Internal problem ID [1796]
Internal file name [OUTPUT/1797_Sunday_June_05_2022_02_31_48_AM_78808701/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.2, Regular singular points, the method of Frobenius. Page 214
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left ({\mathrm e}^{t}-1\right ) y^{\prime \prime }+y^{\prime } {\mathrm e}^{t}+y=0} \] With the expansion point for the power series method at \(t = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left ({\mathrm e}^{t}-1\right ) y^{\prime \prime }+y^{\prime } {\mathrm e}^{t}+y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(t) y^{\prime } + q(t) y &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {{\mathrm e}^{t}}{{\mathrm e}^{t}-1}\\ q(t) &= \frac {1}{{\mathrm e}^{t}-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 \(t = 0\) is regular singular point, then Frobenius power series is used. Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right ) \left ({\mathrm e}^{t}-1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right ) {\mathrm e}^{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Expanding \({\mathrm e}^{t}-1\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} {\mathrm e}^{t}-1 &= t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6} + \dots \\ &= t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6} \end {align*}
Expanding \({\mathrm e}^{t}\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} {\mathrm e}^{t} &= 1+t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6} + \dots \\ &= 1+t +\frac {1}{2} t^{2}+\frac {1}{6} t^{3}+\frac {1}{24} t^{4}+\frac {1}{120} t^{5}+\frac {1}{720} t^{6} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +3} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +5} a_{n} \left (n +r \right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right )}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +3} a_{n} \left (n +r \right )}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +2} a_{n} \left (n +r \right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{1+n +r} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{720} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n +r -5\right ) \left (n +r -6\right ) t^{n +r -1}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{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 (n +r -4\right ) \left (n +r -5\right ) t^{n +r -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{24} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (-3+n +r \right ) \left (n +r -4\right ) t^{n +r -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) \left (-3+n +r \right ) t^{n +r -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )}{2} &= \moverset {\infty }{\munderset {n =1}{\sum }}\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +5} a_{n} \left (n +r \right )}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {a_{n -6} \left (n +r -6\right ) t^{n +r -1}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right )}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n +r -5\right ) t^{n +r -1}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +3} a_{n} \left (n +r \right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) t^{n +r -1}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +2} a_{n} \left (n +r \right )}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (-3+n +r \right ) t^{n +r -1}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{1+n +r} a_{n} \left (n +r \right )}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) t^{n +r -1}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n +r -5\right ) \left (n +r -6\right ) t^{n +r -1}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) \left (n +r -5\right ) t^{n +r -1}}{120}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (-3+n +r \right ) \left (n +r -4\right ) t^{n +r -1}}{24}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) \left (-3+n +r \right ) t^{n +r -1}}{6}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{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 ) t^{n +r -1}}{720}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {a_{n -5} \left (n +r -5\right ) t^{n +r -1}}{120}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n +r -4\right ) t^{n +r -1}}{24}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {a_{n -3} \left (-3+n +r \right ) t^{n +r -1}}{6}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {a_{n -2} \left (n +r -2\right ) t^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} t^{-1+r} = 0 \] Or \[ \left (t^{-1+r} r \left (-1+r \right )+r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ t^{-1+r} r^{2} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ t^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution \(y_{1}\left (t \right )\). 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 \[ a_{1} = \frac {-r^{2}-r -2}{2 \left (1+r \right )^{2}} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{4}+4 r^{3}+17 r^{2}+26 r +24}{12 \left (1+r \right )^{2} \left (2+r \right )^{2}} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {-5 r^{4}-30 r^{3}-79 r^{2}-102 r -72}{12 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2}} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {-r^{8}-16 r^{7}-46 r^{6}+344 r^{5}+3251 r^{4}+12056 r^{3}+24876 r^{2}+28656 r +17280}{720 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {-r^{8}-20 r^{7}-246 r^{6}-1940 r^{5}-9429 r^{4}-28040 r^{3}-50804 r^{2}-53520 r -28800}{240 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}} \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} \frac {a_{n -5} \left (n +r -5\right ) \left (n +r -6\right )}{720}+\frac {a_{n -4} \left (n +r -4\right ) \left (n +r -5\right )}{120}+\frac {a_{n -3} \left (-3+n +r \right ) \left (n +r -4\right )}{24}+\frac {a_{n -2} \left (n +r -2\right ) \left (-3+n +r \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 -6} \left (n +r -6\right )}{720}+\frac {a_{n -5} \left (n +r -5\right )}{120}+\frac {a_{n -4} \left (n +r -4\right )}{24}+\frac {a_{n -3} \left (-3+n +r \right )}{6}+\frac {a_{n -2} \left (n +r -2\right )}{2}+a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -5}+6 n^{2} a_{n -4}+30 n^{2} a_{n -3}+120 n^{2} a_{n -2}+360 n^{2} a_{n -1}+2 n r a_{n -5}+12 n r a_{n -4}+60 n r a_{n -3}+240 n r a_{n -2}+720 n r a_{n -1}+r^{2} a_{n -5}+6 r^{2} a_{n -4}+30 r^{2} a_{n -3}+120 r^{2} a_{n -2}+360 r^{2} a_{n -1}+n a_{n -6}-5 n a_{n -5}-24 n a_{n -4}-90 n a_{n -3}-240 n a_{n -2}-360 n a_{n -1}+r a_{n -6}-5 r a_{n -5}-24 r a_{n -4}-90 r a_{n -3}-240 r a_{n -2}-360 r a_{n -1}-6 a_{n -6}+720 a_{n -1}}{720 \left (n^{2}+2 n r +r^{2}\right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (-a_{n -5}-6 a_{n -4}-30 a_{n -3}-120 a_{n -2}-360 a_{n -1}\right ) n^{2}+\left (-a_{n -6}+5 a_{n -5}+24 a_{n -4}+90 a_{n -3}+240 a_{n -2}+360 a_{n -1}\right ) n +6 a_{n -6}-720 a_{n -1}}{720 n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-r -2}{2 \left (1+r \right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {r^{4}+4 r^{3}+17 r^{2}+26 r +24}{12 \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {-5 r^{4}-30 r^{3}-79 r^{2}-102 r -72}{12 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(-{\frac {1}{6}}\) |
| \(a_{4}\) | \(\frac {-r^{8}-16 r^{7}-46 r^{6}+344 r^{5}+3251 r^{4}+12056 r^{3}+24876 r^{2}+28656 r +17280}{720 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {-r^{8}-20 r^{7}-246 r^{6}-1940 r^{5}-9429 r^{4}-28040 r^{3}-50804 r^{2}-53520 r -28800}{240 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(-{\frac {1}{120}}\) |
Using the above table, then the first solution \(y_{1}\left (t \right )\) becomes \begin{align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \\ &= 1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (t \right ) = y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-r^{2}-r -2}{2 \left (1+r \right )^{2}}\) | \(-1\) | \(\frac {-r +3}{2 \left (1+r \right )^{3}}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {r^{4}+4 r^{3}+17 r^{2}+26 r +24}{12 \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(\frac {1}{2}\) | \(\frac {r^{4}-7 r^{3}-27 r^{2}-53 r -46}{6 \left (1+r \right )^{3} \left (2+r \right )^{3}}\) | \(-{\frac {23}{24}}\) |
| \(b_{3}\) | \(\frac {-5 r^{4}-30 r^{3}-79 r^{2}-102 r -72}{12 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(-{\frac {1}{6}}\) | \(\frac {5 r^{6}+45 r^{5}+193 r^{4}+504 r^{3}+864 r^{2}+951 r +486}{6 \left (1+r \right )^{3} \left (2+r \right )^{3} \left (3+r \right )^{3}}\) | \(\frac {3}{8}\) |
| \(b_{4}\) | \(\frac {-r^{8}-16 r^{7}-46 r^{6}+344 r^{5}+3251 r^{4}+12056 r^{3}+24876 r^{2}+28656 r +17280}{720 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2}}\) | \(\frac {1}{24}\) | \(\frac {-r^{10}-52 r^{9}-753 r^{8}-5964 r^{7}-31287 r^{6}-116490 r^{5}-312803 r^{4}-599366 r^{3}-794412 r^{2}-664488 r -260064}{180 \left (1+r \right )^{3} \left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3}}\) | \(-{\frac {301}{2880}}\) |
| \(b_{5}\) | \(\frac {-r^{8}-20 r^{7}-246 r^{6}-1940 r^{5}-9429 r^{4}-28040 r^{3}-50804 r^{2}-53520 r -28800}{240 \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(-{\frac {1}{120}}\) | \(\frac {r^{12}+30 r^{11}+557 r^{10}+7240 r^{9}+64365 r^{8}+393930 r^{7}+1694723 r^{6}+5216220 r^{5}+11579434 r^{4}+18427460 r^{3}+20359800 r^{2}+14195760 r +4680000}{120 \left (1+r \right )^{3} \left (2+r \right )^{3} \left (3+r \right )^{3} \left (4+r \right )^{3} \left (5+r \right )^{3}}\) | \(\frac {13}{576}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (t \right )&=y_{1}\left (t \right ) \ln \left (t \right )+b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}\dots \\ &= \left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {3 t}{2}-\frac {23 t^{2}}{24}+\frac {3 t^{3}}{8}-\frac {301 t^{4}}{2880}+\frac {13 t^{5}}{576}+O\left (t^{6}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) + c_{2} \left (\left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {3 t}{2}-\frac {23 t^{2}}{24}+\frac {3 t^{3}}{8}-\frac {301 t^{4}}{2880}+\frac {13 t^{5}}{576}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (\left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {3 t}{2}-\frac {23 t^{2}}{24}+\frac {3 t^{3}}{8}-\frac {301 t^{4}}{2880}+\frac {13 t^{5}}{576}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (\left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {3 t}{2}-\frac {23 t^{2}}{24}+\frac {3 t^{3}}{8}-\frac {301 t^{4}}{2880}+\frac {13 t^{5}}{576}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (\left (1-t +\frac {t^{2}}{2}-\frac {t^{3}}{6}+\frac {t^{4}}{24}-\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {3 t}{2}-\frac {23 t^{2}}{24}+\frac {3 t^{3}}{8}-\frac {301 t^{4}}{2880}+\frac {13 t^{5}}{576}+O\left (t^{6}\right )\right ) \] 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: [t = ln(t)] Linear ODE actually solved: u(t)+(2*t^2-t)*diff(u(t),t)+(t^3-t^2)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.25 (sec). Leaf size: 59
Order:=6; dsolve((exp(t)-1)*diff(y(t),t$2)+exp(t)*diff(y(t),t)+y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = \left (c_{2} \ln \left (t \right )+c_{1} \right ) \left (1-t +\frac {1}{2} t^{2}-\frac {1}{6} t^{3}+\frac {1}{24} t^{4}-\frac {1}{120} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\left (\frac {3}{2} t -\frac {23}{24} t^{2}+\frac {3}{8} t^{3}-\frac {301}{2880} t^{4}+\frac {13}{576} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 113
AsymptoticDSolveValue[(Exp[t]-1)*y''[t]+Exp[t]*y'[t]+y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (-\frac {t^5}{120}+\frac {t^4}{24}-\frac {t^3}{6}+\frac {t^2}{2}-t+1\right )+c_2 \left (\frac {13 t^5}{576}-\frac {301 t^4}{2880}+\frac {3 t^3}{8}-\frac {23 t^2}{24}+\left (-\frac {t^5}{120}+\frac {t^4}{24}-\frac {t^3}{6}+\frac {t^2}{2}-t+1\right ) \log (t)+\frac {3 t}{2}\right ) \]