Internal problem ID [1813]
Internal file name [OUTPUT/1814_Sunday_June_05_2022_02_33_27_AM_86548608/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: 21.
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]]
\[ \boxed {t y^{\prime \prime }+t y^{\prime }+2 y=0} \] 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. \[ t y^{\prime \prime }+t y^{\prime }+2 y = 0 \] 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) &= 1\\ q(t) &= \frac {2}{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 \[ t y^{\prime \prime }+t y^{\prime }+2 y = 0 \] 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 )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+2 \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 }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 }}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 }}2 a_{n} t^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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 =0}{\sum }}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 )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right ) = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Or \[ t^{-1+r} a_{0} r \left (-1+r \right ) = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ t^{-1+r} r \left (-1+r \right ) = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r \left (-1+r \right ) = 0 \] 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 \[ t^{-1+r} r \left (-1+r \right ) = 0 \] 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} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n +r +1\right )}{\left (n +r \right ) \left (n +r -1\right )}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n +2\right )}{\left (n +1\right ) n}\tag {5} \] 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 \[ a_{1}=\frac {-2-r}{\left (1+r \right ) r} \] Which for the root \(r = 1\) becomes \[ a_{1}=-{\frac {3}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right ) r}\) | \(-{\frac {3}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3+r}{\left (1+r \right )^{2} r} \] Which for the root \(r = 1\) becomes \[ a_{2}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right ) r}\) | \(-{\frac {3}{2}}\) |
| \(a_{2}\) | \(\frac {3+r}{\left (1+r \right )^{2} r}\) | \(1\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-4-r}{\left (1+r \right )^{2} r \left (2+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {5}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right ) r}\) | \(-{\frac {3}{2}}\) |
| \(a_{2}\) | \(\frac {3+r}{\left (1+r \right )^{2} r}\) | \(1\) |
| \(a_{3}\) | \(\frac {-4-r}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(-{\frac {5}{12}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5+r}{\left (1+r \right )^{2} r \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right ) r}\) | \(-{\frac {3}{2}}\) |
| \(a_{2}\) | \(\frac {3+r}{\left (1+r \right )^{2} r}\) | \(1\) |
| \(a_{3}\) | \(\frac {-4-r}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(-{\frac {5}{12}}\) |
| \(a_{4}\) | \(\frac {5+r}{\left (1+r \right )^{2} r \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-6-r}{\left (1+r \right )^{2} r \left (2+r \right ) \left (3+r \right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {7}{240}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2-r}{\left (1+r \right ) r}\) | \(-{\frac {3}{2}}\) |
| \(a_{2}\) | \(\frac {3+r}{\left (1+r \right )^{2} r}\) | \(1\) |
| \(a_{3}\) | \(\frac {-4-r}{\left (1+r \right )^{2} r \left (2+r \right )}\) | \(-{\frac {5}{12}}\) |
| \(a_{4}\) | \(\frac {5+r}{\left (1+r \right )^{2} r \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{8}\) |
| \(a_{5}\) | \(\frac {-6-r}{\left (1+r \right )^{2} r \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}\) | \(-{\frac {7}{240}}\) |
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-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+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 {-2-r}{\left (1+r \right ) r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-2-r}{\left (1+r \right ) r}&= \lim _{r\rightarrow 0}\frac {-2-r}{\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 \(t y^{\prime \prime }+t y^{\prime }+2 y = 0\) gives \[ t \left (C y_{1}^{\prime \prime }\left (t \right ) \ln \left (t \right )+\frac {2 C y_{1}^{\prime }\left (t \right )}{t}-\frac {C y_{1}\left (t \right )}{t^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )^{2}}{t^{2}}-\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t^{2}}\right )\right )+t \left (C y_{1}^{\prime }\left (t \right ) \ln \left (t \right )+\frac {C y_{1}\left (t \right )}{t}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )\right )+2 C y_{1}\left (t \right ) \ln \left (t \right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (t \right ) t +y_{1}^{\prime }\left (t \right ) t +2 y_{1}\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 )+y_{1}\left (t \right )\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 )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )+2 \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 \[ y_{1}^{\prime \prime }\left (t \right ) t +y_{1}^{\prime }\left (t \right ) t +2 y_{1}\left (t \right ) = 0 \] 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 )+y_{1}\left (t \right )\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 )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} t^{n +r_{2}} \left (n +r_{2}\right )}{t}\right )+2 \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 +\left (t -1\right ) \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^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\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 ) t^{2}+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 (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n} a_{n} \left (n +1\right )\right ) t +\left (t -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +1}\right )\right ) C}{t}+\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n -1} b_{n} n \right ) t^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{-2+n} b_{n} n \left (n -1\right )\right ) t^{2}+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 }}2 C \,t^{n} a_{n} \left (n +1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}C \,t^{n +1} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n} b_{n} n \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,t^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 }}2 C \,t^{n} a_{n} \left (n +1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}C \,t^{n +1} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}C a_{-2+n} t^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C a_{n} t^{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} t^{n -1}\right ) \\ \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 }}2 b_{n} t^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 b_{n -1} t^{n -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 \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}2 C a_{n -1} n \,t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}C a_{-2+n} t^{n -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-C a_{n -1} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} t^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,t^{n -1} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 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 \[ C +2 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-2 \] For \(n=2\), Eq (2B) gives \[ \left (a_{0}+3 a_{1}\right ) C +3 b_{1}+2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 7+2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=-{\frac {7}{2}} \] For \(n=3\), Eq (2B) gives \[ \left (a_{1}+5 a_{2}\right ) C +4 b_{2}+6 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -21+6 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}={\frac {7}{2}} \] For \(n=4\), Eq (2B) gives \[ \left (a_{2}+7 a_{3}\right ) C +5 b_{3}+12 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {64}{3}+12 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-{\frac {16}{9}} \] For \(n=5\), Eq (2B) gives \[ \left (a_{3}+9 a_{4}\right ) C +6 b_{4}+20 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -\frac {145}{12}+20 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {29}{48}} \] 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 \(C=-2\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (t \right )= \left (-2\right )\eslowast \left (t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right )\right ) \ln \left (t \right )+1-\frac {7 t^{2}}{2}+\frac {7 t^{3}}{2}-\frac {16 t^{4}}{9}+\frac {29 t^{5}}{48}+O\left (t^{6}\right ) \] 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-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right ) + c_{2} \left (\left (-2\right )\eslowast \left (t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right )\right ) \ln \left (t \right )+1-\frac {7 t^{2}}{2}+\frac {7 t^{3}}{2}-\frac {16 t^{4}}{9}+\frac {29 t^{5}}{48}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right )+c_{2} \left (-2 t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right ) \ln \left (t \right )+1-\frac {7 t^{2}}{2}+\frac {7 t^{3}}{2}-\frac {16 t^{4}}{9}+\frac {29 t^{5}}{48}+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-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right )+c_{2} \left (-2 t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right ) \ln \left (t \right )+1-\frac {7 t^{2}}{2}+\frac {7 t^{3}}{2}-\frac {16 t^{4}}{9}+\frac {29 t^{5}}{48}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right )+c_{2} \left (-2 t \left (1-\frac {3 t}{2}+t^{2}-\frac {5 t^{3}}{12}+\frac {t^{4}}{8}-\frac {7 t^{5}}{240}+O\left (t^{6}\right )\right ) \ln \left (t \right )+1-\frac {7 t^{2}}{2}+\frac {7 t^{3}}{2}-\frac {16 t^{4}}{9}+\frac {29 t^{5}}{48}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t \left (\frac {d}{d t}y^{\prime }\right )+t y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-y^{\prime }-\frac {2 y}{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} \\ {} & {} & \frac {d}{d t}y^{\prime }+y^{\prime }+\frac {2 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 )=1, P_{3}\left (t \right )=\frac {2}{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}}}=0 \\ {} & \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 \left (\frac {d}{d t}y^{\prime }\right )+t y^{\prime }+2 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} t \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot \left (\frac {d}{d t}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot \left (\frac {d}{d t}y^{\prime }\right )=\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 \left (\frac {d}{d t}y^{\prime }\right )=\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 (-1+r \right ) t^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r \right )+a_{k} \left (k +r +2\right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\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 +r \right )+a_{k} \left (k +r +2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r +2\right )}{\left (k +1+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +3\right )}{\left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}, a_{k +1}=-\frac {a_{k} \left (k +3\right )}{\left (k +2\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k +1}\right ), a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +1\right ) k}, b_{k +1}=-\frac {b_{k} \left (k +3\right )}{\left (k +2\right ) \left (k +1\right )}\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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 60
Order:=6; dsolve(t*diff(y(t),t$2)+t*diff(y(t),t)+2*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} t \left (1-\frac {3}{2} t +t^{2}-\frac {5}{12} t^{3}+\frac {1}{8} t^{4}-\frac {7}{240} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+c_{2} \left (\ln \left (t \right ) \left (\left (-2\right ) t +3 t^{2}-2 t^{3}+\frac {5}{6} t^{4}-\frac {1}{4} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\left (1-t -2 t^{2}+\frac {5}{2} t^{3}-\frac {49}{36} t^{4}+\frac {23}{48} t^{5}+\operatorname {O}\left (t^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.024 (sec). Leaf size: 83
AsymptoticDSolveValue[t*y''[t]+t*y'[t]+2*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (\frac {1}{6} t \left (5 t^3-12 t^2+18 t-12\right ) \log (t)+\frac {1}{36} \left (-79 t^4+162 t^3-180 t^2+36 t+36\right )\right )+c_2 \left (\frac {t^5}{8}-\frac {5 t^4}{12}+t^3-\frac {3 t^2}{2}+t\right ) \]