Internal problem ID [6576]
Internal file name [OUTPUT/5824_Sunday_June_05_2022_03_56_01_PM_62230208/index.tex
]
Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL,
WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th
edition.
Section: CHAPTER 6 SERIES SOLUTIONS OF LINEAR EQUATIONS. Section 6.2
SOLUTIONS ABOUT ORDINARY POINTS. EXERCISES 6.2. Page 246
Problem number: 25 expansion at 0.
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]]
\[ \boxed {\cos \left (x \right ) y^{\prime \prime }+y^{\prime }+5 y=0} \] 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 {y^{\prime }+5 y}{\cos \left (x \right )}\\ 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 \\ &= -5 \sec \left (x \right )^{2} \left (\left (\cos \left (x \right )+\frac {\sin \left (x \right )}{5}-\frac {1}{5}\right ) y^{\prime }+y \left (\sin \left (x \right )-1\right )\right )\\ 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 {5 \sec \left (x \right )^{2} \left (\frac {\left (-3+11 \cos \left (x \right )^{2}+\left (8-9 \sin \left (x \right )\right ) \cos \left (x \right )+3 \sin \left (x \right )\right ) y^{\prime }}{5}+\left (\cos \left (x \right )^{2}+\left (\sin \left (x \right )+3\right ) \cos \left (x \right )+8 \sin \left (x \right )+2\right ) y\right )}{\cos \left (x \right )+\sin \left (x \right )+1}\\ 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 {5 \sec \left (x \right )^{3} \left (\frac {14 \left (\frac {8 \sin \left (x \right )^{2}}{7}+\left (\cos \left (x \right )-\frac {6}{7}\right ) \sin \left (x \right )-\frac {32 \cos \left (x \right )}{7}-\frac {2}{7}\right ) \sin \left (x \right ) y^{\prime }}{5}+\left (-22+\left (-\cos \left (x \right )^{2}-14 \cos \left (x \right )+2\right ) \sin \left (x \right )+\cos \left (x \right )^{3}+27 \cos \left (x \right )^{2}+4 \cos \left (x \right )\right ) y\right )}{\cos \left (x \right )+\sin \left (x \right )+1}\\ 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 {5 \sec \left (x \right )^{4} \left (\left (\frac {21 \cos \left (x \right )^{4}}{5}+\left (-\frac {19 \sin \left (x \right )}{5}+\frac {276}{5}\right ) \cos \left (x \right )^{3}+\left (-3 \sin \left (x \right )-18\right ) \cos \left (x \right )^{2}+\left (27 \sin \left (x \right )-57\right ) \cos \left (x \right )-12 \sin \left (x \right )+12\right ) y^{\prime }+\left (\cos \left (x \right )^{4}+\left (\sin \left (x \right )+41\right ) \cos \left (x \right )^{3}+\left (70 \sin \left (x \right )-40\right ) \cos \left (x \right )^{2}+\left (70 \sin \left (x \right )-50\right ) \cos \left (x \right )-90 \sin \left (x \right )+30\right ) y\right )}{\cos \left (x \right )+\sin \left (x \right )+1}\\ F_5 &= \frac {d F_4}{dx} \\ &= \frac {\partial F_{4}}{\partial x}+ \frac {\partial F_{4}}{\partial y} y^{\prime }+ \frac {\partial F_{4}}{\partial y^{\prime }} F_4 \\ &= \frac {5 \sec \left (x \right )^{5} \left (\left (-\frac {24 \cos \left (x \right )^{5}}{5}+\left (-\frac {26 \sin \left (x \right )}{5}-\frac {194}{5}\right ) \cos \left (x \right )^{4}+\left (-184 \sin \left (x \right )+211\right ) \cos \left (x \right )^{3}+\left (5 \sin \left (x \right )-31\right ) \cos \left (x \right )^{2}+\left (324 \sin \left (x \right )-204\right ) \cos \left (x \right )-72 \sin \left (x \right )+72\right ) y^{\prime }+\left (\cos \left (x \right )^{5}+\left (-\sin \left (x \right )+161\right ) \cos \left (x \right )^{4}+\left (-100 \sin \left (x \right )+470\right ) \cos \left (x \right )^{3}+\left (160 \sin \left (x \right )-680\right ) \cos \left (x \right )^{2}+\left (330 \sin \left (x \right )-510\right ) \cos \left (x \right )-240 \sin \left (x \right )+480\right ) y\right )}{\cos \left (x \right )+\sin \left (x \right )+1}\\ F_6 &= \frac {d F_5}{dx} \\ &= \frac {\partial F_{5}}{\partial x}+ \frac {\partial F_{5}}{\partial y} y^{\prime }+ \frac {\partial F_{5}}{\partial y^{\prime }} F_5 \\ &= \frac {5 \left (\left (\frac {31 \cos \left (x \right )^{6}}{5}+\left (\frac {2668}{5}-\frac {29 \sin \left (x \right )}{5}\right ) \cos \left (x \right )^{5}+\left (\frac {2067}{5}-\frac {837 \sin \left (x \right )}{5}\right ) \cos \left (x \right )^{4}+\left (-2750+1086 \sin \left (x \right )\right ) \cos \left (x \right )^{3}+\left (76+76 \sin \left (x \right )\right ) \cos \left (x \right )^{2}+\left (2208-1608 \sin \left (x \right )\right ) \cos \left (x \right )-504+504 \sin \left (x \right )\right ) y^{\prime }+y \left (\cos \left (x \right )^{6}+\left (\sin \left (x \right )+223\right ) \cos \left (x \right )^{5}+\left (348 \sin \left (x \right )-418\right ) \cos \left (x \right )^{4}+\left (2300 \sin \left (x \right )-2780\right ) \cos \left (x \right )^{3}+\left (-2980 \sin \left (x \right )+2420\right ) \cos \left (x \right )^{2}+\left (-3840 \sin \left (x \right )+2640\right ) \cos \left (x \right )+3120 \sin \left (x \right )-1920\right )\right ) \sec \left (x \right )^{6}}{\cos \left (x \right )+\sin \left (x \right )+1} \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 &= -y^{\prime }\left (0\right )-5 y \left (0\right )\\ F_1 &= -4 y^{\prime }\left (0\right )+5 y \left (0\right )\\ F_2 &= 15 y \left (0\right )+8 y^{\prime }\left (0\right )\\ F_3 &= -25 y \left (0\right )\\ F_4 &= 45 y \left (0\right )+9 y^{\prime }\left (0\right )\\ F_5 &= -195 y \left (0\right )+11 y^{\prime }\left (0\right )\\ F_6 &= 415 y \left (0\right )-42 y^{\prime }\left (0\right ) \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[ y = \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}+\frac {83}{8064} x^{8}\right ) y \left (0\right )+\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}-\frac {1}{960} x^{8}\right ) y^{\prime }\left (0\right )+O\left (x^{8}\right ) \] 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 {\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 )}{\cos \left (x \right )}\tag {1} \end {align*}
Expanding \(\cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} \cos \left (x \right ) &= \frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 + \dots \\ &= \frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1 \end {align*}
Hence the ODE in Eq (1) becomes \[ \left (\frac {1}{40320} x^{8}-\frac {1}{720} x^{6}+\frac {1}{24} x^{4}-\frac {1}{2} x^{2}+1\right ) \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0 \] Expanding the first term in (1) gives \[ \frac {x^{8}}{40320}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{6}}{720}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\frac {x^{4}}{24}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )-\frac {x^{2}}{2}\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+1\cdot \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right )+5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right ) = 0 \] Which simplifies to \begin{equation} \tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +6} a_{n} \left (n -1\right )}{40320}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\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 }}\frac {n \,x^{n +6} a_{n} \left (n -1\right )}{40320} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {\left (n -6\right ) a_{n -6} \left (n -7\right ) x^{n}}{40320} \\ \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n \,x^{n +4} a_{n} \left (n -1\right )}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right ) \\ \moverset {\infty }{\munderset {n =2}{\sum }}\frac {n \,x^{n +2} a_{n} \left (n -1\right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24} \\ \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 =1}{\sum }}n a_{n} x^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +1\right ) 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 =8}{\sum }}\frac {\left (n -6\right ) a_{n -6} \left (n -7\right ) x^{n}}{40320}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right ) x^{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {\left (n -2\right ) a_{n -2} \left (n -3\right ) x^{n}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {n a_{n} x^{n} \left (n -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +1\right ) a_{n +1} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 a_{n} x^{n}\right ) = 0 \end{equation} \(n=0\) gives \[ 2 a_{2}+a_{1}+5 a_{0}=0 \] \[ a_{2} = -\frac {5 a_{0}}{2}-\frac {a_{1}}{2} \] \(n=1\) gives \[ 6 a_{3}+2 a_{2}+5 a_{1}=0 \] Which after substituting earlier equations, simplifies to \[ a_{3} = \frac {5 a_{0}}{6}-\frac {2 a_{1}}{3} \] \(n=2\) gives \[ 4 a_{2}+12 a_{4}+3 a_{3} = 0 \] Which after substituting earlier equations, simplifies to \[ -\frac {15 a_{0}}{2}-4 a_{1}+12 a_{4} = 0 \] Or \[ a_{4} = \frac {5 a_{0}}{8}+\frac {a_{1}}{3} \] \(n=3\) gives \[ 2 a_{3}+20 a_{5}+4 a_{4} = 0 \] Which after substituting earlier equations, simplifies to \[ \frac {25 a_{0}}{6}+20 a_{5} = 0 \] Or \[ a_{5} = -\frac {5 a_{0}}{24} \] \(n=4\) gives \[ \frac {a_{2}}{12}-a_{4}+30 a_{6}+5 a_{5} = 0 \] Which after substituting earlier equations, simplifies to \[ -\frac {15 a_{0}}{8}-\frac {3 a_{1}}{8}+30 a_{6} = 0 \] Or \[ a_{6} = \frac {a_{0}}{16}+\frac {a_{1}}{80} \] \(n=5\) gives \[ \frac {a_{3}}{4}-5 a_{5}+42 a_{7}+6 a_{6} = 0 \] Which after substituting earlier equations, simplifies to \[ \frac {13 a_{0}}{8}-\frac {11 a_{1}}{120}+42 a_{7} = 0 \] Or \[ a_{7} = -\frac {13 a_{0}}{336}+\frac {11 a_{1}}{5040} \] \(n=6\) gives \[ -\frac {a_{2}}{360}+\frac {a_{4}}{2}-10 a_{6}+56 a_{8}+7 a_{7} = 0 \] Which after substituting earlier equations, simplifies to \[ -\frac {83 a_{0}}{144}+\frac {7 a_{1}}{120}+56 a_{8} = 0 \] Or \[ a_{8} = \frac {83 a_{0}}{8064}-\frac {a_{1}}{960} \] \(n=7\) gives \[ -\frac {a_{3}}{120}+\frac {5 a_{5}}{6}-16 a_{7}+72 a_{9}+8 a_{8} = 0 \] Which after substituting earlier equations, simplifies to \[ \frac {25 a_{0}}{48}-\frac {19 a_{1}}{504}+72 a_{9} = 0 \] Or \[ a_{9} = -\frac {25 a_{0}}{3456}+\frac {19 a_{1}}{36288} \] For \(8\le n\), the recurrence equation is \begin{equation} \tag{4} \frac {\left (n -6\right ) a_{n -6} \left (n -7\right )}{40320}-\frac {\left (n -4\right ) a_{n -4} \left (n -5\right )}{720}+\frac {\left (n -2\right ) a_{n -2} \left (n -3\right )}{24}-\frac {n a_{n} \left (n -1\right )}{2}+\left (n +2\right ) a_{n +2} \left (n +1\right )+\left (n +1\right ) a_{n +1}+5 a_{n} = 0 \end{equation} Solving for \(a_{n +2}\), gives \begin{align*} \tag{5} a_{n +2}&= \frac {20160 n^{2} a_{n}-n^{2} a_{n -6}+56 n^{2} a_{n -4}-1680 n^{2} a_{n -2}-20160 n a_{n}+13 n a_{n -6}-504 n a_{n -4}+8400 n a_{n -2}-40320 n a_{n +1}-201600 a_{n}-42 a_{n -6}+1120 a_{n -4}-10080 a_{n -2}-40320 a_{n +1}}{40320 \left (n +2\right ) \left (n +1\right )} \\ &= \frac {\left (20160 n^{2}-20160 n -201600\right ) a_{n}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-n^{2}+13 n -42\right ) a_{n -6}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (56 n^{2}-504 n +1120\right ) a_{n -4}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-1680 n^{2}+8400 n -10080\right ) a_{n -2}}{40320 \left (n +2\right ) \left (n +1\right )}+\frac {\left (-40320 n -40320\right ) a_{n +1}}{40320 \left (n +2\right ) \left (n +1\right )} \\ \end{align*} 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 \[ y = a_{0}+a_{1} x +\left (-\frac {5 a_{0}}{2}-\frac {a_{1}}{2}\right ) x^{2}+\left (\frac {5 a_{0}}{6}-\frac {2 a_{1}}{3}\right ) x^{3}+\left (\frac {5 a_{0}}{8}+\frac {a_{1}}{3}\right ) x^{4}-\frac {5 a_{0} x^{5}}{24}+\left (\frac {a_{0}}{16}+\frac {a_{1}}{80}\right ) x^{6}+\left (-\frac {13 a_{0}}{336}+\frac {11 a_{1}}{5040}\right ) x^{7}+\dots \] Collecting terms, the solution becomes \begin{equation} \tag{3} y = \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}\right ) a_{0}+\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}\right ) a_{1}+O\left (x^{8}\right ) \end{equation} At \(x = 0\) the solution above becomes \[ y = \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}\right ) c_{1} +\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}\right ) c_{2} +O\left (x^{8}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}+\frac {83}{8064} x^{8}\right ) y \left (0\right )+\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}-\frac {1}{960} x^{8}\right ) y^{\prime }\left (0\right )+O\left (x^{8}\right ) \\ \tag{2} y &= \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}\right ) c_{1} +\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}\right ) c_{2} +O\left (x^{8}\right ) \\ \end{align*}
Verification of solutions
\[ y = \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}+\frac {83}{8064} x^{8}\right ) y \left (0\right )+\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}-\frac {1}{960} x^{8}\right ) y^{\prime }\left (0\right )+O\left (x^{8}\right ) \] Verified OK.
\[ y = \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}\right ) c_{1} +\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}\right ) c_{2} +O\left (x^{8}\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5`[0, u]
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 69
Order:=8; dsolve(cos(x)*diff(y(x),x$2)+diff(y(x),x)+5*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (1-\frac {5}{2} x^{2}+\frac {5}{6} x^{3}+\frac {5}{8} x^{4}-\frac {5}{24} x^{5}+\frac {1}{16} x^{6}-\frac {13}{336} x^{7}\right ) y \left (0\right )+\left (x -\frac {1}{2} x^{2}-\frac {2}{3} x^{3}+\frac {1}{3} x^{4}+\frac {1}{80} x^{6}+\frac {11}{5040} x^{7}\right ) D\left (y \right )\left (0\right )+O\left (x^{8}\right ) \]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 91
AsymptoticDSolveValue[Cos[x]*y''[x]+y'[x]+5*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (\frac {11 x^7}{5040}+\frac {x^6}{80}+\frac {x^4}{3}-\frac {2 x^3}{3}-\frac {x^2}{2}+x\right )+c_1 \left (-\frac {13 x^7}{336}+\frac {x^6}{16}-\frac {5 x^5}{24}+\frac {5 x^4}{8}+\frac {5 x^3}{6}-\frac {5 x^2}{2}+1\right ) \]