Internal problem ID [1816]
Internal file name [OUTPUT/1817_Sunday_June_05_2022_02_33_39_AM_60814982/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: 24.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[_Bessel]
\[ \boxed {t^{2} y^{\prime \prime }+t y^{\prime }+\left (t^{2}-v^{2}\right ) y=0} \] 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. \[ t^{2} y^{\prime \prime }+t y^{\prime }+\left (t^{2}-v^{2}\right ) y = 0 \] 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}{t}\\ q(t) &= \frac {t^{2}-v^{2}}{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 \[ t^{2} y^{\prime \prime }+t y^{\prime }+\left (t^{2}-v^{2}\right ) y = 0 \] 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 )+t \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (t^{2}-v^{2}\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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} v^{2} a_{n}\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 }}t^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} t^{n +r} \\ \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} t^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} v^{2} a_{n}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+t^{n +r} a_{n} \left (n +r \right )-t^{n +r} v^{2} a_{n} = 0 \] When \(n = 0\) the above becomes \[ t^{r} a_{0} r \left (-1+r \right )+t^{r} a_{0} r -t^{r} v^{2} a_{0} = 0 \] Or \[ \left (t^{r} r \left (-1+r \right )+t^{r} r -t^{r} v^{2}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-v^{2}\right ) t^{r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2}-v^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= v\\ r_2 &= -v \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-v^{2}\right ) t^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([v, -v]\).
Assuming the roots differ by non-integer Since \(r_1 - r_2 = 2 v\) is not 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 ) &= 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 +v}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -v} \end {align*}
We start by finding \(y_{1}\left (t \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -2}-a_{n} v^{2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -2}}{n^{2}+2 n r +r^{2}-v^{2}}\tag {4} \] Which for the root \(r = v\) becomes \[ a_{n} = -\frac {a_{n -2}}{n \left (n +2 v \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = v\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=-\frac {1}{r^{2}-v^{2}+4 r +4} \] Which for the root \(r = v\) becomes \[ a_{2}=-\frac {1}{4 v +4} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(-\frac {1}{4 v +4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(-\frac {1}{4 v +4}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (r^{2}-v^{2}+4 r +4\right ) \left (r^{2}-v^{2}+8 r +16\right )} \] Which for the root \(r = v\) becomes \[ a_{4}=\frac {1}{32 \left (v +1\right ) \left (v +2\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(-\frac {1}{4 v +4}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{\left (r^{2}-v^{2}+4 r +4\right ) \left (r^{2}-v^{2}+8 r +16\right )}\) | \(\frac {1}{32 \left (v +1\right ) \left (v +2\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(-\frac {1}{4 v +4}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{\left (r^{2}-v^{2}+4 r +4\right ) \left (r^{2}-v^{2}+8 r +16\right )}\) | \(\frac {1}{32 \left (v +1\right ) \left (v +2\right )}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin {align*} y_{1}\left (t \right )&= t^{v} \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^{v} \left (1-\frac {t^{2}}{4 v +4}+\frac {t^{4}}{32 \left (v +1\right ) \left (v +2\right )}+O\left (t^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Eq (2B) derived above is now 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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n} \left (n +r \right )+b_{n -2}-v^{2} b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -2}}{n^{2}+2 n r +r^{2}-v^{2}}\tag {4} \] Which for the root \(r = -v\) becomes \[ b_{n} = -\frac {b_{n -2}}{n \left (n -2 v \right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -v\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=-\frac {1}{r^{2}-v^{2}+4 r +4} \] Which for the root \(r = -v\) becomes \[ b_{2}=\frac {1}{4 v -4} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(\frac {1}{4 v -4}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(\frac {1}{4 v -4}\) |
| \(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{\left (r^{2}-v^{2}+4 r +4\right ) \left (r^{2}-v^{2}+8 r +16\right )} \] Which for the root \(r = -v\) becomes \[ b_{4}=\frac {1}{32 \left (v -1\right ) \left (v -2\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(\frac {1}{4 v -4}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{\left (r^{2}-v^{2}+4 r +4\right ) \left (r^{2}-v^{2}+8 r +16\right )}\) | \(\frac {1}{32 \left (v -1\right ) \left (v -2\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(0\) | \(0\) |
| \(b_{2}\) | \(-\frac {1}{r^{2}-v^{2}+4 r +4}\) | \(\frac {1}{4 v -4}\) |
| \(b_{3}\) | \(0\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{\left (r^{2}-v^{2}+4 r +4\right ) \left (r^{2}-v^{2}+8 r +16\right )}\) | \(\frac {1}{32 \left (v -1\right ) \left (v -2\right )}\) |
| \(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (t \right )\) is \begin {align*} y_{2}\left (t \right )&= t^{v} \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 ) \\ &= t^{-v} \left (1+\frac {t^{2}}{4 v -4}+\frac {t^{4}}{32 \left (v -1\right ) \left (v -2\right )}+O\left (t^{6}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} t^{v} \left (1-\frac {t^{2}}{4 v +4}+\frac {t^{4}}{32 \left (v +1\right ) \left (v +2\right )}+O\left (t^{6}\right )\right ) + c_{2} t^{-v} \left (1+\frac {t^{2}}{4 v -4}+\frac {t^{4}}{32 \left (v -1\right ) \left (v -2\right )}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t^{v} \left (1-\frac {t^{2}}{4 v +4}+\frac {t^{4}}{32 \left (v +1\right ) \left (v +2\right )}+O\left (t^{6}\right )\right )+c_{2} t^{-v} \left (1+\frac {t^{2}}{4 v -4}+\frac {t^{4}}{32 \left (v -1\right ) \left (v -2\right )}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t^{v} \left (1-\frac {t^{2}}{4 v +4}+\frac {t^{4}}{32 \left (v +1\right ) \left (v +2\right )}+O\left (t^{6}\right )\right )+c_{2} t^{-v} \left (1+\frac {t^{2}}{4 v -4}+\frac {t^{4}}{32 \left (v -1\right ) \left (v -2\right )}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t^{v} \left (1-\frac {t^{2}}{4 v +4}+\frac {t^{4}}{32 \left (v +1\right ) \left (v +2\right )}+O\left (t^{6}\right )\right )+c_{2} t^{-v} \left (1+\frac {t^{2}}{4 v -4}+\frac {t^{4}}{32 \left (v -1\right ) \left (v -2\right )}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{2} y^{\prime \prime }+t y^{\prime }+\left (t^{2}-v^{2}\right ) 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 {\left (t^{2}-v^{2}\right ) y}{t^{2}}-\frac {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 {y^{\prime }}{t}+\frac {\left (t^{2}-v^{2}\right ) 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 {1}{t}, P_{3}\left (t \right )=\frac {t^{2}-v^{2}}{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}}}=-v^{2} \\ {} & \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 y^{\prime }+\left (t^{2}-v^{2}\right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & t^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & t^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} t^{k +r} \\ {} & \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^{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 (r +v \right ) \left (r -v \right ) t^{r}+a_{1} \left (1+r +v \right ) \left (1+r -v \right ) t^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +v \right ) \left (k +r -v \right )+a_{k -2}\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (r +v \right ) \left (r -v \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{v , -v \right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r +v \right ) \left (1+r -v \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +v \right ) \left (k +r -v \right )+a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +2+r +v \right ) \left (k +2+r -v \right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+r +v \right ) \left (k +2+r -v \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =v \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2+2 v \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =v \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +v}, a_{k +2}=-\frac {a_{k}}{\left (k +2+2 v \right ) \left (k +2\right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-v \\ {} & {} & a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +2-2 v \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-v \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k -v}, a_{k +2}=-\frac {a_{k}}{\left (k +2\right ) \left (k +2-2 v \right )}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +v}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k -v}\right ), a_{k +2}=-\frac {a_{k}}{\left (k +2+2 v \right ) \left (k +2\right )}, a_{1}=0, b_{k +2}=-\frac {b_{k}}{\left (k +2\right ) \left (k +2-2 v \right )}, b_{1}=0\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: 0.015 (sec). Leaf size: 81
Order:=6; dsolve(t^2*diff(y(t),t$2)+t*diff(y(t),t)+(t^2-v^2)*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} t^{-v} \left (1+\frac {1}{-4+4 v} t^{2}+\frac {1}{32} \frac {1}{\left (v -2\right ) \left (v -1\right )} t^{4}+\operatorname {O}\left (t^{6}\right )\right )+c_{2} t^{v} \left (1-\frac {1}{4 v +4} t^{2}+\frac {1}{32} \frac {1}{\left (v +2\right ) \left (v +1\right )} t^{4}+\operatorname {O}\left (t^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 160
AsymptoticDSolveValue[t^2*y''[t]+t*y'[t]+(t^2-v^2)*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_2 \left (\frac {t^4}{\left (-v^2-v+(1-v) (2-v)+2\right ) \left (-v^2-v+(3-v) (4-v)+4\right )}-\frac {t^2}{-v^2-v+(1-v) (2-v)+2}+1\right ) t^{-v}+c_1 \left (\frac {t^4}{\left (-v^2+v+(v+1) (v+2)+2\right ) \left (-v^2+v+(v+3) (v+4)+4\right )}-\frac {t^2}{-v^2+v+(v+1) (v+2)+2}+1\right ) t^v \]