13.33 problem 33

Internal problem ID [1274]
Internal file name [OUTPUT/1275_Sunday_June_05_2022_02_07_39_AM_40863679/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: 33.
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]]

\[ \boxed {y^{\prime \prime }-3 y^{\prime } x +\left (2 x^{2}+5\right ) y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = -2] \end {align*}

With the expansion point for the power series method at \(x = 0\).

13.33.1 Existence and uniqueness analysis

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) &=-3 x\\ q(x) &=2 x^{2}+5\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-3 y^{\prime } x +\left (2 x^{2}+5\right ) y = 0 \end {align*}

The domain of \(p(x)=-3 x\) 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 &= -2 y x^{2}+3 y^{\prime } x -5 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 \\ &= -6 y x^{3}+7 y^{\prime } x^{2}-19 y x -2 y^{\prime }\\ 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 \\ &= \left (15 x^{3}-11 x \right ) y^{\prime }+\left (-14 x^{4}-49 x^{2}-9\right ) y\\ 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 \\ &= \left (31 x^{4}-37 x^{2}-20\right ) y^{\prime }+\left (-30 x^{5}-109 x^{3}-43 x \right ) y\\ 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 \\ &= \left (63 x^{5}-96 x^{3}-177 x \right ) y^{\prime }+\left (-62 x^{6}-231 x^{4}-102 x^{2}+57\right ) y \end {align*}

And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = 1\) and \(y^{\prime }\left (0\right ) = -2\) gives \begin {align*} F_0 &= -5\\ F_1 &= 4\\ F_2 &= -9\\ F_3 &= 40\\ F_4 &= 57 \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y = -2 x +1-\frac {5 x^{2}}{2}+\frac {2 x^{3}}{3}-\frac {3 x^{4}}{8}+\frac {x^{5}}{3}+\frac {19 x^{6}}{240}+O\left (x^{6}\right ) \] \[ y = -2 x +1-\frac {5 x^{2}}{2}+\frac {2 x^{3}}{3}-\frac {3 x^{4}}{8}+\frac {x^{5}}{3}+\frac {19 x^{6}}{240}+O\left (x^{6}\right ) \] 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} = -2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) x^{2}+3 \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) x -5 \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 }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 n \,x^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 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 \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 =0}{\sum }}2 x^{n +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} 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 }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 n \,x^{n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 a_{n} x^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 2 a_{2}+5 a_{0}=0 \] \[ a_{2} = -\frac {5 a_{0}}{2} \] \(n=1\) gives \[ 6 a_{3}+2 a_{1}=0 \] Which after substituting earlier equations, simplifies to \[ a_{3} = -\frac {a_{1}}{3} \] For \(2\le n\), the recurrence equation is \begin{equation} \tag{4} \left (n +2\right ) a_{n +2} \left (n +1\right )-3 n a_{n}+2 a_{n -2}+5 a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= \frac {3 n a_{n}-5 a_{n}-2 a_{n -2}}{\left (n +2\right ) \left (n +1\right )} \\ &= \frac {\left (3 n -5\right ) a_{n}}{\left (n +2\right ) \left (n +1\right )}-\frac {2 a_{n -2}}{\left (n +2\right ) \left (n +1\right )} \\ \end{align*} For \(n = 2\) the recurrence equation gives \[ 12 a_{4}-a_{2}+2 a_{0} = 0 \] Which after substituting the earlier terms found becomes \[ a_{4} = -\frac {3 a_{0}}{8} \] For \(n = 3\) the recurrence equation gives \[ 20 a_{5}-4 a_{3}+2 a_{1} = 0 \] Which after substituting the earlier terms found becomes \[ a_{5} = -\frac {a_{1}}{6} \] For \(n = 4\) the recurrence equation gives \[ 30 a_{6}-7 a_{4}+2 a_{2} = 0 \] Which after substituting the earlier terms found becomes \[ a_{6} = \frac {19 a_{0}}{240} \] For \(n = 5\) the recurrence equation gives \[ 42 a_{7}-10 a_{5}+2 a_{3} = 0 \] Which after substituting the earlier terms found becomes \[ a_{7} = -\frac {a_{1}}{42} \] 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 \[ y = a_{0}+a_{1} x -\frac {5}{2} a_{0} x^{2}-\frac {1}{3} a_{1} x^{3}-\frac {3}{8} a_{0} x^{4}-\frac {1}{6} a_{1} x^{5}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (1-\frac {5}{2} x^{2}-\frac {3}{8} x^{4}\right ) a_{0}+\left (x -\frac {1}{3} x^{3}-\frac {1}{6} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y = \left (1-\frac {5}{2} x^{2}-\frac {3}{8} x^{4}\right ) c_{1} +\left (x -\frac {1}{3} x^{3}-\frac {1}{6} x^{5}\right ) c_{2} +O\left (x^{6}\right ) \] \[ y = 1-\frac {5 x^{2}}{2}-\frac {3 x^{4}}{8}-2 x +\frac {2 x^{3}}{3}+\frac {x^{5}}{3}+O\left (x^{6}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -2 x +1-\frac {5 x^{2}}{2}+\frac {2 x^{3}}{3}-\frac {3 x^{4}}{8}+\frac {x^{5}}{3}+\frac {19 x^{6}}{240}+O\left (x^{6}\right ) \\ \tag{2} y &= 1-\frac {5 x^{2}}{2}-\frac {3 x^{4}}{8}-2 x +\frac {2 x^{3}}{3}+\frac {x^{5}}{3}+O\left (x^{6}\right ) \\ \end{align*}

Verification of solutions

\[ y = -2 x +1-\frac {5 x^{2}}{2}+\frac {2 x^{3}}{3}-\frac {3 x^{4}}{8}+\frac {x^{5}}{3}+\frac {19 x^{6}}{240}+O\left (x^{6}\right ) \] Verified OK.

\[ y = 1-\frac {5 x^{2}}{2}-\frac {3 x^{4}}{8}-2 x +\frac {2 x^{3}}{3}+\frac {x^{5}}{3}+O\left (x^{6}\right ) \] Verified OK.

13.33.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y^{\prime }=-2 y x^{2}+3 y^{\prime } x -5 y, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-2\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 }=\left (-2 x^{2}-5\right ) y+3 y^{\prime } x \\ \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 }-3 y^{\prime } x +\left (2 x^{2}+5\right ) 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 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 =\max \left (0, -m \right )}{\sum }}a_{k} x^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} x^{k} \\ {} & \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 ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}+5 a_{0}+\left (6 a_{3}+2 a_{1}\right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-a_{k} \left (3 k -5\right )+2 a_{k -2}\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}+5 a_{0}=0, 6 a_{3}+2 a_{1}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=-\frac {5 a_{0}}{2}, a_{3}=-\frac {a_{1}}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}-3 a_{k} k +5 a_{k}+2 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+3 k +8\right ) a_{k +4}-3 a_{k +2} \left (k +2\right )+5 a_{k +2}+2 a_{k}=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 +4}=\frac {3 k a_{k +2}-2 a_{k}+a_{k +2}}{k^{2}+7 k +12}, a_{2}=-\frac {5 a_{0}}{2}, a_{3}=-\frac {a_{1}}{3}\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 
      -> Trying to convert hypergeometric functions to elementary form... 
      <- elementary form could result into a too large expression - returning special function form of solution, free of uncomputed 
   <- Kovacics algorithm successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 20

Order:=6; 
dsolve([diff(y(x),x$2)-3*x*diff(y(x),x)+(5+2*x^2)*y(x)=0,y(0) = 1, D(y)(0) = -2],y(x),type='series',x=0);
 

\[ y \left (x \right ) = 1-2 x -\frac {5}{2} x^{2}+\frac {2}{3} x^{3}-\frac {3}{8} x^{4}+\frac {1}{3} x^{5}+\operatorname {O}\left (x^{6}\right ) \]

✓ Solution by Mathematica

Time used: 0.001 (sec). Leaf size: 36

AsymptoticDSolveValue[{y''[x]-3*x*y'[x]+(5+2*x^2)*y[x]==0,{y[0]==1,y'[0]==-2}},y[x],{x,0,5}]
 

\[ y(x)\to \frac {x^5}{3}-\frac {3 x^4}{8}+\frac {2 x^3}{3}-\frac {5 x^2}{2}-2 x+1 \]