Internal problem ID [6448]
Internal file name [OUTPUT/5696_Sunday_June_05_2022_03_47_30_PM_15278874/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.4. REGULAR
SINGULAR POINTS. Page 175
Problem number: 2(d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{3} y^{\prime \prime }+\sin \left (x \right ) y=0} \] 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. \[ x^{3} y^{\prime \prime }+\sin \left (x \right ) y = 0 \] 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) &= 0\\ q(x) &= \frac {\sin \left (x \right )}{x^{3}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{3} y^{\prime \prime }+\sin \left (x \right ) y = 0 \] 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} x^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\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 \(8\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7}+\frac {1}{362880} x^{9} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}-\frac {1}{5040} x^{7}+\frac {1}{362880} x^{9} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +7} a_{n}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +9} a_{n}}{362880}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(1+n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{1+n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n}}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} x^{1+n +r}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{1+n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +7} a_{n}}{5040}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{1+n +r}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +9} a_{n}}{362880} &= \moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{1+n +r}}{362880} \\ \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 \(1+n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} x^{1+n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{1+n +r}}{120}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} x^{1+n +r}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{1+n +r}}{362880}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{1+n +r} a_{n} = 0 \] When \(n = 0\) the above becomes \[ x^{1+r} a_{0} r \left (-1+r \right )+x^{1+r} a_{0} = 0 \] Or \[ \left (x^{1+r} r \left (-1+r \right )+x^{1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-r +1\right ) x^{1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {1}{2}+\frac {i \sqrt {3}}{2}\\ r_2 &= \frac {1}{2}-\frac {i \sqrt {3}}{2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-r +1\right ) x^{1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {1}{2}+\frac {i \sqrt {3}}{2}, \frac {1}{2}-\frac {i \sqrt {3}}{2}\right ]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}+\frac {i \sqrt {3}}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}-\frac {i \sqrt {3}}{2}} \end {align*}
\(y_{1}\left (x \right )\) is found first. 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 \[ a_{2} = \frac {1}{6 r^{2}+18 r +18} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = 0 \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-3 r^{2}-9 r +1}{360 \left (r^{2}+3 r +3\right ) \left (r^{2}+7 r +13\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = 0 \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {3 r^{4}+30 r^{3}+69 r^{2}-30 r -149}{15120 \left (r^{2}+3 r +3\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+11 r +31\right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = 0 \] For \(8\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n}-\frac {a_{n -2}}{6}+\frac {a_{n -4}}{120}-\frac {a_{n -6}}{5040}+\frac {a_{n -8}}{362880} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {-60480 a_{n -2}+3024 a_{n -4}-72 a_{n -6}+a_{n -8}}{362880 \left (n^{2}+2 n r +r^{2}-n -r +1\right )}\tag {4} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{n} = \frac {60480 a_{n -2}-3024 a_{n -4}+72 a_{n -6}-a_{n -8}}{362880 n \left (i \sqrt {3}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {1}{6 r^{2}+18 r +18}\) | \(\frac {1}{12 i \sqrt {3}+24}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {-3 r^{2}-9 r +1}{360 \left (r^{2}+3 r +3\right ) \left (r^{2}+7 r +13\right )}\) | \(\frac {-3 i \sqrt {3}-1}{1440 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right )}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(\frac {3 r^{4}+30 r^{3}+69 r^{2}-30 r -149}{15120 \left (r^{2}+3 r +3\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+11 r +31\right )}\) | \(\frac {9 i \sqrt {3}-115}{362880 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+6\right )}\) |
| \(a_{7}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{12 i \sqrt {3}+24}+\frac {\left (-3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right )}+\frac {\left (9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (x \right )\) is found by taking the complex conjugate of \(y_{1}\left (x \right )\) which gives \[ y_{2}\left (x \right )= x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{-12 i \sqrt {3}+24}+\frac {\left (3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right )}+\frac {\left (-9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right ) \left (-i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) \] 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^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{12 i \sqrt {3}+24}+\frac {\left (-3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right )}+\frac {\left (9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) + c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{-12 i \sqrt {3}+24}+\frac {\left (3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right )}+\frac {\left (-9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right ) \left (-i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{12 i \sqrt {3}+24}+\frac {\left (-3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right )}+\frac {\left (9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{-12 i \sqrt {3}+24}+\frac {\left (3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right )}+\frac {\left (-9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right ) \left (-i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{12 i \sqrt {3}+24}+\frac {\left (-3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right )}+\frac {\left (9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{-12 i \sqrt {3}+24}+\frac {\left (3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right )}+\frac {\left (-9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right ) \left (-i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{12 i \sqrt {3}+24}+\frac {\left (-3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right )}+\frac {\left (9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (i \sqrt {3}+2\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1+\frac {x^{2}}{-12 i \sqrt {3}+24}+\frac {\left (3 i \sqrt {3}-1\right ) x^{4}}{1440 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right )}+\frac {\left (-9 i \sqrt {3}-115\right ) x^{6}}{362880 \left (-i \sqrt {3}+2\right ) \left (-i \sqrt {3}+4\right ) \left (-i \sqrt {3}+6\right )}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime }\right )+\sin \left (x \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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\sin \left (x \right ) y}{x^{3}} \\ \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 }+\frac {\sin \left (x \right ) y}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d x}y^{\prime }\right )+\sin \left (x \right ) y=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )+\sin \left (x \right ) y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )-\frac {d}{d t}y \left (t \right )\right )+\sin \left (x \right ) y \left (t \right )=0 \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}y \left (t \right )=-\frac {\sin \left (x \right ) y \left (t \right )}{x}+\frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\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 t}\frac {d}{d t}y \left (t \right )+\frac {\sin \left (x \right ) y \left (t \right )}{x}-\frac {d}{d t}y \left (t \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {\sin \left (x \right )}{x}-r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {r^{2} x -r x +\sin \left (x \right )}{x}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\frac {x +\sqrt {x^{2}-4 x \sin \left (x \right )}}{2 x}, -\frac {-x +\sqrt {x^{2}-4 x \sin \left (x \right )}}{2 x}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\frac {\left (x +\sqrt {x^{2}-4 x \sin \left (x \right )}\right ) t}{2 x}} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-\frac {\left (-x +\sqrt {x^{2}-4 x \sin \left (x \right )}\right ) t}{2 x}} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} {\mathrm e}^{\frac {\left (x +\sqrt {x^{2}-4 x \sin \left (x \right )}\right ) t}{2 x}}+c_{2} {\mathrm e}^{-\frac {\left (-x +\sqrt {x^{2}-4 x \sin \left (x \right )}\right ) t}{2 x}} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=c_{1} {\mathrm e}^{\frac {\left (x +\sqrt {x^{2}-4 x \sin \left (x \right )}\right ) \ln \left (x \right )}{2 x}}+c_{2} {\mathrm e}^{-\frac {\left (-x +\sqrt {x^{2}-4 x \sin \left (x \right )}\right ) \ln \left (x \right )}{2 x}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=x^{\frac {x +\sqrt {x \left (-4 \sin \left (x \right )+x \right )}}{2 x}} c_{1} +c_{2} x^{-\frac {-x +\sqrt {x \left (-4 \sin \left (x \right )+x \right )}}{2 x}} \end {array} \]
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 -> 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 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 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 <- unable to find a useful change of variables trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.172 (sec). Leaf size: 427
Order:=8; dsolve(x^3*diff(y(x),x$2)+sin(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (c_{2} x^{\frac {i \sqrt {3}}{2}} \left (1+\frac {1}{12 i \sqrt {3}+24} x^{2}+\frac {1}{1440} \frac {-3 i \sqrt {3}-1}{\left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+2\right )} x^{4}+\frac {1}{362880} \frac {9 i \sqrt {3}-115}{\left (i \sqrt {3}+6\right ) \left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+2\right )} x^{6}+\operatorname {O}\left (x^{8}\right )\right )+c_{1} x^{-\frac {i \sqrt {3}}{2}} \left (1-\frac {1}{12 i \sqrt {3}-24} x^{2}+\frac {-3 \sqrt {3}-i}{7200 i+8640 \sqrt {3}} x^{4}+\frac {9 \sqrt {3}-115 i}{4354560 i+14878080 \sqrt {3}} x^{6}+\operatorname {O}\left (x^{8}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 410
AsymptoticDSolveValue[x^3*y''[x]+Sin[x]*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {\left (\frac {1}{5040}-\frac {1}{720 \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right )}+\frac {\frac {1}{36 \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right )}-\frac {1}{120}}{6 \left (1+\left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right )\right )}\right ) x^6}{1+\left (5-(-1)^{2/3}\right ) \left (6-(-1)^{2/3}\right )}+\frac {\left (\frac {1}{36 \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right )}-\frac {1}{120}\right ) x^4}{1+\left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right )}+\frac {x^2}{6 \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right )}+1\right ) x^{-(-1)^{2/3}}+c_2 \left (\frac {\left (\frac {1}{5040}-\frac {1}{720 \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right )}+\frac {\frac {1}{36 \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right )}-\frac {1}{120}}{6 \left (1+\left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right )\right )}\right ) x^6}{1+\left (5+\sqrt [3]{-1}\right ) \left (6+\sqrt [3]{-1}\right )}+\frac {\left (\frac {1}{36 \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right )}-\frac {1}{120}\right ) x^4}{1+\left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right )}+\frac {x^2}{6 \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right )}+1\right ) x^{\sqrt [3]{-1}} \]