Internal problem ID [6475]
Internal file name [OUTPUT/5723_Sunday_June_05_2022_03_49_11_PM_96073013/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 4.6. Gauss’s
Hypergeometric Equation. Page 187
Problem number: 5.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _homogeneous]]
\[ \boxed {\left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }+\frac {y^{\prime }}{2}+{\mathrm e}^{x} y=0} \] 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. \[ \left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }+\frac {y^{\prime }}{2}+{\mathrm e}^{x} y = 0 \] 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 {1}{2 \left ({\mathrm e}^{x}-1\right )}\\ q(x) &= -\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}-1}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([2 i \pi Z, 2 i \pi Z]\)
Irregular singular points : \([\infty ]\)
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} \left (1-{\mathrm e}^{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 )+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )}{2}+{\mathrm e}^{x} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(1-{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} 1-{\mathrm e}^{x} &= -x -\frac {1}{2} x^{2}-\frac {1}{6} x^{3}-\frac {1}{24} x^{4}-\frac {1}{120} x^{5}-\frac {1}{720} x^{6}-\frac {1}{5040} x^{7}-\frac {1}{40320} x^{8} + \dots \\ &= -x -\frac {1}{2} x^{2}-\frac {1}{6} x^{3}-\frac {1}{24} x^{4}-\frac {1}{120} x^{5}-\frac {1}{720} x^{6}-\frac {1}{5040} x^{7}-\frac {1}{40320} x^{8} \end {align*}
Expanding \({\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(8\) terms gives \begin {align*} {\mathrm e}^{x} &= 1+x +\frac {1}{2} x^{2}+\frac {1}{6} x^{3}+\frac {1}{24} x^{4}+\frac {1}{120} x^{5}+\frac {1}{720} x^{6}+\frac {1}{5040} x^{7}+\frac {1}{40320} x^{8} + \dots \\ &= 1+x +\frac {1}{2} x^{2}+\frac {1}{6} x^{3}+\frac {1}{24} x^{4}+\frac {1}{120} x^{5}+\frac {1}{720} x^{6}+\frac {1}{5040} x^{7}+\frac {1}{40320} x^{8} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +6} a_{n} \left (n +r \right ) \left (n +r -1\right )}{40320}\right )+\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 =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{720}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\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^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{24}\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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {\left (n +r \right ) a_{n} x^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +3} a_{n}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +4} a_{n}}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n}}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +7} a_{n}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{40320}\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 +6} a_{n} \left (n +r \right ) \left (n +r -1\right )}{40320}\right ) &= \moverset {\infty }{\munderset {n =7}{\sum }}\left (-\frac {a_{n -7} \left (n +r -7\right ) \left (n +r -8\right ) x^{n +r -1}}{40320}\right ) \\ \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 +r -7\right ) x^{n +r -1}}{5040}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{720}\right ) &= \moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} \left (-5+n +r \right ) \left (n +r -6\right ) x^{n +r -1}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (n +r -4\right ) \left (-5+n +r \right ) x^{n +r -1}}{120}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{24}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} \left (n +r -3\right ) \left (n +r -4\right ) x^{n +r -1}}{24}\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 ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}}{2}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +2} a_{n}}{2} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} x^{n +r -1}}{2} \\ \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 }}\frac {x^{n +r +4} a_{n}}{24} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +5} a_{n}}{120} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} x^{n +r -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +6} a_{n}}{720} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} x^{n +r -1}}{720} \\ \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} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {x^{n +r +8} a_{n}}{40320} &= \moverset {\infty }{\munderset {n =9}{\sum }}\frac {a_{n -9} x^{n +r -1}}{40320} \\ \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 =7}{\sum }}\left (-\frac {a_{n -7} \left (n +r -7\right ) \left (n +r -8\right ) x^{n +r -1}}{40320}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n +r -6\right ) \left (n +r -7\right ) x^{n +r -1}}{5040}\right )+\moverset {\infty }{\munderset {n =5}{\sum }}\left (-\frac {a_{n -5} \left (-5+n +r \right ) \left (n +r -6\right ) x^{n +r -1}}{720}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-\frac {a_{n -4} \left (n +r -4\right ) \left (-5+n +r \right ) x^{n +r -1}}{120}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-\frac {a_{n -3} \left (n +r -3\right ) \left (n +r -4\right ) x^{n +r -1}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}}{6}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}}{2}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {\left (n +r \right ) a_{n} x^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} x^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} x^{n +r -1}}{6}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} x^{n +r -1}}{24}\right )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} x^{n +r -1}}{120}\right )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {a_{n -7} x^{n +r -1}}{720}\right )+\left (\moverset {\infty }{\munderset {n =8}{\sum }}\frac {a_{n -8} x^{n +r -1}}{5040}\right )+\left (\moverset {\infty }{\munderset {n =9}{\sum }}\frac {a_{n -9} x^{n +r -1}}{40320}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {\left (n +r \right ) a_{n} x^{n +r -1}}{2} = 0 \] When \(n = 0\) the above becomes \[ -x^{-1+r} a_{0} r \left (-1+r \right )+\frac {r a_{0} x^{-1+r}}{2} = 0 \] Or \[ \left (-x^{-1+r} r \left (-1+r \right )+\frac {r \,x^{-1+r}}{2}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (\frac {3}{2}-r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r^{2}+\frac {3}{2} r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {3}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (\frac {3}{2}-r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {3}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {3}{2}}\) is not 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 ) &= 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 {3}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \[ a_{1} = \frac {-r +2}{2 r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{2}-2 r +3}{12 r^{2}-3} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {-4 r^{2}+5 r +12}{96 r^{3}+144 r^{2}-24 r -36} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-4 r^{4}+24 r^{3}-44 r^{2}+39 r +225}{2880 r^{4}+11520 r^{3}+10080 r^{2}-2880 r -2700} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {8 r^{4}-6 r^{3}-158 r^{2}+291 r +630}{360 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {32 r^{6}-96 r^{5}+224 r^{4}+948 r^{3}-8311 r^{2}+16653 r +39690}{15120 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ a_{7} = \frac {-128 r^{6}-864 r^{5}+2944 r^{4}-1176 r^{3}-51308 r^{2}+125943 r +249480}{60480 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right )} \] Substituting \(n = 8\) in Eq. (2B) gives \[ a_{8} = \frac {224 r^{9}+4128 r^{8}+28272 r^{7}+70968 r^{6}+47004 r^{5}+213042 r^{4}-779417 r^{3}-3196848 r^{2}+15434577 r +24324300}{453600 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right ) \left (2 r^{2}+29 r +104\right )} \] For \(9\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {a_{n -7} \left (n +r -7\right ) \left (n +r -8\right )}{40320}-\frac {a_{n -6} \left (n +r -6\right ) \left (n +r -7\right )}{5040}-\frac {a_{n -5} \left (-5+n +r \right ) \left (n +r -6\right )}{720}-\frac {a_{n -4} \left (n +r -4\right ) \left (-5+n +r \right )}{120}-\frac {a_{n -3} \left (n +r -3\right ) \left (n +r -4\right )}{24}-\frac {a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )}{6}-\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{2}-a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {a_{n} \left (n +r \right )}{2}+a_{n -1}+a_{n -2}+\frac {a_{n -3}}{2}+\frac {a_{n -4}}{6}+\frac {a_{n -5}}{24}+\frac {a_{n -6}}{120}+\frac {a_{n -7}}{720}+\frac {a_{n -8}}{5040}+\frac {a_{n -9}}{40320} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -7}+8 n^{2} a_{n -6}+56 n^{2} a_{n -5}+336 n^{2} a_{n -4}+1680 n^{2} a_{n -3}+6720 n^{2} a_{n -2}+20160 n^{2} a_{n -1}+2 n r a_{n -7}+16 n r a_{n -6}+112 n r a_{n -5}+672 n r a_{n -4}+3360 n r a_{n -3}+13440 n r a_{n -2}+40320 n r a_{n -1}+r^{2} a_{n -7}+8 r^{2} a_{n -6}+56 r^{2} a_{n -5}+336 r^{2} a_{n -4}+1680 r^{2} a_{n -3}+6720 r^{2} a_{n -2}+20160 r^{2} a_{n -1}-15 n a_{n -7}-104 n a_{n -6}-616 n a_{n -5}-3024 n a_{n -4}-11760 n a_{n -3}-33600 n a_{n -2}-60480 n a_{n -1}-15 r a_{n -7}-104 r a_{n -6}-616 r a_{n -5}-3024 r a_{n -4}-11760 r a_{n -3}-33600 r a_{n -2}-60480 r a_{n -1}-a_{n -9}-8 a_{n -8}}{20160 \left (2 n^{2}+4 n r +2 r^{2}-3 n -3 r \right )}\tag {4} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{n} = \frac {\left (-80640 a_{n -1}-4 a_{n -7}-32 a_{n -6}-224 a_{n -5}-1344 a_{n -4}-6720 a_{n -3}-26880 a_{n -2}\right ) n^{2}+\left (48 a_{n -7}+320 a_{n -6}+1792 a_{n -5}+8064 a_{n -4}+26880 a_{n -3}+53760 a_{n -2}\right ) n +181440 a_{n -1}+4 a_{n -9}+32 a_{n -8}+81 a_{n -7}+552 a_{n -6}+3192 a_{n -5}+15120 a_{n -4}+55440 a_{n -3}+141120 a_{n -2}}{161280 n^{2}+241920 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {3}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r +2}{2 r -1}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {r^{2}-2 r +3}{12 r^{2}-3}\) | \(\frac {3}{32}\) |
| \(a_{3}\) | \(\frac {-4 r^{2}+5 r +12}{96 r^{3}+144 r^{2}-24 r -36}\) | \(\frac {7}{384}\) |
| \(a_{4}\) | \(\frac {-4 r^{4}+24 r^{3}-44 r^{2}+39 r +225}{2880 r^{4}+11520 r^{3}+10080 r^{2}-2880 r -2700}\) | \(\frac {109}{30720}\) |
| \(a_{5}\) | \(\frac {8 r^{4}-6 r^{3}-158 r^{2}+291 r +630}{360 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right )}\) | \(\frac {13}{24576}\) |
| \(a_{6}\) | \(\frac {32 r^{6}-96 r^{5}+224 r^{4}+948 r^{3}-8311 r^{2}+16653 r +39690}{15120 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right )}\) | \(\frac {4439}{61931520}\) |
| \(a_{7}\) | \(\frac {-128 r^{6}-864 r^{5}+2944 r^{4}-1176 r^{3}-51308 r^{2}+125943 r +249480}{60480 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right )}\) | \(\frac {2069}{247726080}\) |
| \(a_{8}\) | \(\frac {224 r^{9}+4128 r^{8}+28272 r^{7}+70968 r^{6}+47004 r^{5}+213042 r^{4}-779417 r^{3}-3196848 r^{2}+15434577 r +24324300}{453600 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right ) \left (2 r^{2}+29 r +104\right )}\) | \(\frac {685613}{753087283200}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{{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^{{3}/{2}} \left (1+\frac {x}{4}+\frac {3 x^{2}}{32}+\frac {7 x^{3}}{384}+\frac {109 x^{4}}{30720}+\frac {13 x^{5}}{24576}+\frac {4439 x^{6}}{61931520}+\frac {2069 x^{7}}{247726080}+\frac {685613 x^{8}}{753087283200}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now 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. (2B) gives \[ b_{1} = \frac {-r +2}{2 r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {r^{2}-2 r +3}{12 r^{2}-3} \] Substituting \(n = 3\) in Eq. (2B) gives \[ b_{3} = \frac {-4 r^{2}+5 r +12}{96 r^{3}+144 r^{2}-24 r -36} \] Substituting \(n = 4\) in Eq. (2B) gives \[ b_{4} = \frac {-4 r^{4}+24 r^{3}-44 r^{2}+39 r +225}{2880 r^{4}+11520 r^{3}+10080 r^{2}-2880 r -2700} \] Substituting \(n = 5\) in Eq. (2B) gives \[ b_{5} = \frac {8 r^{4}-6 r^{3}-158 r^{2}+291 r +630}{360 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right )} \] Substituting \(n = 6\) in Eq. (2B) gives \[ b_{6} = \frac {32 r^{6}-96 r^{5}+224 r^{4}+948 r^{3}-8311 r^{2}+16653 r +39690}{15120 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right )} \] Substituting \(n = 7\) in Eq. (2B) gives \[ b_{7} = \frac {-128 r^{6}-864 r^{5}+2944 r^{4}-1176 r^{3}-51308 r^{2}+125943 r +249480}{60480 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right )} \] Substituting \(n = 8\) in Eq. (2B) gives \[ b_{8} = \frac {224 r^{9}+4128 r^{8}+28272 r^{7}+70968 r^{6}+47004 r^{5}+213042 r^{4}-779417 r^{3}-3196848 r^{2}+15434577 r +24324300}{453600 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right ) \left (2 r^{2}+29 r +104\right )} \] For \(9\le n\) the recursive equation is \begin{equation} \tag{3} -\frac {b_{n -7} \left (n +r -7\right ) \left (n +r -8\right )}{40320}-\frac {b_{n -6} \left (n +r -6\right ) \left (n +r -7\right )}{5040}-\frac {b_{n -5} \left (-5+n +r \right ) \left (n +r -6\right )}{720}-\frac {b_{n -4} \left (n +r -4\right ) \left (-5+n +r \right )}{120}-\frac {b_{n -3} \left (n +r -3\right ) \left (n +r -4\right )}{24}-\frac {b_{n -2} \left (n +r -2\right ) \left (n +r -3\right )}{6}-\frac {b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{2}-b_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {\left (n +r \right ) b_{n}}{2}+b_{n -1}+b_{n -2}+\frac {b_{n -3}}{2}+\frac {b_{n -4}}{6}+\frac {b_{n -5}}{24}+\frac {b_{n -6}}{120}+\frac {b_{n -7}}{720}+\frac {b_{n -8}}{5040}+\frac {b_{n -9}}{40320} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {n^{2} b_{n -7}+8 n^{2} b_{n -6}+56 n^{2} b_{n -5}+336 n^{2} b_{n -4}+1680 n^{2} b_{n -3}+6720 n^{2} b_{n -2}+20160 n^{2} b_{n -1}+2 n r b_{n -7}+16 n r b_{n -6}+112 n r b_{n -5}+672 n r b_{n -4}+3360 n r b_{n -3}+13440 n r b_{n -2}+40320 n r b_{n -1}+r^{2} b_{n -7}+8 r^{2} b_{n -6}+56 r^{2} b_{n -5}+336 r^{2} b_{n -4}+1680 r^{2} b_{n -3}+6720 r^{2} b_{n -2}+20160 r^{2} b_{n -1}-15 n b_{n -7}-104 n b_{n -6}-616 n b_{n -5}-3024 n b_{n -4}-11760 n b_{n -3}-33600 n b_{n -2}-60480 n b_{n -1}-15 r b_{n -7}-104 r b_{n -6}-616 r b_{n -5}-3024 r b_{n -4}-11760 r b_{n -3}-33600 r b_{n -2}-60480 r b_{n -1}-b_{n -9}-8 b_{n -8}}{20160 \left (2 n^{2}+4 n r +2 r^{2}-3 n -3 r \right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {\left (-20160 b_{n -1}-b_{n -7}-8 b_{n -6}-56 b_{n -5}-336 b_{n -4}-1680 b_{n -3}-6720 b_{n -2}\right ) n^{2}+\left (60480 b_{n -1}+15 b_{n -7}+104 b_{n -6}+616 b_{n -5}+3024 b_{n -4}+11760 b_{n -3}+33600 b_{n -2}\right ) n +b_{n -9}+8 b_{n -8}}{40320 n^{2}-60480 n}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r +2}{2 r -1}\) | \(-2\) |
| \(b_{2}\) | \(\frac {r^{2}-2 r +3}{12 r^{2}-3}\) | \(-1\) |
| \(b_{3}\) | \(\frac {-4 r^{2}+5 r +12}{96 r^{3}+144 r^{2}-24 r -36}\) | \(-{\frac {1}{3}}\) |
| \(b_{4}\) | \(\frac {-4 r^{4}+24 r^{3}-44 r^{2}+39 r +225}{2880 r^{4}+11520 r^{3}+10080 r^{2}-2880 r -2700}\) | \(-{\frac {1}{12}}\) |
| \(b_{5}\) | \(\frac {8 r^{4}-6 r^{3}-158 r^{2}+291 r +630}{360 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right )}\) | \(-{\frac {1}{60}}\) |
| \(b_{6}\) | \(\frac {32 r^{6}-96 r^{5}+224 r^{4}+948 r^{3}-8311 r^{2}+16653 r +39690}{15120 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right )}\) | \(-{\frac {1}{360}}\) |
| \(b_{7}\) | \(\frac {-128 r^{6}-864 r^{5}+2944 r^{4}-1176 r^{3}-51308 r^{2}+125943 r +249480}{60480 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right )}\) | \(-{\frac {1}{2520}}\) |
| \(b_{8}\) | \(\frac {224 r^{9}+4128 r^{8}+28272 r^{7}+70968 r^{6}+47004 r^{5}+213042 r^{4}-779417 r^{3}-3196848 r^{2}+15434577 r +24324300}{453600 \left (16 r^{4}+64 r^{3}+56 r^{2}-16 r -15\right ) \left (2 r +7\right ) \left (2 r +9\right ) \left (2 r +11\right ) \left (2 r^{2}+29 r +104\right )}\) | \(-{\frac {1}{20160}}\) |
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}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= 1-2 x -x^{2}-\frac {x^{3}}{3}-\frac {x^{4}}{12}-\frac {x^{5}}{60}-\frac {x^{6}}{360}-\frac {x^{7}}{2520}-\frac {x^{8}}{20160}+O\left (x^{8}\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}/{2}} \left (1+\frac {x}{4}+\frac {3 x^{2}}{32}+\frac {7 x^{3}}{384}+\frac {109 x^{4}}{30720}+\frac {13 x^{5}}{24576}+\frac {4439 x^{6}}{61931520}+\frac {2069 x^{7}}{247726080}+\frac {685613 x^{8}}{753087283200}+O\left (x^{8}\right )\right ) + c_{2} \left (1-2 x -x^{2}-\frac {x^{3}}{3}-\frac {x^{4}}{12}-\frac {x^{5}}{60}-\frac {x^{6}}{360}-\frac {x^{7}}{2520}-\frac {x^{8}}{20160}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{{3}/{2}} \left (1+\frac {x}{4}+\frac {3 x^{2}}{32}+\frac {7 x^{3}}{384}+\frac {109 x^{4}}{30720}+\frac {13 x^{5}}{24576}+\frac {4439 x^{6}}{61931520}+\frac {2069 x^{7}}{247726080}+\frac {685613 x^{8}}{753087283200}+O\left (x^{8}\right )\right )+c_{2} \left (1-2 x -x^{2}-\frac {x^{3}}{3}-\frac {x^{4}}{12}-\frac {x^{5}}{60}-\frac {x^{6}}{360}-\frac {x^{7}}{2520}-\frac {x^{8}}{20160}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{{3}/{2}} \left (1+\frac {x}{4}+\frac {3 x^{2}}{32}+\frac {7 x^{3}}{384}+\frac {109 x^{4}}{30720}+\frac {13 x^{5}}{24576}+\frac {4439 x^{6}}{61931520}+\frac {2069 x^{7}}{247726080}+\frac {685613 x^{8}}{753087283200}+O\left (x^{8}\right )\right )+c_{2} \left (1-2 x -x^{2}-\frac {x^{3}}{3}-\frac {x^{4}}{12}-\frac {x^{5}}{60}-\frac {x^{6}}{360}-\frac {x^{7}}{2520}-\frac {x^{8}}{20160}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{{3}/{2}} \left (1+\frac {x}{4}+\frac {3 x^{2}}{32}+\frac {7 x^{3}}{384}+\frac {109 x^{4}}{30720}+\frac {13 x^{5}}{24576}+\frac {4439 x^{6}}{61931520}+\frac {2069 x^{7}}{247726080}+\frac {685613 x^{8}}{753087283200}+O\left (x^{8}\right )\right )+c_{2} \left (1-2 x -x^{2}-\frac {x^{3}}{3}-\frac {x^{4}}{12}-\frac {x^{5}}{60}-\frac {x^{6}}{360}-\frac {x^{7}}{2520}-\frac {x^{8}}{20160}+O\left (x^{8}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.281 (sec). Leaf size: 52
Order:=8; dsolve((1-exp(x))*diff(y(x),x$2)+1/2*diff(y(x),x)+exp(x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {3}{2}} \left (1+\frac {1}{4} x +\frac {3}{32} x^{2}+\frac {7}{384} x^{3}+\frac {109}{30720} x^{4}+\frac {13}{24576} x^{5}+\frac {4439}{61931520} x^{6}+\frac {2069}{247726080} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (1-2 x -x^{2}-\frac {1}{3} x^{3}-\frac {1}{12} x^{4}-\frac {1}{60} x^{5}-\frac {1}{360} x^{6}-\frac {1}{2520} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 109
AsymptoticDSolveValue[(1-Exp[x])*y''[x]+1/2*y'[x]+Exp[x]*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 \left (-\frac {x^7}{2520}-\frac {x^6}{360}-\frac {x^5}{60}-\frac {x^4}{12}-\frac {x^3}{3}-x^2-2 x+1\right )+c_1 \left (\frac {2069 x^7}{247726080}+\frac {4439 x^6}{61931520}+\frac {13 x^5}{24576}+\frac {109 x^4}{30720}+\frac {7 x^3}{384}+\frac {3 x^2}{32}+\frac {x}{4}+1\right ) x^{3/2} \]