Internal problem ID [1806]
Internal file name [OUTPUT/1807_Sunday_June_05_2022_02_32_56_AM_82062633/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: 14.
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]]
\[ \boxed {t^{2} y^{\prime \prime }+\left (-t^{2}+t \right ) y^{\prime }-y=0} \] 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. \[ t^{2} y^{\prime \prime }+\left (-t^{2}+t \right ) y^{\prime }-y = 0 \] 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 {t -1}{t}\\ q(t) &= -\frac {1}{t^{2}}\\ \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 \[ t^{2} y^{\prime \prime }+\left (-t^{2}+t \right ) y^{\prime }-y = 0 \] 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} t^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right )+\left (-t^{2}+t \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} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{1+n +r} a_{n} \left (n +r \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 \(n +r\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) t^{n +r}\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 \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) t^{n +r}\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 indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+t^{n +r} a_{n} \left (n +r \right )-a_{n} t^{n +r} = 0 \] When \(n = 0\) the above becomes \[ t^{r} a_{0} r \left (-1+r \right )+t^{r} a_{0} r -a_{0} t^{r} = 0 \] Or \[ \left (t^{r} r \left (-1+r \right )+t^{r} r -t^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-1\right ) t^{r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2}-1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-1\right ) t^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, -1]\).
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 ) &= 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 )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}}{t} \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{1+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 -1}\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 )+a_{n} \left (n +r \right )-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{1+n +r}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {a_{n -1}}{2+n}\tag {5} \] 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 \[ a_{1}=\frac {1}{2+r} \] Which for the root \(r = 1\) becomes \[ a_{1}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+r}\) | \(\frac {1}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{\left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{12}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {1}{60}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{60}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (3+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1}{360}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{60}\) |
| \(a_{4}\) | \(\frac {1}{\left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (3+r \right )}\) | \(\frac {1}{360}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{\left (2+r \right ) \left (5+r \right ) \left (3+r \right ) \left (6+r \right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{5}={\frac {1}{2520}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{2+r}\) | \(\frac {1}{3}\) |
| \(a_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{12}\) |
| \(a_{3}\) | \(\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{60}\) |
| \(a_{4}\) | \(\frac {1}{\left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (3+r \right )}\) | \(\frac {1}{360}\) |
| \(a_{5}\) | \(\frac {1}{\left (2+r \right ) \left (5+r \right ) \left (3+r \right ) \left (6+r \right ) \left (4+r \right )}\) | \(\frac {1}{2520}\) |
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 {t}{3}+\frac {t^{2}}{12}+\frac {t^{3}}{60}+\frac {t^{4}}{360}+\frac {t^{5}}{2520}+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 \(N=2\). 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 {1}{\left (2+r \right ) \left (3+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1}{\left (2+r \right ) \left (3+r \right )}&= \lim _{r\rightarrow -1}\frac {1}{\left (2+r \right ) \left (3+r \right )}\\ &= {\frac {1}{2}} \end {align*}
The limit is \(\frac {1}{2}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -1} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )-b_{n} = 0 \end{equation} Which for for the root \(r = -1\) becomes \begin{equation} \tag{4A} b_{n} \left (n -1\right ) \left (n -2\right )-b_{n -1} \left (n -2\right )+b_{n} \left (n -1\right )-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {b_{n -1}}{1+n +r}\tag {5} \] Which for the root \(r = -1\) becomes \[ b_{n} = \frac {b_{n -1}}{n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {1}{2+r} \] Which for the root \(r = -1\) becomes \[ b_{1}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+r}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{\left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = -1\) becomes \[ b_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+r}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )} \] Which for the root \(r = -1\) becomes \[ b_{3}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+r}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{\left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (3+r \right )} \] Which for the root \(r = -1\) becomes \[ b_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+r}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{6}\) |
| \(b_{4}\) | \(\frac {1}{\left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (3+r \right )}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1}{\left (2+r \right ) \left (5+r \right ) \left (3+r \right ) \left (6+r \right ) \left (4+r \right )} \] Which for the root \(r = -1\) becomes \[ b_{5}={\frac {1}{120}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{2+r}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{\left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {1}{\left (3+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{6}\) |
| \(b_{4}\) | \(\frac {1}{\left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (3+r \right )}\) | \(\frac {1}{24}\) |
| \(b_{5}\) | \(\frac {1}{\left (2+r \right ) \left (5+r \right ) \left (3+r \right ) \left (6+r \right ) \left (4+r \right )}\) | \(\frac {1}{120}\) |
Using the above table, then the solution \(y_{2}\left (t \right )\) is \begin {align*} y_{2}\left (t \right )&= t \left (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 \right ) \\ &= \frac {1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )}{t} \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} t \left (1+\frac {t}{3}+\frac {t^{2}}{12}+\frac {t^{3}}{60}+\frac {t^{4}}{360}+\frac {t^{5}}{2520}+O\left (t^{6}\right )\right ) + \frac {c_{2} \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 )}{t} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t \left (1+\frac {t}{3}+\frac {t^{2}}{12}+\frac {t^{3}}{60}+\frac {t^{4}}{360}+\frac {t^{5}}{2520}+O\left (t^{6}\right )\right )+\frac {c_{2} \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 )}{t} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t \left (1+\frac {t}{3}+\frac {t^{2}}{12}+\frac {t^{3}}{60}+\frac {t^{4}}{360}+\frac {t^{5}}{2520}+O\left (t^{6}\right )\right )+\frac {c_{2} \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 )}{t} \\ \end{align*}
Verification of solutions
\[ y = c_{1} t \left (1+\frac {t}{3}+\frac {t^{2}}{12}+\frac {t^{3}}{60}+\frac {t^{4}}{360}+\frac {t^{5}}{2520}+O\left (t^{6}\right )\right )+\frac {c_{2} \left (1+t +\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right )}{t} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} y^{\prime \prime }+\left (-t^{2}+t \right ) 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}{t^{2}}+\frac {\left (t -1\right ) 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 {\left (t -1\right ) y^{\prime }}{t}-\frac {y}{t^{2}}=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 {t -1}{t}, P_{3}\left (t \right )=-\frac {1}{t^{2}}\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}}}=1 \\ {} & \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}}}=-1 \\ {} & \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^{2} y^{\prime \prime }-t \left (t -1\right ) y^{\prime }-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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & t^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) t^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & t^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) t^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+r \right ) \left (-1+r \right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -1\right )-a_{k -1} \left (k +r -1\right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (a_{k} \left (k +r +1\right )-a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r \right ) \left (a_{k +1} \left (k +2+r \right )-a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +2+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +1} \\ \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}}{k +1}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{k +3} \\ \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}}{k +3}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{1+k}\right ), a_{1+k}=\frac {a_{k}}{1+k}, b_{1+k}=\frac {b_{k}}{k +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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 45
Order:=6; dsolve(t^2*diff(y(t),t$2)+(t-t^2)*diff(y(t),t)-y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} t \left (1+\frac {1}{3} t +\frac {1}{12} t^{2}+\frac {1}{60} t^{3}+\frac {1}{360} t^{4}+\frac {1}{2520} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+\frac {c_{2} \left (-2-2 t -t^{2}-\frac {1}{3} t^{3}-\frac {1}{12} t^{4}-\frac {1}{60} t^{5}+\operatorname {O}\left (t^{6}\right )\right )}{t} \]
✓ Solution by Mathematica
Time used: 0.018 (sec). Leaf size: 64
AsymptoticDSolveValue[t^2*y''[t]+(t-t^2)*y'[t]-y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 \left (\frac {t^3}{24}+\frac {t^2}{6}+\frac {t}{2}+\frac {1}{t}+1\right )+c_2 \left (\frac {t^5}{360}+\frac {t^4}{60}+\frac {t^3}{12}+\frac {t^2}{3}+t\right ) \]