4.21 problem 21

Internal problem ID [7890]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 21
Date solved : Monday, October 21, 2024 at 04:31:27 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }+\left (x -6\right ) 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 &= -\left (x -6\right ) 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 \\ &= \left (-x +6\right ) y^{\prime }-y\\ 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 \\ &= -2 y^{\prime }+\left (x^{2}-12 x +36\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 (\left (x -6\right ) y^{\prime }+4 y\right ) \left (x -6\right )\\ 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 (x -6\right )^{3} y+6 \left (x -6\right ) y^{\prime }+4 y \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 &= 6 y \left (0\right )\\ F_1 &= 6 y^{\prime }\left (0\right )-y \left (0\right )\\ F_2 &= -2 y^{\prime }\left (0\right )+36 y \left (0\right )\\ F_3 &= 36 y^{\prime }\left (0\right )-24 y \left (0\right )\\ F_4 &= 220 y \left (0\right )-36 y^{\prime }\left (0\right ) \end{align*}

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

\[ y = \left (1+3 x^{2}-\frac {1}{6} x^{3}+\frac {3}{2} x^{4}-\frac {1}{5} x^{5}\right ) y \left (0\right )+\left (x +x^{3}-\frac {1}{12} x^{4}+\frac {3}{10} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \]

Since the expansion point \(x = 0\) is an ordinary point, then this can also be solved using the standard power series method. 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 (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (x -6\right ) \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 }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 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 (1+n \right ) x^{n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =0}{\sum }}\left (n +2\right ) a_{n +2} \left (1+n \right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 a_{n} x^{n}\right ) = 0 \end{equation}

\(n=0\) gives

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

\[ a_{2} = 3 a_{0} \]

For \(1\le n\), the recurrence equation is

\begin{equation} \tag{4} \left (n +2\right ) a_{n +2} \left (1+n \right )+a_{n -1}-6 a_{n} = 0 \end{equation}

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

\begin{align*} \tag{5} a_{n +2}&= \frac {-a_{n -1}+6 a_{n}}{\left (n +2\right ) \left (1+n \right )} \\ &= \frac {6 a_{n}}{\left (n +2\right ) \left (1+n \right )}-\frac {a_{n -1}}{\left (n +2\right ) \left (1+n \right )} \\ \end{align*}

For \(n = 1\) the recurrence equation gives

\[ 6 a_{3}+a_{0}-6 a_{1} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{3} = -\frac {a_{0}}{6}+a_{1} \]

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

\[ 12 a_{4}+a_{1}-6 a_{2} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{4} = -\frac {a_{1}}{12}+\frac {3 a_{0}}{2} \]

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

\[ 20 a_{5}+a_{2}-6 a_{3} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{5} = -\frac {a_{0}}{5}+\frac {3 a_{1}}{10} \]

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

\[ 30 a_{6}+a_{3}-6 a_{4} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{6} = \frac {11 a_{0}}{36}-\frac {a_{1}}{20} \]

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

\[ 42 a_{7}+a_{4}-6 a_{5} = 0 \]

Which after substituting the earlier terms found becomes

\[ a_{7} = \frac {113 a_{1}}{2520}-\frac {9 a_{0}}{140} \]

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

\[ y = a_{0}+a_{1} x +3 a_{0} x^{2}+\left (-\frac {a_{0}}{6}+a_{1}\right ) x^{3}+\left (-\frac {a_{1}}{12}+\frac {3 a_{0}}{2}\right ) x^{4}+\left (-\frac {a_{0}}{5}+\frac {3 a_{1}}{10}\right ) x^{5}+\dots \]

Collecting terms, the solution becomes

\begin{equation} \tag{3} y = \left (1+3 x^{2}-\frac {1}{6} x^{3}+\frac {3}{2} x^{4}-\frac {1}{5} x^{5}\right ) a_{0}+\left (x +x^{3}-\frac {1}{12} x^{4}+\frac {3}{10} x^{5}\right ) a_{1}+O\left (x^{6}\right ) \end{equation}

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

\[ y = \left (1+3 x^{2}-\frac {1}{6} x^{3}+\frac {3}{2} x^{4}-\frac {1}{5} x^{5}\right ) c_1 +\left (x +x^{3}-\frac {1}{12} x^{4}+\frac {3}{10} x^{5}\right ) c_2 +O\left (x^{6}\right ) \]

4.21.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\left (x -6\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \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..1 \\ {} & {} & 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} \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}-6 a_{0}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-6 a_{k}+a_{k -1}\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 2 a_{2}-6 a_{0}=0 \\ \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}-6 a_{k}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+3 k +5\right ) a_{k +3}-6 a_{k +1}+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 +3}=-\frac {-6 a_{k +1}+a_{k}}{k^{2}+5 k +6}, 2 a_{2}-6 a_{0}=0\right ] \end {array} \]

4.21.2 Maple trace

Methods for second order ODEs:
 

4.21.3 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 52

dsolve(diff(diff(y(x),x),x)+(x-6)*y(x) = 0,y(x), 
       series,x=0)
 

\[ y = \left (1+3 x^{2}-\frac {1}{6} x^{3}+\frac {3}{2} x^{4}-\frac {1}{5} x^{5}\right ) y \left (0\right )+\left (x +x^{3}-\frac {1}{12} x^{4}+\frac {3}{10} x^{5}\right ) y^{\prime }\left (0\right )+O\left (x^{6}\right ) \]

4.21.4 Mathematica DSolve solution

Solving time : 0.001 (sec)
Leaf size : 57

AsymptoticDSolveValue[{D[y[x],{x,2}]+(x-6)*y[x]==0,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to c_2 \left (\frac {3 x^5}{10}-\frac {x^4}{12}+x^3+x\right )+c_1 \left (-\frac {x^5}{5}+\frac {3 x^4}{2}-\frac {x^3}{6}+3 x^2+1\right ) \]