Internal problem ID [5653]
Internal file name [OUTPUT/4901_Sunday_June_05_2022_03_09_50_PM_22191277/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number: 20.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {3 t \left (t +1\right ) y^{\prime \prime }+t y^{\prime }-y=0} \] 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. \[ \left (3 t^{2}+3 t \right ) y^{\prime \prime }+t y^{\prime }-y = 0 \] 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 {1}{3 t +3}\\ q(t) &= -\frac {1}{3 t \left (t +1\right )}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 3 t \left (t +1\right ) y^{\prime \prime }+t y^{\prime }-y = 0 \] 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} 3 t \left (t +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+t \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} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}3 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}3 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1} \\ \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 }}\left (-a_{n} t^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-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 =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ 3 t^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ 3 t^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 3 t^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ 3 r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 3 t^{-1+r} r \left (-1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 0]\).
Since \(r_1 - r_2 = 1\) 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 ) &= t \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 )+\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 +1}\\ 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}\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} 3 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (3 n^{2}+6 n r +3 r^{2}-8 n -8 r +4\right )}{3 \left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -\frac {a_{n -1} \left (3 n^{2}-2 n -1\right )}{3 \left (n +1\right ) n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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 \[ a_{1}=\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r} \] Which for the root \(r = 1\) becomes \[ a_{1}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9 r^{3}+6 r^{2}-11 r -4}{9 \left (1+r \right )^{2} \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {9 r^{3}+6 r^{2}-11 r -4}{9 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-27 r^{4}-81 r^{3}-9 r^{2}+89 r +28}{27 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {9 r^{3}+6 r^{2}-11 r -4}{9 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-27 r^{4}-81 r^{3}-9 r^{2}+89 r +28}{27 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r^{5}+513 r^{4}+837 r^{3}-177 r^{2}-974 r -280}{81 \left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {9 r^{3}+6 r^{2}-11 r -4}{9 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-27 r^{4}-81 r^{3}-9 r^{2}+89 r +28}{27 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {81 r^{5}+513 r^{4}+837 r^{3}-177 r^{2}-974 r -280}{81 \left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )^{2}}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-243 r^{6}-2592 r^{5}-9180 r^{4}-10350 r^{3}+5223 r^{2}+13502 r +3640}{243 \left (5+r \right ) \left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}\) | \(0\) |
| \(a_{2}\) | \(\frac {9 r^{3}+6 r^{2}-11 r -4}{9 \left (1+r \right )^{2} \left (2+r \right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-27 r^{4}-81 r^{3}-9 r^{2}+89 r +28}{27 \left (3+r \right ) \left (2+r \right )^{2} \left (1+r \right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {81 r^{5}+513 r^{4}+837 r^{3}-177 r^{2}-974 r -280}{81 \left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )^{2}}\) | \(0\) |
| \(a_{5}\) | \(\frac {-243 r^{6}-2592 r^{5}-9180 r^{4}-10350 r^{3}+5223 r^{2}+13502 r +3640}{243 \left (5+r \right ) \left (3+r \right ) \left (2+r \right ) \left (1+r \right ) \left (4+r \right )^{2}}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin {align*} y_{1}\left (t \right )&= t \left (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 \right ) \\ &= t \left (1+O\left (t^{6}\right )\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 \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). 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_{1} \\ &= \frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}&= \lim _{r\rightarrow 0}\frac {-3 r^{2}+2 r +1}{3 \left (1+r \right ) r}\\ &= \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 \(3 t \left (t +1\right ) y^{\prime \prime }+t y^{\prime }-y = 0\) gives \[ 3 t \left (t +1\right ) \left (C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+t \left (C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )\right )-C y_{1}\left (t \right ) \ln \left (t \right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (3 \left (t +1\right ) t y_{1}^{\prime \prime }\left (t \right )+y_{1}^{\prime }\left (t \right ) t -y_{1}\left (t \right )\right ) \ln \left (t \right )+3 \left (t +1\right ) t \left (\frac {2 y_{1}^{\prime }\left (t \right )}{t}-\frac {y_{1}\left (t \right )}{t^{2}}\right )+y_{1}\left (t \right )\right ) C +3 \left (t +1\right ) 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 )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )-\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 \[ 3 \left (t +1\right ) t y_{1}^{\prime \prime }\left (t \right )+y_{1}^{\prime }\left (t \right ) t -y_{1}\left (t \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (3 \left (t +1\right ) t \left (\frac {2 y_{1}^{\prime }\left (t \right )}{t}-\frac {y_{1}\left (t \right )}{t^{2}}\right )+y_{1}\left (t \right )\right ) C +3 \left (t +1\right ) 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 )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )-\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 (6 t \left (t +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )+\left (-3-2 t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r_{1}}\right )\right ) C}{t}+\frac {3 \left (t^{3}+t^{2}\right ) \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) t^{2}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) t}{t} = 0 \end{equation} Since \(r_{1} = 1\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (6 t \left (t +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n} a_{n} \left (n +1\right )\right )+\left (-3-2 t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +1}\right )\right ) C}{t}+\frac {3 \left (t^{3}+t^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-2+n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n -1} b_{n} n \right ) t^{2}-\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) t}{t} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}6 C \,t^{n +1} a_{n} \left (n +1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 C \,t^{n} a_{n} \left (n +1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 C a_{n} t^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,t^{n +1} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 t^{n} b_{n} n \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 n \,t^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n} b_{n} n \right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} t^{n}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}6 C \,t^{n +1} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}6 C a_{-2+n} \left (n -1\right ) t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 C \,t^{n} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}6 C a_{n -1} n \,t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 C a_{n} t^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 C a_{n -1} t^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,t^{n +1} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{-2+n} t^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 t^{n} b_{n} n \left (n -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} \left (-2+n \right ) t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}t^{n} b_{n} n &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-b_{n} t^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} t^{n -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 \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}6 C a_{-2+n} \left (n -1\right ) t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 C a_{n -1} n \,t^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-3 C a_{n -1} t^{n -1}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-2 C a_{-2+n} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 \left (n -1\right ) b_{n -1} \left (-2+n \right ) t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 n \,t^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} t^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-b_{n -1} t^{n -1}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=N\), where \(N=1\) which is the difference between the two roots, we are free to choose \(b_{1} = 0\). Hence for \(n=1\), Eq (2B) gives \[ 3 C -1 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C={\frac {1}{3}} \] For \(n=2\), Eq (2B) gives \[ \left (4 a_{0}+9 a_{1}\right ) C +6 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {4}{3}+6 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {2}{9}} \] For \(n=3\), Eq (2B) gives \[ \left (10 a_{1}+15 a_{2}\right ) C +7 b_{2}+18 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {14}{9}+18 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {7}{81}} \] For \(n=4\), Eq (2B) gives \[ \left (16 a_{2}+21 a_{3}\right ) C +20 b_{3}+36 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {140}{81}+36 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {35}{729}} \] For \(n=5\), Eq (2B) gives \[ \left (22 a_{3}+27 a_{4}\right ) C +39 b_{4}+60 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {455}{243}+60 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {91}{2916}} \] 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 \(C={\frac {1}{3}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (t \right )= \frac {1}{3}\eslowast \left (t \left (1+O\left (t^{6}\right )\right )\right ) \ln \left (t \right )+1-\frac {2 t^{2}}{9}+\frac {7 t^{3}}{81}-\frac {35 t^{4}}{729}+\frac {91 t^{5}}{2916}+O\left (t^{6}\right ) \] 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} t \left (1+O\left (t^{6}\right )\right ) + c_{2} \left (\frac {1}{3}\eslowast \left (t \left (1+O\left (t^{6}\right )\right )\right ) \ln \left (t \right )+1-\frac {2 t^{2}}{9}+\frac {7 t^{3}}{81}-\frac {35 t^{4}}{729}+\frac {91 t^{5}}{2916}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t \left (1+O\left (t^{6}\right )\right )+c_{2} \left (\frac {t \left (1+O\left (t^{6}\right )\right ) \ln \left (t \right )}{3}+1-\frac {2 t^{2}}{9}+\frac {7 t^{3}}{81}-\frac {35 t^{4}}{729}+\frac {91 t^{5}}{2916}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t \left (1+O\left (t^{6}\right )\right )+c_{2} \left (\frac {t \left (1+O\left (t^{6}\right )\right ) \ln \left (t \right )}{3}+1-\frac {2 t^{2}}{9}+\frac {7 t^{3}}{81}-\frac {35 t^{4}}{729}+\frac {91 t^{5}}{2916}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t \left (1+O\left (t^{6}\right )\right )+c_{2} \left (\frac {t \left (1+O\left (t^{6}\right )\right ) \ln \left (t \right )}{3}+1-\frac {2 t^{2}}{9}+\frac {7 t^{3}}{81}-\frac {35 t^{4}}{729}+\frac {91 t^{5}}{2916}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 t \left (t +1\right ) y^{\prime \prime }+t y^{\prime }-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 {y}{3 \left (t +1\right ) t}-\frac {y^{\prime }}{3 \left (t +1\right )} \\ \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 {y^{\prime }}{3 \left (t +1\right )}-\frac {y}{3 \left (t +1\right ) t}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (t \right )=\frac {1}{3 \left (t +1\right )}, P_{3}\left (t \right )=-\frac {1}{3 t \left (t +1\right )}\right ] \\ {} & \circ & \left (t +1\right )\cdot P_{2}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =-1 \\ {} & {} & \left (\left (t +1\right )\cdot P_{2}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}-1}}}=\frac {1}{3} \\ {} & \circ & \left (t +1\right )^{2}\cdot P_{3}\left (t \right )\textrm {is analytic at}\hspace {3pt} t =-1 \\ {} & {} & \left (\left (t +1\right )^{2}\cdot P_{3}\left (t \right )\right )\bigg | {\mstack {}{_{t \hiderel {=}-1}}}=0 \\ {} & \circ & t =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} t_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 t \left (t +1\right ) y^{\prime \prime }+t y^{\prime }-y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} t =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (3 u^{2}-3 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (u -1\right ) \left (\frac {d}{d u}y \left (u \right )\right )-y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-2+3 r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (3 k +3 r +1\right )+a_{k} \left (3 k +3 r +1\right ) \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-2+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +r +\frac {1}{3}\right ) \left (\left (-k -r -1\right ) a_{k +1}+a_{k} \left (k +r -1\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r -1\right )}{k +1+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -1\right )}{k +1} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (-u +1\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =t +1 \\ {} & {} & \left [y=-a_{0} t \right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -\frac {1}{3}\right )}{k +\frac {5}{3}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2}{3}}, a_{k +1}=\frac {a_{k} \left (k -\frac {1}{3}\right )}{k +\frac {5}{3}}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =t +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (t +1\right )^{k +\frac {2}{3}}, a_{k +1}=\frac {a_{k} \left (k -\frac {1}{3}\right )}{k +\frac {5}{3}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=-a_{0} t +\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (t +1\right )^{k +\frac {2}{3}}\right ), b_{k +1}=\frac {b_{k} \left (k -\frac {1}{3}\right )}{k +\frac {5}{3}}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 42
Order:=6; dsolve(3*t*(1+t)*diff(y(t),t$2)+t*diff(y(t),t)-y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} t \left (1+\operatorname {O}\left (t^{6}\right )\right )+\left (\frac {1}{3} t +\operatorname {O}\left (t^{6}\right )\right ) \ln \left (t \right ) c_{2} +\left (1-\frac {1}{3} t -\frac {2}{9} t^{2}+\frac {7}{81} t^{3}-\frac {35}{729} t^{4}+\frac {91}{2916} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.05 (sec). Leaf size: 43
AsymptoticDSolveValue[3*t*(1+t)*y''[t]+t*y'[t]-y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (\frac {1}{729} \left (-35 t^4+63 t^3-162 t^2+243 t+729\right )+\frac {1}{3} t \log (t)\right )+c_2 t \]