16.9 problem Problem 9

Internal problem ID [3555]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.2. page 739
Problem number : Problem 9
Date solved : Thursday, October 17, 2024 at 03:58:53 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

\begin{align*} \left (x^{2}-3\right ) y^{\prime \prime }-3 x y^{\prime }-5 y&=0 \end{align*}

Using series expansion around \(x=0\)

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 x y^{\prime }+5 y}{x^{2}-3}\\ 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 {11 y^{\prime } x^{2}+5 y x -24 y^{\prime }}{\left (x^{2}-3\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 {16 y^{\prime } x^{3}+40 y x^{2}-57 x y^{\prime }-135 y}{\left (x^{2}-3\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 (40 x^{4}-285 x^{2}+576\right ) y^{\prime }+\left (-80 x^{3}+285 x \right ) y}{\left (x^{2}-3\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 (-120 x^{5}+900 x^{3}-2025 x \right ) y^{\prime }+600 \left (x^{4}-\frac {9}{2} x^{2}+\frac {27}{8}\right ) y}{\left (x^{2}-3\right )^{5}} \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 {5 y \left (0\right )}{3}\\ F_1 &= -\frac {8 y^{\prime }\left (0\right )}{3}\\ F_2 &= 5 y \left (0\right )\\ F_3 &= \frac {64 y^{\prime }\left (0\right )}{9}\\ F_4 &= -\frac {25 y \left (0\right )}{3} \end{align*}

Substituting all the above in (7) and simplifying gives the solution as

\[ y = \left (1-\frac {5}{6} x^{2}+\frac {5}{24} x^{4}\right ) y \left (0\right )+\left (x -\frac {4}{9} x^{3}+\frac {8}{135} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \]

Since the expansion point \(x = 0\) is an ordinary point, then this can also be solved using the standard power series method. The ode is normalized to be

\[ \left (x^{2}-3\right ) y^{\prime \prime }-3 x y^{\prime }-5 y = 0 \]

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 (x^{2}-3\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-3 x \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )-5 \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 }}x^{n} a_{n} n \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 n \left (n -1\right ) a_{n} x^{n -2}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}\left (-3 n \left (n -1\right ) a_{n} x^{n -2}\right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 \left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right ) \\ \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 }}x^{n} a_{n} n \left (n -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 \left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 n a_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n}\right ) = 0 \end{equation}

\(n=0\) gives

\[ -6 a_{2}-5 a_{0}=0 \]

\[ a_{2} = -\frac {5 a_{0}}{6} \]

\(n=1\) gives

\[ -18 a_{3}-8 a_{1}=0 \]

Which after substituting earlier equations, simplifies to

\[ a_{3} = -\frac {4 a_{1}}{9} \]

For \(2\le n\), the recurrence equation is

\begin{equation} \tag{4} n a_{n} \left (n -1\right )-3 \left (n +2\right ) a_{n +2} \left (n +1\right )-3 n a_{n}-5 a_{n} = 0 \end{equation}

Solving for \(a_{n +2}\), gives

\begin{equation} \tag{5} a_{n +2} = \frac {\left (n -5\right ) a_{n}}{3 n +6} \end{equation}

For \(n = 2\) the recurrence equation gives

\[ -9 a_{2}-36 a_{4} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{4} = \frac {5 a_{0}}{24} \]

For \(n = 3\) the recurrence equation gives

\[ -8 a_{3}-60 a_{5} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{5} = \frac {8 a_{1}}{135} \]

For \(n = 4\) the recurrence equation gives

\[ -5 a_{4}-90 a_{6} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{6} = -\frac {5 a_{0}}{432} \]

For \(n = 5\) the recurrence equation gives

\[ -126 a_{7} = 0 \]

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

\[ y = a_{0}+a_{1} x -\frac {5}{6} a_{0} x^{2}-\frac {4}{9} a_{1} x^{3}+\frac {5}{24} a_{0} x^{4}+\frac {8}{135} a_{1} x^{5}+\dots \]

Collecting terms, the solution becomes

\begin{equation} \tag{3} y = \left (1-\frac {5}{6} x^{2}+\frac {5}{24} x^{4}\right ) a_{0}+\left (x -\frac {4}{9} x^{3}+\frac {8}{135} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation}

At \(x = 0\) the solution above becomes

\[ y = \left (1-\frac {5}{6} x^{2}+\frac {5}{24} x^{4}\right ) c_1 +\left (x -\frac {4}{9} x^{3}+\frac {8}{135} x^{5}\right ) c_2 +O\left (x^{6}\right ) \]

16.9.1 Maple step by step solution

16.9.2 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 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful`
 

16.9.3 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 39

dsolve((x^2-3)*diff(diff(y(x),x),x)-3*x*diff(y(x),x)-5*y(x) = 0,y(x), 
       series,x=0)
 

\[ y \left (x \right ) = \left (1-\frac {5}{6} x^{2}+\frac {5}{24} x^{4}\right ) y \left (0\right )+\left (x -\frac {4}{9} x^{3}+\frac {8}{135} x^{5}\right ) D\left (y \right )\left (0\right )+O\left (x^{6}\right ) \]

16.9.4 Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 42

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

\[ y(x)\to c_2 \left (\frac {8 x^5}{135}-\frac {4 x^3}{9}+x\right )+c_1 \left (\frac {5 x^4}{24}-\frac {5 x^2}{6}+1\right ) \]