +1}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_{0}}{\partial x}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= \frac {\left (234 x^{2}+60 x -21\right ) y^{\prime }+\left (252 x +36\right ) y}{\left (3 x^{2}+x +1\right )^{2}}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_{1}}{\partial x}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= \frac {\left (-4158 x^{3}-1548 x^{2}+1143 x +180\right ) y^{\prime }-5076 y \left (x^{2}+\frac {13}{47} x -\frac {4}{47}\right )}{\left (3 x^{2}+x +1\right )^{3}}\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_{2}}{\partial x}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \frac {\left (84564 x^{4}+40824 x^{3}-47304 x^{2}-14580 x +675\right ) y^{\prime }+110808 y \left (x^{3}+\frac {23}{57} x^{2}-\frac {89}{342} x -\frac {5}{114}\right )}{\left (3 x^{2}+x +1\right )^{4}}\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_{3}}{\partial x}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= \frac {\left (-1950804 x^{5}-1148904 x^{4}+1845828 x^{3}+837864 x^{2}-81729 x -23490\right ) y^{\prime }-2676888 y \left (x^{4}+\frac {241}{459} x^{3}-\frac {1457}{2754} x^{2}-\frac {481}{2754} x +\frac {1}{153}\right )}{\left (3 x^{2}+x +1\right )^{5}} \end {align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = 0\) and \(y^{\prime }\left (0\right ) = 1\) gives \begin {align*} F_0 &= -2\\ F_1 &= -21\\ F_2 &= 180\\ F_3 &= 675\\ F_4 &= -23490 \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[
y = -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}-\frac {261 x^{6}}{8}+O\left (x^{6}\right )
\] \[
y = -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}-\frac {261 x^{6}}{8}+O\left (x^{6}\right )
\] Since the expansion
point \(x = 0\) is an ordinary, we can also solve this using standard power series The ode is
normalized to be \[ \left (3 x^{2}+x +1\right ) y^{\prime \prime }+\left (2+15 x \right ) y^{\prime }+12 y = 0 \] Let the solution be represented as power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} \] Then
\begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} \end {align*}
Substituting the above back into the ode gives \begin {align*} \left (3 x^{2}+x +1\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (2+15 x \right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+12 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0\tag {1} \end {align*}
Which simplifies to \begin{equation}
\tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}3 x^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \,x^{n -1} a_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}15 n a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 a_{n} x^{n}\right ) = 0
\end{equation} The next step is to make all powers of \(x\) be \(n\) in each summation term.
Going over each summation term above with power of \(x\) in it which is not already \(x^{n}\) and
adjusting the power and the corresponding index gives \begin{align*}
\moverset {\infty }{\munderset {n =2}{\sum }}n \,x^{n -1} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n +1\right ) a_{n +1} n \,x^{n} \\
\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n} \\
\moverset {\infty }{\munderset {n =1}{\sum }}2 n a_{n} x^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +1\right ) a_{n +1} x^{n} \\
\end{align*} Substituting all the above in Eq (2)
gives the following equation where now all powers of \(x\) are the same and equal to \(n\). \begin{equation}
\tag{3} \left (\moverset {\infty }{\munderset {n =2}{\sum }}3 x^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n +1\right ) a_{n +1} n \,x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +1\right ) a_{n +1} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}15 n a_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 a_{n} x^{n}\right ) = 0
\end{equation} \(n=0\) gives \[
2 a_{2}+2 a_{1}+12 a_{0}=0
\] \[
a_{2} = -6 a_{0}-a_{1}
\] \(n=1\)
gives \[
6 a_{2}+6 a_{3}+27 a_{1}=0
\] Which after substituting earlier equations, simplifies to \[
a_{3} = 6 a_{0}-\frac {7 a_{1}}{2}
\] For \(2\le n\), the recurrence equation
is \begin{equation}
\tag{4} 3 n a_{n} \left (n -1\right )+\left (n +1\right ) a_{n +1} n +\left (n +2\right ) a_{n +2} \left (n +1\right )+2 \left (n +1\right ) a_{n +1}+15 n a_{n}+12 a_{n} = 0
\end{equation} Solving for \(a_{n +2}\), gives \begin{align*}
\tag{5} a_{n +2}&= -\frac {3 n a_{n}+n a_{n +1}+6 a_{n}+a_{n +1}}{n +1} \\
&= -\frac {\left (3 n +6\right ) a_{n}}{n +1}-a_{n +1} \\
\end{align*} For \(n = 2\) the recurrence equation gives \[
48 a_{2}+12 a_{3}+12 a_{4} = 0
\] Which after substituting the earlier
terms found becomes \[
a_{4} = 18 a_{0}+\frac {15 a_{1}}{2}
\] For \(n = 3\) the recurrence equation gives \[
75 a_{3}+20 a_{4}+20 a_{5} = 0
\] Which after substituting the earlier
terms found becomes \[
a_{5} = -\frac {81 a_{0}}{2}+\frac {45 a_{1}}{8}
\] For \(n = 4\) the recurrence equation gives \[
108 a_{4}+30 a_{5}+30 a_{6} = 0
\] Which after substituting
the earlier terms found becomes \[
a_{6} = -\frac {243 a_{0}}{10}-\frac {261 a_{1}}{8}
\] For \(n = 5\) the recurrence equation gives \[
147 a_{5}+42 a_{6}+42 a_{7} = 0
\] Which after
substituting the earlier terms found becomes \[
a_{7} = \frac {3321 a_{0}}{20}+\frac {207 a_{1}}{16}
\] And so on. Therefore the solution is
\begin {align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ &= a_{3} x^{3}+a_{2} x^{2}+a_{1} x +a_{0} + \dots \end {align*}
Substituting the values for \(a_{n}\) found above, the solution becomes \[
y = a_{0}+a_{1} x +\left (-6 a_{0}-a_{1}\right ) x^{2}+\left (6 a_{0}-\frac {7 a_{1}}{2}\right ) x^{3}+\left (18 a_{0}+\frac {15 a_{1}}{2}\right ) x^{4}+\left (-\frac {81 a_{0}}{2}+\frac {45 a_{1}}{8}\right ) x^{5}+\dots
\]
Collecting terms, the solution becomes \begin{equation}
\tag{3} y = \left (1-6 x^{2}+6 x^{3}+18 x^{4}-\frac {81}{2} x^{5}\right ) a_{0}+\left (-x^{2}+x -\frac {7}{2} x^{3}+\frac {15}{2} x^{4}+\frac {45}{8} x^{5}\right ) a_{1}+O\left (x^{6}\right )
\end{equation} At \(x = 0\) the solution above becomes \[
y = \left (1-6 x^{2}+6 x^{3}+18 x^{4}-\frac {81}{2} x^{5}\right ) c_{1} +\left (-x^{2}+x -\frac {7}{2} x^{3}+\frac {15}{2} x^{4}+\frac {45}{8} x^{5}\right ) c_{2} +O\left (x^{6}\right )
\] \[
y = -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}+O\left (x^{6}\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}-\frac {261 x^{6}}{8}+O\left (x^{6}\right ) \\
\tag{2} y &= -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}+O\left (x^{6}\right ) \\
\end{align*} Verification of solutions
\[
y = -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}-\frac {261 x^{6}}{8}+O\left (x^{6}\right )
\] Verified OK.
\[
y = -x^{2}+x -\frac {7 x^{3}}{2}+\frac {15 x^{4}}{2}+\frac {45 x^{5}}{8}+O\left (x^{6}\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (3 x^{2}+x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2+15 x \right ) y^{\prime }+12 y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\right ] \\ \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 {12 y}{3 x^{2}+x +1}-\frac {\left (2+15 x \right ) y^{\prime }}{3 x^{2}+x +1} \\ \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 {\left (2+15 x \right ) y^{\prime }}{3 x^{2}+x +1}+\frac {12 y}{3 x^{2}+x +1}=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+15 x}{3 x^{2}+x +1}, P_{3}\left (x \right )=\frac {12}{3 x^{2}+x +1}\right ] \\ {} & \circ & \left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6} \\ {} & {} & \left (\left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6}}}}=0 \\ {} & \circ & \left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6} \\ {} & {} & \left (\left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6}}}}=0 \\ {} & \circ & x =-\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6}\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}=-\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (3 x^{2}+x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2+15 x \right ) y^{\prime }+12 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {\mathrm {I} \sqrt {11}}{6}-\frac {1}{6}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (3 u^{2}-\mathrm {I} u \sqrt {11}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-\frac {1}{2}+15 u -\frac {5 \,\mathrm {I} \sqrt {11}}{2}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+12 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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}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}{d u}\frac {d}{d u}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} \\ {} & {} & \frac {\mathrm {I} \sqrt {11}\, \left (\mathrm {I} \sqrt {11}-33-22 r \right ) r a_{0} u^{-1+r}}{22}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (\frac {\mathrm {I} \sqrt {11}\, \left (\mathrm {I} \sqrt {11}-22 k -55-22 r \right ) \left (k +1+r \right ) a_{k +1}}{22}+3 a_{k} \left (k +r +2\right )^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {\mathrm {I}}{22} \sqrt {11}\, \left (\mathrm {I} \sqrt {11}-33-22 r \right ) r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 a_{k} \left (k +r +2\right )^{2}-\left (k +1+r \right ) a_{k +1} \left (\frac {1}{2}+\mathrm {I} \left (k +r +\frac {5}{2}\right ) \sqrt {11}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {6 a_{k} \left (k^{2}+2 k r +r^{2}+4 k +4 r +4\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+4 \,\mathrm {I} k r \sqrt {11}+2 \,\mathrm {I} \sqrt {11}\, r^{2}+7 \,\mathrm {I} k \sqrt {11}+7 \,\mathrm {I} r \sqrt {11}+5 \,\mathrm {I} \sqrt {11}+k +r +1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {6 a_{k} \left (k^{2}+4 k +4\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+1+7 \,\mathrm {I} k \sqrt {11}+5 \,\mathrm {I} \sqrt {11}+k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=\frac {6 a_{k} \left (k^{2}+4 k +4\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+1+7 \,\mathrm {I} k \sqrt {11}+5 \,\mathrm {I} \sqrt {11}+k}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )^{k}, a_{k +1}=\frac {6 a_{k} \left (k^{2}+4 k +4\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+1+7 \,\mathrm {I} k \sqrt {11}+5 \,\mathrm {I} \sqrt {11}+k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22} \\ {} & {} & a_{k +1}=\frac {6 a_{k} \left (k^{2}+2 k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )+\left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+4 k -2+\frac {2 \,\mathrm {I} \sqrt {11}}{11}\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+4 \,\mathrm {I} k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+2 \,\mathrm {I} \sqrt {11}\, \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+7 \,\mathrm {I} k \sqrt {11}+7 \,\mathrm {I} \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+\frac {111 \,\mathrm {I} \sqrt {11}}{22}+k -\frac {1}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}}, a_{k +1}=\frac {6 a_{k} \left (k^{2}+2 k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )+\left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+4 k -2+\frac {2 \,\mathrm {I} \sqrt {11}}{11}\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+4 \,\mathrm {I} k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+2 \,\mathrm {I} \sqrt {11}\, \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+7 \,\mathrm {I} k \sqrt {11}+7 \,\mathrm {I} \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+\frac {111 \,\mathrm {I} \sqrt {11}}{22}+k -\frac {1}{2}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )^{k -\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}}, a_{k +1}=\frac {6 a_{k} \left (k^{2}+2 k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )+\left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+4 k -2+\frac {2 \,\mathrm {I} \sqrt {11}}{11}\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+4 \,\mathrm {I} k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+2 \,\mathrm {I} \sqrt {11}\, \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+7 \,\mathrm {I} k \sqrt {11}+7 \,\mathrm {I} \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+\frac {111 \,\mathrm {I} \sqrt {11}}{22}+k -\frac {1}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (\frac {\mathrm {I} \sqrt {11}}{6}+x +\frac {1}{6}\right )^{k -\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}}\right ), a_{k +1}=\frac {6 a_{k} \left (k^{2}+4 k +4\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+1+7 \,\mathrm {I} k \sqrt {11}+5 \,\mathrm {I} \sqrt {11}+k}, b_{k +1}=\frac {6 b_{k} \left (k^{2}+2 k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )+\left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+4 k -2+\frac {2 \,\mathrm {I} \sqrt {11}}{11}\right )}{2 \,\mathrm {I} \sqrt {11}\, k^{2}+4 \,\mathrm {I} k \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+2 \,\mathrm {I} \sqrt {11}\, \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right )^{2}+7 \,\mathrm {I} k \sqrt {11}+7 \,\mathrm {I} \left (-\frac {3}{2}+\frac {\mathrm {I} \sqrt {11}}{22}\right ) \sqrt {11}+\frac {111 \,\mathrm {I} \sqrt {11}}{22}+k -\frac {1}{2}}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 18
\[
y \left (x \right ) = x -x^{2}-\frac {7}{2} x^{3}+\frac {15}{2} x^{4}+\frac {45}{8} x^{5}+\operatorname {O}\left (x^{6}\right )
\]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 31
\[
y(x)\to \frac {45 x^5}{8}+\frac {15 x^4}{2}-\frac {7 x^3}{2}-x^2+x
\]
13.4.2 Maple step by step solution
`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
Solution has integrals. Trying a special function solution free of integrals...
-> 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
<- hyper3 successful: received ODE is equivalent to the 2F1 ODE
<- hypergeometric successful
<- special function solution successful
-> Trying to convert hypergeometric functions to elementary form...
<- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals
<- Kovacics algorithm successful`
Order:=6;
dsolve([(1+x+3*x^2)*diff(y(x),x$2)+(2+15*x)*diff(y(x),x)+12*y(x)=0,y(0) = 0, D(y)(0) = 1],y(x),type='series',x=0);
AsymptoticDSolveValue[{(1+x+3*x^2)*y''[x]+(2+15*x)*y'[x]+12*y[x]==0,{y[0]==0,y'[0]==1}},y[x],{x,0,5}]