Internal problem ID [426]
Internal file name [OUTPUT/426_Sunday_June_05_2022_01_41_05_AM_20911788/index.tex]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Chapter 11 Power series methods. Section 11.2 Power series solutions. Page
624
Problem number: problem 11.
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]]
\[ \boxed {5 y^{\prime \prime }-2 x y^{\prime }+10 y=0} \] 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 {2 x y^{\prime }}{5}-2 y\\ 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 {4 y^{\prime } x^{2}}{25}-\frac {4 x y}{5}-\frac {8 y^{\prime }}{5}\\ 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 {4 \left (2 x^{3}-35 x \right ) y^{\prime }}{125}+\frac {4 \left (-2 x^{2}+15\right ) y}{25}\\ 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 {8 \left (2 x^{4}-45 x^{2}+100\right ) y^{\prime }}{625}+\frac {8 \left (-2 x^{3}+25 x \right ) y}{125}\\ 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 {8 \left (4 x^{5}-100 x^{3}+375 x \right ) y^{\prime }}{3125}-\frac {32 \left (x^{4}-15 x^{2}+\frac {75}{4}\right ) y}{625} \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 &= -2 y \left (0\right )\\ F_1 &= -\frac {8 y^{\prime }\left (0\right )}{5}\\ F_2 &= \frac {12 y \left (0\right )}{5}\\ F_3 &= \frac {32 y^{\prime }\left (0\right )}{25}\\ F_4 &= -\frac {24 y \left (0\right )}{25} \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[ y = \left (1-x^{2}+\frac {1}{10} x^{4}-\frac {1}{750} x^{6}\right ) y \left (0\right )+\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \] Since the expansion point \(x = 0\) is an ordinary, we can also solve this using standard power series 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*} \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} = \frac {2 x \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )}{5}-2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\tag {1} \end {align*}
Which simplifies to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}5 n \left (n -1\right ) a_{n} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 n \,x^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 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 }}5 n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}5 \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 =0}{\sum }}5 \left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 n \,x^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 a_{n} x^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 10 a_{2}+10 a_{0}=0 \] \[ a_{2} = -a_{0} \] For \(1\le n\), the recurrence equation is \begin{equation} \tag{4} 5 \left (n +2\right ) a_{n +2} \left (n +1\right )-2 n a_{n}+10 a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{equation} \tag{5} a_{n +2} = \frac {2 a_{n} \left (n -5\right )}{5 \left (n +2\right ) \left (n +1\right )} \end{equation} For \(n = 1\) the recurrence equation gives \[ 30 a_{3}+8 a_{1} = 0 \] Which after substituting the earlier terms found becomes \[ a_{3} = -\frac {4 a_{1}}{15} \] For \(n = 2\) the recurrence equation gives \[ 60 a_{4}+6 a_{2} = 0 \] Which after substituting the earlier terms found becomes \[ a_{4} = \frac {a_{0}}{10} \] For \(n = 3\) the recurrence equation gives \[ 100 a_{5}+4 a_{3} = 0 \] Which after substituting the earlier terms found becomes \[ a_{5} = \frac {4 a_{1}}{375} \] For \(n = 4\) the recurrence equation gives \[ 150 a_{6}+2 a_{4} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = -\frac {a_{0}}{750} \] For \(n = 5\) the recurrence equation gives \[ 210 a_{7} = 0 \] Which after substituting the earlier terms found becomes \[ a_{7} = 0 \] 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 \[ y = a_{0}+a_{1} x -a_{0} x^{2}-\frac {4}{15} a_{1} x^{3}+\frac {1}{10} a_{0} x^{4}+\frac {4}{375} a_{1} x^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (1-x^{2}+\frac {1}{10} x^{4}\right ) a_{0}+\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y = \left (1-x^{2}+\frac {1}{10} x^{4}\right ) c_{1} +\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (1-x^{2}+\frac {1}{10} x^{4}-\frac {1}{750} x^{6}\right ) y \left (0\right )+\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \\ \tag{2} y &= \left (1-x^{2}+\frac {1}{10} x^{4}\right ) c_{1} +\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \\ \end{align*}
Verification of solutions
\[ y = \left (1-x^{2}+\frac {1}{10} x^{4}-\frac {1}{750} x^{6}\right ) y \left (0\right )+\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \] Verified OK.
\[ y = \left (1-x^{2}+\frac {1}{10} x^{4}\right ) c_{1} +\left (x -\frac {4}{15} x^{3}+\frac {4}{375} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {2 x y^{\prime }}{5}-2 y \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \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 {2 x y^{\prime }}{5}+2 y=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 5 \frac {d}{d x}y^{\prime }-2 x y^{\prime }+10 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \square & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite DE with series expansions}\hspace {3pt} \\ {} & {} & \moverset {\infty }{\munderset {k =0}{\sum }}\left (5 a_{k +2} \left (k +2\right ) \left (k +1\right )-2 a_{k} \left (k -5\right )\right ) x^{k}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 5 \left (k^{2}+3 k +2\right ) a_{k +2}-2 a_{k} \left (k -5\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {2 a_{k} \left (k -5\right )}{5 \left (k^{2}+3 k +2\right )}\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 Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Kummer successful <- special function solution successful Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 34
Order:=6; dsolve(5*diff(y(x),x$2)-2*x*diff(y(x),x)+10*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (1-x^{2}+\frac {1}{10} x^{4}\right ) y \left (0\right )+\left (\frac {4}{375} x^{5}-\frac {4}{15} x^{3}+x \right ) D\left (y \right )\left (0\right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.001 (sec). Leaf size: 40
AsymptoticDSolveValue[5*y''[x]-2*x*y'[x]+10*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {4 x^5}{375}-\frac {4 x^3}{15}+x\right )+c_1 \left (\frac {x^4}{10}-x^2+1\right ) \]