Internal problem ID [1823]
Internal file name [OUTPUT/1824_Sunday_June_05_2022_02_34_10_AM_45666397/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.3, The method of Frobenius. Equal roots, and roots differering by an integer.
Page 223
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. Difference is integer"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {t y^{\prime \prime }+3 y^{\prime }-3 y=0} \] 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. \[ t y^{\prime \prime }+3 y^{\prime }-3 y = 0 \] 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 {3}{t}\\ q(t) &= -\frac {3}{t}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ t y^{\prime \prime }+3 y^{\prime }-3 y = 0 \] 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} t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )-3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \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 }}3 \left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 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 }}\left (-3 a_{n} t^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} t^{n +r -1}\right ) \\ \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 =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 }}3 \left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 \left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right )+3 r a_{0} t^{-1+r} = 0 \] Or \[ \left (t^{-1+r} r \left (-1+r \right )+3 r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,t^{-1+r} \left (2+r \right ) = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,t^{-1+r} \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -2]\).
Since \(r_1 - r_2 = 2\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= C y_{1}\left (t \right ) \ln \left (t \right )+t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ y_{2}\left (t \right ) &= C y_{1}\left (t \right ) \ln \left (t \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}}{t^{2}} \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ y_{2}\left (t \right ) &= C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -2}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n} \left (n +r \right )-3 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {3 a_{n -1}}{n^{2}+2 n r +r^{2}+2 n +2 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {3 a_{n -1}}{n \left (n +2\right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {3}{r^{2}+4 r +3} \] Which for the root \(r = 0\) becomes \[ a_{1}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3}{r^{2}+4 r +3}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {3}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3}{r^{2}+4 r +3}\) | \(1\) |
| \(a_{2}\) | \(\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )}\) | \(\frac {3}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {27}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right ) \left (2+r \right ) \left (r +5\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {3}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3}{r^{2}+4 r +3}\) | \(1\) |
| \(a_{2}\) | \(\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )}\) | \(\frac {3}{8}\) |
| \(a_{3}\) | \(\frac {27}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right ) \left (2+r \right ) \left (r +5\right )}\) | \(\frac {3}{40}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right )^{2} \left (2+r \right ) \left (r +5\right ) \left (r +6\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {3}{320}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3}{r^{2}+4 r +3}\) | \(1\) |
| \(a_{2}\) | \(\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )}\) | \(\frac {3}{8}\) |
| \(a_{3}\) | \(\frac {27}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right ) \left (2+r \right ) \left (r +5\right )}\) | \(\frac {3}{40}\) |
| \(a_{4}\) | \(\frac {81}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right )^{2} \left (2+r \right ) \left (r +5\right ) \left (r +6\right )}\) | \(\frac {3}{320}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {243}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right )^{2} \left (2+r \right ) \left (r +5\right )^{2} \left (r +6\right ) \left (7+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {9}{11200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3}{r^{2}+4 r +3}\) | \(1\) |
| \(a_{2}\) | \(\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )}\) | \(\frac {3}{8}\) |
| \(a_{3}\) | \(\frac {27}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right ) \left (2+r \right ) \left (r +5\right )}\) | \(\frac {3}{40}\) |
| \(a_{4}\) | \(\frac {81}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right )^{2} \left (2+r \right ) \left (r +5\right ) \left (r +6\right )}\) | \(\frac {3}{320}\) |
| \(a_{5}\) | \(\frac {243}{\left (r +3\right )^{2} \left (r +1\right ) \left (r +4\right )^{2} \left (2+r \right ) \left (r +5\right )^{2} \left (r +6\right ) \left (7+r \right )}\) | \(\frac {9}{11200}\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \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 {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{2} \\ &= \frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )}&= \lim _{r\rightarrow -2}\frac {9}{\left (r^{2}+4 r +3\right ) \left (r^{2}+6 r +8\right )}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (t \right ) = C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d t}y_{2}\left (t \right ) &= C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right ) \\ &= C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d t^{2}}y_{2}\left (t \right ) &= C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(t y^{\prime \prime }+3 y^{\prime }-3 y = 0\) gives \[ t \left (C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+3 C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {3 C y_{1}\left (t \right )}{t}+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )-3 C y_{1}\left (t \right ) \ln \left (t \right )-3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (t \right ) t -3 y_{1}\left (t \right )+3 y_{1}^{\prime }\left (t \right )\right ) \ln \left (t \right )+t \left (\frac {2 y_{1}^{\prime }\left (t \right )}{t}-\frac {y_{1}\left (t \right )}{t^{2}}\right )+\frac {3 y_{1}\left (t \right )}{t}\right ) C +t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )-3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (t \right )\) is a solution to the ode, then \[ y_{1}^{\prime \prime }\left (t \right ) t -3 y_{1}\left (t \right )+3 y_{1}^{\prime }\left (t \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (t \left (\frac {2 y_{1}^{\prime }\left (t \right )}{t}-\frac {y_{1}\left (t \right )}{t^{2}}\right )+\frac {3 y_{1}\left (t \right )}{t}\right ) C +t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )-3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) t +2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r_{1}}\right )\right ) C}{t}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) t^{2}-3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) t +3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) t}{t} = 0 \end{equation} Since \(r_{1} = 0\) and \(r_{2} = -2\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n -1} a_{n} n \right ) t +2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\right ) C}{t}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-4+n} b_{n} \left (n -2\right ) \left (-3+n \right )\right ) t^{2}-3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -2}\right ) t +3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-3+n} b_{n} \left (n -2\right )\right ) t}{t} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,t^{n -1} a_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,t^{n -1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-3+n} b_{n} \left (n^{2}-5 n +6\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 b_{n} t^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 t^{-3+n} b_{n} \left (n -2\right )\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(-3+n\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{-3+n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,t^{n -1} a_{n} n &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C \left (n -2\right ) a_{n -2} t^{-3+n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,t^{n -1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} t^{-3+n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 b_{n} t^{n -2}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 b_{n -1} t^{-3+n}\right ) \\ \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 \(-3+n\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C \left (n -2\right ) a_{n -2} t^{-3+n}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 C a_{n -2} t^{-3+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-3+n} b_{n} \left (n^{2}-5 n +6\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 b_{n -1} t^{-3+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 t^{-3+n} b_{n} \left (n -2\right )\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ -b_{1}-3 b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -b_{1}-3 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=-3 \] For \(n=N\), where \(N=2\) which is the difference between the two roots, we are free to choose \(b_{2} = 0\). Hence for \(n=2\), Eq (2B) gives \[ 2 C +9 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {9}{2}} \] For \(n=3\), Eq (2B) gives \[ 4 C a_{1}-3 b_{2}+3 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 3 b_{3}-18 = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=6 \] For \(n=4\), Eq (2B) gives \[ 6 C a_{2}-3 b_{3}+8 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 8 b_{4}-\frac {225}{8} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {225}{64}} \] For \(n=5\), Eq (2B) gives \[ 8 C a_{3}-3 b_{4}+15 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 15 b_{5}-\frac {4239}{320} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {1413}{1600}} \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (t \right ) = C y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) \] Using the above value found for \(C=-{\frac {9}{2}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (t \right )= -\frac {9}{2}\eslowast \left (1+t +\frac {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {1-3 t +6 t^{3}+\frac {225 t^{4}}{64}+\frac {1413 t^{5}}{1600}+O\left (t^{6}\right )}{t^{2}} \] 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 {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right )\right ) + c_{2} \left (-\frac {9}{2}\eslowast \left (1+t +\frac {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right )\right ) \ln \left (t \right )+\frac {1-3 t +6 t^{3}+\frac {225 t^{4}}{64}+\frac {1413 t^{5}}{1600}+O\left (t^{6}\right )}{t^{2}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+t +\frac {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right )\right )+c_{2} \left (\left (-\frac {9}{2}-\frac {9 t}{2}-\frac {27 t^{2}}{16}-\frac {27 t^{3}}{80}-\frac {27 t^{4}}{640}-\frac {81 t^{5}}{22400}-\frac {9 O\left (t^{6}\right )}{2}\right ) \ln \left (t \right )+\frac {1-3 t +6 t^{3}+\frac {225 t^{4}}{64}+\frac {1413 t^{5}}{1600}+O\left (t^{6}\right )}{t^{2}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+t +\frac {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right )\right )+c_{2} \left (\left (-\frac {9}{2}-\frac {9 t}{2}-\frac {27 t^{2}}{16}-\frac {27 t^{3}}{80}-\frac {27 t^{4}}{640}-\frac {81 t^{5}}{22400}-\frac {9 O\left (t^{6}\right )}{2}\right ) \ln \left (t \right )+\frac {1-3 t +6 t^{3}+\frac {225 t^{4}}{64}+\frac {1413 t^{5}}{1600}+O\left (t^{6}\right )}{t^{2}}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+t +\frac {3 t^{2}}{8}+\frac {3 t^{3}}{40}+\frac {3 t^{4}}{320}+\frac {9 t^{5}}{11200}+O\left (t^{6}\right )\right )+c_{2} \left (\left (-\frac {9}{2}-\frac {9 t}{2}-\frac {27 t^{2}}{16}-\frac {27 t^{3}}{80}-\frac {27 t^{4}}{640}-\frac {81 t^{5}}{22400}-\frac {9 O\left (t^{6}\right )}{2}\right ) \ln \left (t \right )+\frac {1-3 t +6 t^{3}+\frac {225 t^{4}}{64}+\frac {1413 t^{5}}{1600}+O\left (t^{6}\right )}{t^{2}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t y^{\prime \prime }+3 y^{\prime }-3 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {3 y}{t}-\frac {3 y^{\prime }}{t} \\ \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} \\ {} & {} & y^{\prime \prime }+\frac {3 y^{\prime }}{t}-\frac {3 y}{t}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {3}{t}, P_{3}\left (t \right )=-\frac {3}{t}\right ] \\ {} & \circ & t \cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t \cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=3 \\ {} & \circ & t^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =0 \\ {} & {} & \left (t^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}0}}}=0 \\ {} & \circ & t =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t y^{\prime \prime }+3 y^{\prime }-3 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & t \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (2+r \right ) t^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )-3 a_{k}\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )-3 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {3 a_{k}}{\left (k +1+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +1}=\frac {3 a_{k}}{\left (k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {3 a_{k}}{\left (k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {3 a_{k}}{\left (k +1\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +1}=\frac {3 a_{k}}{\left (k +1\right ) \left (k +3\right )}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- 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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 62
Order:=6; dsolve(t*diff(y(t),t$2)+3*diff(y(t),t)-3*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = \frac {c_{1} \left (1+t +\frac {3}{8} t^{2}+\frac {3}{40} t^{3}+\frac {3}{320} t^{4}+\frac {9}{11200} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) t^{2}+c_{2} \left (\ln \left (t \right ) \left (9 t^{2}+9 t^{3}+\frac {27}{8} t^{4}+\frac {27}{40} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\left (-2+6 t -12 t^{3}-\frac {225}{32} t^{4}-\frac {1413}{800} t^{5}+\operatorname {O}\left (t^{6}\right )\right )\right )}{t^{2}} \]
✓ Solution by Mathematica
Time used: 0.021 (sec). Leaf size: 78
AsymptoticDSolveValue[t*y''[t]+3*y'[t]-3*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_2 \left (\frac {3 t^4}{320}+\frac {3 t^3}{40}+\frac {3 t^2}{8}+t+1\right )+c_1 \left (\frac {279 t^4+528 t^3+144 t^2-192 t+64}{64 t^2}-\frac {9}{16} \left (3 t^2+8 t+8\right ) \log (t)\right ) \]