Internal problem ID [2924]
Internal file name [OUTPUT/2416_Sunday_June_05_2022_03_07_24_AM_191559/index.tex]
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 11.4. page
758
Problem number: 9.
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]]
\[ \boxed {x^{2} y^{\prime \prime }-x \cos \left (x \right ) y^{\prime }+5 y \,{\mathrm e}^{2 x}=0} \] 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. \[ x^{2} y^{\prime \prime }-x \cos \left (x \right ) y^{\prime }+5 y \,{\mathrm e}^{2 x} = 0 \] 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 {\cos \left (x \right )}{x}\\ q(x) &= \frac {5 \,{\mathrm e}^{2 x}}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }-x \cos \left (x \right ) y^{\prime }+5 y \,{\mathrm e}^{2 x} = 0 \] 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^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )-x \cos \left (x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) {\mathrm e}^{2 x} = 0 \end{equation} Expanding \(-\cos \left (x \right ) x\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} -\cos \left (x \right ) x &= -x +\frac {1}{2} x^{3}-\frac {1}{24} x^{5}+\frac {1}{720} x^{7} + \dots \\ &= -x +\frac {1}{2} x^{3}-\frac {1}{24} x^{5}+\frac {1}{720} x^{7} \end {align*}
Expanding \(5 \,{\mathrm e}^{2 x}\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} 5 \,{\mathrm e}^{2 x} &= 5+10 x +10 x^{2}+\frac {20}{3} x^{3}+\frac {10}{3} x^{4}+\frac {4}{3} x^{5}+\frac {4}{9} x^{6} + \dots \\ &= 5+10 x +10 x^{2}+\frac {20}{3} x^{3}+\frac {10}{3} x^{4}+\frac {4}{3} x^{5}+\frac {4}{9} x^{6} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n} \left (n +r \right )}{720}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n} \left (n +r \right )}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n} \left (n +r \right )}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {20 x^{n +r +3} a_{n}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {10 x^{n +r +4} a_{n}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {4 x^{n +r +5} a_{n}}{3}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {4 x^{n +r +6} a_{n}}{9}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n} \left (n +r \right )}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n -6+r \right ) x^{n +r}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n} \left (n +r \right )}{24}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{24}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n} \left (n +r \right )}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}10 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {20 x^{n +r +3} a_{n}}{3} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {20 a_{n -3} x^{n +r}}{3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {10 x^{n +r +4} a_{n}}{3} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {10 a_{n -4} x^{n +r}}{3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {4 x^{n +r +5} a_{n}}{3} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {4 a_{n -5} x^{n +r}}{3} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {4 x^{n +r +6} a_{n}}{9} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {4 a_{n -6} x^{n +r}}{9} \\ \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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} 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 -6+r \right ) x^{n +r}}{720}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (n -4+r \right ) x^{n +r}}{24}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) x^{n +r}}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}10 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}10 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {20 a_{n -3} x^{n +r}}{3}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {10 a_{n -4} x^{n +r}}{3}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {4 a_{n -5} x^{n +r}}{3}\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {4 a_{n -6} x^{n +r}}{9}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+5 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +5 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} r +5 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-2 r +5\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-2 r +5 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1+2 i\\ r_2 &= 1-2 i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-2 r +5\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1+2 i, 1-2 i]\).
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 +1+2 i}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +1-2 i} \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 \[ a_{1} = -\frac {10}{r^{2}+4} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {-r^{3}-20 r^{2}-4 r +120}{2 \left (r^{2}+4\right ) \left (r^{2}+2 r +5\right )} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {-\frac {10}{3} r^{3}+155 r^{2}+\frac {605}{3} r -\frac {20}{3} r^{4}-\frac {625}{3}}{\left (r^{2}+4\right ) \left (r^{2}+2 r +5\right ) \left (r^{2}+4 r +8\right )} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {r^{7}-68 r^{6}-179 r^{5}+4980 r^{4}+18204 r^{3}-8672 r^{2}-67056 r -25520}{24 \left (r^{2}+4\right ) \left (r^{2}+2 r +5\right ) \left (r^{2}+4 r +8\right ) \left (r^{2}+6 r +13\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {-16 r^{8}-122 r^{7}+1871 r^{6}+16327 r^{5}+12206 r^{4}-150483 r^{3}-334731 r^{2}-69372 r +174270}{12 \left (r^{2}+4\right ) \left (r^{2}+2 r +5\right ) \left (r^{2}+4 r +8\right ) \left (r^{2}+6 r +13\right ) \left (r^{2}+8 r +20\right )} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {a_{n -6} \left (n -6+r \right )}{720}-\frac {a_{n -4} \left (n -4+r \right )}{24}+\frac {a_{n -2} \left (n +r -2\right )}{2}-a_{n} \left (n +r \right )+5 a_{n}+10 a_{n -1}+10 a_{n -2}+\frac {20 a_{n -3}}{3}+\frac {10 a_{n -4}}{3}+\frac {4 a_{n -5}}{3}+\frac {4 a_{n -6}}{9} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -6}-30 n a_{n -4}+360 n a_{n -2}+r a_{n -6}-30 r a_{n -4}+360 r a_{n -2}+314 a_{n -6}+960 a_{n -5}+2520 a_{n -4}+4800 a_{n -3}+6480 a_{n -2}+7200 a_{n -1}}{720 \left (n^{2}+2 n r +r^{2}-2 n -2 r +5\right )}\tag {4} \] Which for the root \(r = 1+2 i\) becomes \[ a_{n} = \frac {\left (-a_{n -6}+30 a_{n -4}-360 a_{n -2}\right ) n +\left (-315-2 i\right ) a_{n -6}+\left (-2490+60 i\right ) a_{n -4}+\left (-6840-720 i\right ) a_{n -2}-960 a_{n -5}-4800 a_{n -3}-7200 a_{n -1}}{720 n \left (n +4 i\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1+2 i\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {10}{r^{2}+4}\) | \(-\frac {10}{17}+\frac {40 i}{17}\) |
| \(a_{2}\) | \(\frac {-r^{3}-20 r^{2}-4 r +120}{2 \left (r^{2}+4\right ) \left (r^{2}+2 r +5\right )}\) | \(-\frac {365}{136}-\frac {13 i}{17}\) |
| \(a_{3}\) | \(\frac {-\frac {10}{3} r^{3}+155 r^{2}+\frac {605}{3} r -\frac {20}{3} r^{4}-\frac {625}{3}}{\left (r^{2}+4\right ) \left (r^{2}+2 r +5\right ) \left (r^{2}+4 r +8\right )}\) | \(\frac {223}{1020}-\frac {1723 i}{765}\) |
| \(a_{4}\) | \(\frac {r^{7}-68 r^{6}-179 r^{5}+4980 r^{4}+18204 r^{3}-8672 r^{2}-67056 r -25520}{24 \left (r^{2}+4\right ) \left (r^{2}+2 r +5\right ) \left (r^{2}+4 r +8\right ) \left (r^{2}+6 r +13\right )}\) | \(\frac {114911}{78336}-\frac {24835 i}{78336}\) |
| \(a_{5}\) | \(\frac {-16 r^{8}-122 r^{7}+1871 r^{6}+16327 r^{5}+12206 r^{4}-150483 r^{3}-334731 r^{2}-69372 r +174270}{12 \left (r^{2}+4\right ) \left (r^{2}+2 r +5\right ) \left (r^{2}+4 r +8\right ) \left (r^{2}+6 r +13\right ) \left (r^{2}+8 r +20\right )}\) | \(\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{1+2 i} \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^{1+2 i} \left (1+\left (-\frac {10}{17}+\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}-\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}-\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}-\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\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 \[ y_{2}\left (x \right )= x^{1-2 i} \left (1+\left (-\frac {10}{17}-\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}+\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}+\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}+\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}-\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right ) \] 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^{1+2 i} \left (1+\left (-\frac {10}{17}+\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}-\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}-\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}-\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right ) + c_{2} x^{1-2 i} \left (1+\left (-\frac {10}{17}-\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}+\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}+\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}+\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}-\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{1+2 i} \left (1+\left (-\frac {10}{17}+\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}-\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}-\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}-\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{1-2 i} \left (1+\left (-\frac {10}{17}-\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}+\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}+\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}+\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}-\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{1+2 i} \left (1+\left (-\frac {10}{17}+\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}-\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}-\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}-\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{1-2 i} \left (1+\left (-\frac {10}{17}-\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}+\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}+\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}+\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}-\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{1+2 i} \left (1+\left (-\frac {10}{17}+\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}-\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}-\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}-\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{1-2 i} \left (1+\left (-\frac {10}{17}-\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}+\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}+\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}+\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}-\frac {1112267 i}{1605888}\right ) x^{5}+O\left (x^{6}\right )\right ) \] 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 -> 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 -> 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: 0.047 (sec). Leaf size: 71
Order:=6; dsolve(x^2*diff(y(x),x$2)-x*cos(x)*diff(y(x),x)+5*exp(2*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{1-2 i} \left (1+\left (-\frac {10}{17}-\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}+\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}+\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}+\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}-\frac {1112267 i}{1605888}\right ) x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{1+2 i} \left (1+\left (-\frac {10}{17}+\frac {40 i}{17}\right ) x +\left (-\frac {365}{136}-\frac {13 i}{17}\right ) x^{2}+\left (\frac {223}{1020}-\frac {1723 i}{765}\right ) x^{3}+\left (\frac {114911}{78336}-\frac {24835 i}{78336}\right ) x^{4}+\left (\frac {4041077}{8029440}+\frac {1112267 i}{1605888}\right ) x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 94
AsymptoticDSolveValue[x^2*y''[x]-x*Cos[x]*y'[x]+5*Exp[2*x]*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \left (\frac {11}{391680}+\frac {7 i}{391680}\right ) c_1 \left ((32064-31693 i) x^4-(30784+60608 i) x^3-(80352-23904 i) x^2+(23040+69120 i) x+(25344-16128 i)\right ) x^{1+2 i}+\left (\frac {7}{391680}+\frac {11 i}{391680}\right ) c_2 \left ((31693-32064 i) x^4+(60608+30784 i) x^3-(23904-80352 i) x^2-(69120+23040 i) x+(16128-25344 i)\right ) x^{1-2 i} \]