problem 3.24 (f)

Internal problem ID [5490]
Internal ﬁle name [OUTPUT/4738_Sunday_June_05_2022_03_04_29_PM_85478439/index.tex]

Book: Advanced Mathematical Methods for Scientists and Engineers, Bender and Orszag. Springer October 29, 1999
Section: Chapter 3. APPROXIMATE SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS. page 136
Problem number: 3.24 (f).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second order series method. Regular singular point. Diﬀerence is integer"

\[ \boxed {\sin \left (x \right ) y^{\prime \prime }-2 \cos \left (x \right ) y^{\prime }-\sin \left (x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).

The type of the expansion point is ﬁrst determined. This is done on the homogeneous part of the ODE. \[ \sin \left (x \right ) y^{\prime \prime }-2 \cos \left (x \right ) y^{\prime }-\sin \left (x \right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}

Where \begin {align*} p(x) &= -\frac {2 \cos \left (x \right )}{\sin \left (x \right )}\\ q(x) &= -1\\ \end {align*}

Combining everything together gives the following summary of singularities for the ode as

Since \(x = 0\) is regular singular point, then Frobenius power series is used. Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )-2 \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-\sin \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7} \end {align*}

Expanding \(-2 \cos \left (x \right )\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} -2 \cos \left (x \right ) &= -2+x^{2}-\frac {1}{12} x^{4}+\frac {1}{360} x^{6} + \dots \\ &= -2+x^{2}-\frac {1}{12} x^{4}+\frac {1}{360} x^{6} \end {align*}

Expanding \(-\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the ﬁrst \(6\) terms gives \begin {align*} -\sin \left (x \right ) &= -x +\frac {1}{6} x^{3}-\frac {1}{120} x^{5}+\frac {1}{5040} x^{7} + \dots \\ &= -x +\frac {1}{6} x^{3}-\frac {1}{120} x^{5}+\frac {1}{5040} x^{7} \end {align*}

Which simpliﬁes to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right )}{360}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n} \left (n +r \right )}{12}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n}}{6}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +7} a_{n}}{5040}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n} \left (n +r \right ) \left (n +r -1\right )}{5040}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right ) x^{n +r -1}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right ) x^{n +r -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n} \left (n +r \right )}{360} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n +r -6\right ) x^{n +r -1}}{360} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n} \left (n +r \right )}{12}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (-4+n +r \right ) x^{n +r -1}}{12}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n}}{6} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +5} a_{n}}{120}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r -1}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +7} a_{n}}{5040} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r -1}}{5040} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right ) x^{n +r -1}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right ) x^{n +r -1}}{120}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n +r -6\right ) x^{n +r -1}}{360}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (-4+n +r \right ) x^{n +r -1}}{12}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r -1}}{6}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{n +r -1}}{120}\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r -1}}{5040}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )-2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )-2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simpliﬁes to \[ r \,x^{-1+r} \left (-3+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= 0 \end {align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, 0]\).

Since \(r_1 - r_2 = 3\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}

Or \begin {align*} y_{1}\left (x \right ) &= x^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}

Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +3}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}

Where \(C\) above can be zero. We start by ﬁnding \(y_{1}\). Eq (2B) derived above is now used to ﬁnd all \(a_{n}\) coeﬃcients. The case \(n = 0\) is skipped since it was used to ﬁnd the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r -6}{12+6 r} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = 0 \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {7 r^{3}-63 r^{2}+146 r +120}{360 \left (2+r \right ) \left (r +4\right ) \left (r +1\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = 0 \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {31 r^{5}-248 r^{4}+497 r^{3}+1508 r^{2}-6324 r -3024}{15120 \left (2+r \right ) \left (r +4\right ) \left (r +1\right ) \left (6+r \right ) \left (3+r \right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = 0 \] For \(8\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {a_{n -6} \left (n +r -6\right ) \left (n -7+r \right )}{5040}+\frac {a_{n -4} \left (-4+n +r \right ) \left (n -5+r \right )}{120}-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )}{6}+a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {a_{n -6} \left (n +r -6\right )}{360}-\frac {a_{n -4} \left (-4+n +r \right )}{12}+a_{n -2} \left (n +r -2\right )-2 a_{n} \left (n +r \right )-a_{n -2}+\frac {a_{n -4}}{6}-\frac {a_{n -6}}{120}+\frac {a_{n -8}}{5040} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -6}-42 n^{2} a_{n -4}+840 n^{2} a_{n -2}+2 n r a_{n -6}-84 n r a_{n -4}+1680 n r a_{n -2}+r^{2} a_{n -6}-42 r^{2} a_{n -4}+840 r^{2} a_{n -2}-27 n a_{n -6}+798 n a_{n -4}-9240 n a_{n -2}-27 r a_{n -6}+798 r a_{n -4}-9240 r a_{n -2}-a_{n -8}+168 a_{n -6}-3360 a_{n -4}+20160 a_{n -2}}{5040 n^{2}+10080 n r +5040 r^{2}-15120 n -15120 r}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = \frac {\left (a_{n -6}-42 a_{n -4}+840 a_{n -2}\right ) n^{2}+\left (-21 a_{n -6}+546 a_{n -4}-4200 a_{n -2}\right ) n -a_{n -8}+96 a_{n -6}-1344 a_{n -4}}{5040 n \left (n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 3\) and after as more terms are found using the above recursive equation.


\(n\)	\(a_{n ,r}\)	\(a_{n}\)

\(a_{0}\)	\(1\)	\(1\)

\(a_{1}\)	\(0\)	\(0\)

\(a_{2}\)	\(\frac {r -6}{12+6 r}\)	\(-{\frac {1}{10}}\)

\(a_{3}\)	\(0\)	\(0\)

\(a_{4}\)	\(\frac {7 r^{3}-63 r^{2}+146 r +120}{360 \left (2+r \right ) \left (r +4\right ) \left (r +1\right )}\)	\(\frac {1}{280}\)

\(a_{5}\)	\(0\)	\(0\)

\(a_{6}\)	\(\frac {31 r^{5}-248 r^{4}+497 r^{3}+1508 r^{2}-6324 r -3024}{15120 \left (2+r \right ) \left (r +4\right ) \left (r +1\right ) \left (6+r \right ) \left (3+r \right )}\)	\(-{\frac {1}{15120}}\)

\(a_{7}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{3} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \right ) \\ &= x^{3} \left (1-\frac {x^{2}}{10}+\frac {x^{4}}{280}-\frac {x^{6}}{15120}+O\left (x^{6}\right )\right ) \end {align*}

Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the diﬀerence between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by ﬁnding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{3} \\ &= 0 \end {align*}

Therefore \begin {align*} \lim _{r\rightarrow r_{2}}0&= \lim _{r\rightarrow 0}0\\ &= 0 \end {align*}

The limit is \(0\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}

Eq (3) derived above is used to ﬁnd all \(b_{n}\) coeﬃcients. The case \(n = 0\) is skipped since it was used to ﬁnd the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives \[ b_{1} = 0 \] Substituting \(n = 2\) in Eq(3) gives \[ b_{2} = \frac {r -6}{12+6 r} \] Substituting \(n = 3\) in Eq(3) gives \[ b_{3} = 0 \] Substituting \(n = 4\) in Eq(3) gives \[ b_{4} = \frac {7 r^{3}-63 r^{2}+146 r +120}{360 \left (2+r \right ) \left (r^{2}+5 r +4\right )} \] Substituting \(n = 5\) in Eq(3) gives \[ b_{5} = 0 \] Substituting \(n = 6\) in Eq(3) gives \[ b_{6} = \frac {31 r^{5}-248 r^{4}+497 r^{3}+1508 r^{2}-6324 r -3024}{15120 \left (2+r \right ) \left (r^{2}+5 r +4\right ) \left (r^{2}+9 r +18\right )} \] Substituting \(n = 7\) in Eq(3) gives \[ b_{7} = 0 \] For \(8\le n\) the recursive equation is \begin{equation} \tag{4} -\frac {b_{n -6} \left (n +r -6\right ) \left (n -7+r \right )}{5040}+\frac {b_{n -4} \left (-4+n +r \right ) \left (n -5+r \right )}{120}-\frac {b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )}{6}+b_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {b_{n -6} \left (n +r -6\right )}{360}-\frac {b_{n -4} \left (-4+n +r \right )}{12}+b_{n -2} \left (n +r -2\right )-2 \left (n +r \right ) b_{n}-b_{n -2}+\frac {b_{n -4}}{6}-\frac {b_{n -6}}{120}+\frac {b_{n -8}}{5040} = 0 \end{equation} Which for for the root \(r = 0\) becomes \begin{equation} \tag{4A} -\frac {b_{n -6} \left (n -6\right ) \left (n -7\right )}{5040}+\frac {b_{n -4} \left (n -4\right ) \left (n -5\right )}{120}-\frac {b_{n -2} \left (n -2\right ) \left (n -3\right )}{6}+b_{n} n \left (n -1\right )+\frac {b_{n -6} \left (n -6\right )}{360}-\frac {b_{n -4} \left (n -4\right )}{12}+b_{n -2} \left (n -2\right )-2 n b_{n}-b_{n -2}+\frac {b_{n -4}}{6}-\frac {b_{n -6}}{120}+\frac {b_{n -8}}{5040} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {n^{2} b_{n -6}-42 n^{2} b_{n -4}+840 n^{2} b_{n -2}+2 n r b_{n -6}-84 n r b_{n -4}+1680 n r b_{n -2}+r^{2} b_{n -6}-42 r^{2} b_{n -4}+840 r^{2} b_{n -2}-27 n b_{n -6}+798 n b_{n -4}-9240 n b_{n -2}-27 r b_{n -6}+798 r b_{n -4}-9240 r b_{n -2}-b_{n -8}+168 b_{n -6}-3360 b_{n -4}+20160 b_{n -2}}{5040 n^{2}+10080 n r +5040 r^{2}-15120 n -15120 r}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {n^{2} b_{n -6}-42 n^{2} b_{n -4}+840 n^{2} b_{n -2}-27 n b_{n -6}+798 n b_{n -4}-9240 n b_{n -2}-b_{n -8}+168 b_{n -6}-3360 b_{n -4}+20160 b_{n -2}}{5040 n^{2}-15120 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.


\(n\)	\(b_{n ,r}\)	\(b_{n}\)

\(b_{0}\)	\(1\)	\(1\)

\(b_{1}\)	\(0\)	\(0\)

\(b_{2}\)	\(\frac {r -6}{12+6 r}\)	\(-{\frac {1}{2}}\)

\(b_{3}\)	\(0\)	\(0\)

\(b_{4}\)	\(\frac {7 r^{3}-63 r^{2}+146 r +120}{360 \left (2+r \right ) \left (r +4\right ) \left (r +1\right )}\)	\(\frac {1}{24}\)

\(b_{5}\)	\(0\)	\(0\)

\(b_{6}\)	\(\frac {31 r^{5}-248 r^{4}+497 r^{3}+1508 r^{2}-6324 r -3024}{15120 \left (2+r \right ) \left (r +4\right ) \left (r +1\right ) \left (6+r \right ) \left (3+r \right )}\)	\(-{\frac {1}{720}}\)

\(b_{7}\)	\(0\)	\(0\)

Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= 1-\frac {x^{2}}{2}+\frac {x^{4}}{24}-\frac {x^{6}}{720}+O\left (x^{6}\right ) \end {align*}

Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{3} \left (1-\frac {x^{2}}{10}+\frac {x^{4}}{280}-\frac {x^{6}}{15120}+O\left (x^{6}\right )\right ) + c_{2} \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}-\frac {x^{6}}{720}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the ﬁnal solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1-\frac {x^{2}}{10}+\frac {x^{4}}{280}-\frac {x^{6}}{15120}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}-\frac {x^{6}}{720}+O\left (x^{6}\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1-\frac {x^{2}}{10}+\frac {x^{4}}{280}-\frac {x^{6}}{15120}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}-\frac {x^{6}}{720}+O\left (x^{6}\right )\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} x^{3} \left (1-\frac {x^{2}}{10}+\frac {x^{4}}{280}-\frac {x^{6}}{15120}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x^{2}}{2}+\frac {x^{4}}{24}-\frac {x^{6}}{720}+O\left (x^{6}\right )\right ) \] Veriﬁed OK.

Maple trace Kovacic algorithm successful

`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 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 
   <- Kovacics algorithm successful 
   Change of variables used: 
      [x = arccos(t)] 
   Linear ODE actually solved: 
      -u(t)+t*diff(u(t),t)+(-t^2+1)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

\[ y \left (x \right ) = c_{1} x^{3} \left (1-\frac {1}{10} x^{2}+\frac {1}{280} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (12-6 x^{2}+\frac {1}{2} x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \]

\[ y(x)\to c_1 \left (\frac {x^4}{24}-\frac {x^2}{2}+1\right )+c_2 \left (\frac {x^7}{280}-\frac {x^5}{10}+x^3\right ) \]


\(p(x)=-\frac {2 \cos \left (x \right )}{\sin \left (x \right )}\)

singularity	type

\(x = \pi Z\)	\(\text {``regular''}\)

1.11 problem 3.24 (f)