Internal problem ID [5672]
Internal file name [OUTPUT/4920_Sunday_June_05_2022_03_10_33_PM_38014976/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number: 14.
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
[[_Emden, _Fowler]]
\[ \boxed {16 \left (x +1\right )^{2} y^{\prime \prime }+3 y=0} \] With the expansion point for the power series method at \(x = 0\).
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 {3 y}{16 \left (x +1\right )^{2}}\\ 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 (-3 x -3\right ) y^{\prime }+6 y}{16 \left (x +1\right )^{3}}\\ 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 {-\frac {279 y}{256}+\frac {3 \left (x +1\right ) y^{\prime }}{4}}{\left (x +1\right )^{4}}\\ 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 {1080 y+\left (-855 x -855\right ) y^{\prime }}{256 \left (x +1\right )^{5}}\\ 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 {-\frac {83835 y}{4096}+\frac {1125 \left (x +1\right ) y^{\prime }}{64}}{\left (x +1\right )^{6}} \end {align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = y \left (0\right )\) and \(y^{\prime }\left (0\right ) = y^{\prime }\left (0\right )\) gives \begin {align*} F_0 &= -\frac {3 y \left (0\right )}{16}\\ F_1 &= \frac {3 y \left (0\right )}{8}-\frac {3 y^{\prime }\left (0\right )}{16}\\ F_2 &= -\frac {279 y \left (0\right )}{256}+\frac {3 y^{\prime }\left (0\right )}{4}\\ F_3 &= \frac {135 y \left (0\right )}{32}-\frac {855 y^{\prime }\left (0\right )}{256}\\ F_4 &= -\frac {83835 y \left (0\right )}{4096}+\frac {1125 y^{\prime }\left (0\right )}{64} \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[ y = \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}-\frac {1863}{65536} x^{6}\right ) y \left (0\right )+\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}+\frac {25}{1024} x^{6}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \] Since the expansion point \(x = 0\) is an ordinary, we can also solve this using standard power series The ode is normalized to be \[ \left (16 x^{2}+32 x +16\right ) y^{\prime \prime }+3 y = 0 \] 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 (16 x^{2}+32 x +16\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+3 \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 }}16 x^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}32 n \,x^{n -1} a_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}16 n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}32 n \,x^{n -1} a_{n} \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}32 \left (n +1\right ) a_{n +1} n \,x^{n} \\ \moverset {\infty }{\munderset {n =2}{\sum }}16 n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}16 \left (n +2\right ) a_{n +2} \left (n +1\right ) 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 }}16 x^{n} a_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}32 \left (n +1\right ) a_{n +1} n \,x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 \left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 32 a_{2}+3 a_{0}=0 \] \[ a_{2} = -\frac {3 a_{0}}{32} \] \(n=1\) gives \[ 64 a_{2}+96 a_{3}+3 a_{1}=0 \] Which after substituting earlier equations, simplifies to \[ a_{3} = \frac {a_{0}}{16}-\frac {a_{1}}{32} \] For \(2\le n\), the recurrence equation is \begin{equation} \tag{4} 16 n a_{n} \left (n -1\right )+32 \left (n +1\right ) a_{n +1} n +16 \left (n +2\right ) a_{n +2} \left (n +1\right )+3 a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= -\frac {16 n^{2} a_{n}+32 n^{2} a_{n +1}-16 n a_{n}+32 n a_{n +1}+3 a_{n}}{16 \left (n +2\right ) \left (n +1\right )} \\ &= -\frac {\left (16 n^{2}-16 n +3\right ) a_{n}}{16 \left (n +2\right ) \left (n +1\right )}-\frac {\left (32 n^{2}+32 n \right ) a_{n +1}}{16 \left (n +2\right ) \left (n +1\right )} \\ \end{align*} For \(n = 2\) the recurrence equation gives \[ 35 a_{2}+192 a_{3}+192 a_{4} = 0 \] Which after substituting the earlier terms found becomes \[ a_{4} = -\frac {93 a_{0}}{2048}+\frac {a_{1}}{32} \] For \(n = 3\) the recurrence equation gives \[ 99 a_{3}+384 a_{4}+320 a_{5} = 0 \] Which after substituting the earlier terms found becomes \[ a_{5} = \frac {9 a_{0}}{256}-\frac {57 a_{1}}{2048} \] For \(n = 4\) the recurrence equation gives \[ 195 a_{4}+640 a_{5}+480 a_{6} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = -\frac {1863 a_{0}}{65536}+\frac {25 a_{1}}{1024} \] For \(n = 5\) the recurrence equation gives \[ 323 a_{5}+960 a_{6}+672 a_{7} = 0 \] Which after substituting the earlier terms found becomes \[ a_{7} = \frac {777 a_{0}}{32768}-\frac {1409 a_{1}}{65536} \] 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 \[ y = a_{0}+a_{1} x -\frac {3 a_{0} x^{2}}{32}+\left (\frac {a_{0}}{16}-\frac {a_{1}}{32}\right ) x^{3}+\left (-\frac {93 a_{0}}{2048}+\frac {a_{1}}{32}\right ) x^{4}+\left (\frac {9 a_{0}}{256}-\frac {57 a_{1}}{2048}\right ) x^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}\right ) a_{0}+\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y = \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}\right ) c_{1} +\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}-\frac {1863}{65536} x^{6}\right ) y \left (0\right )+\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}+\frac {25}{1024} x^{6}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \\ \tag{2} y &= \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}\right ) c_{1} +\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \\ \end{align*}
Verification of solutions
\[ y = \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}-\frac {1863}{65536} x^{6}\right ) y \left (0\right )+\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}+\frac {25}{1024} x^{6}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \] Verified OK.
\[ y = \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}\right ) c_{1} +\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (16 x^{2}+32 x +16\right ) \left (\frac {d}{d x}y^{\prime }\right )+3 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 {3 y}{16 \left (x^{2}+2 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 {3 y}{16 \left (x^{2}+2 x +1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=0, P_{3}\left (x \right )=\frac {3}{16 \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}}}=0 \\ {} & \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}}}=\frac {3}{16} \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-1\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (16 x^{2}+32 x +16\right ) \left (\frac {d}{d x}y^{\prime }\right )+3 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 \\ {} & {} & 16 u^{2} \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+3 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 DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{2}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u^{2}\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} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (4 k +4 r -1\right ) \left (4 k +4 r -3\right ) u^{k +r}=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (4 k -1\right ) \left (4 k -3\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k}=0 \\ \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}=0\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}, a_{k}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type <- LODE of Euler type successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 49
Order:=6; dsolve(16*(x+1)^2*diff(y(x),x$2)+3*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (1-\frac {3}{32} x^{2}+\frac {1}{16} x^{3}-\frac {93}{2048} x^{4}+\frac {9}{256} x^{5}\right ) y \left (0\right )+\left (x -\frac {1}{32} x^{3}+\frac {1}{32} x^{4}-\frac {57}{2048} x^{5}\right ) D\left (y \right )\left (0\right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 63
AsymptoticDSolveValue[16*(x+1)^2*y''[x]+3*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (-\frac {57 x^5}{2048}+\frac {x^4}{32}-\frac {x^3}{32}+x\right )+c_1 \left (\frac {9 x^5}{256}-\frac {93 x^4}{2048}+\frac {x^3}{16}-\frac {3 x^2}{32}+1\right ) \]