Internal problem ID [1804]
Internal file name [OUTPUT/1805_Sunday_June_05_2022_02_32_49_AM_67982648/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: 12.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {2 t^{2} y^{\prime \prime }+\left (t^{2}-t \right ) y^{\prime }+y=0} \] 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. \[ 2 t^{2} y^{\prime \prime }+\left (t^{2}-t \right ) y^{\prime }+y = 0 \] 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}{2 t}\\ q(t) &= \frac {1}{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 \[ 2 t^{2} y^{\prime \prime }+\left (t^{2}-t \right ) y^{\prime }+y = 0 \] 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} 2 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 }}2 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}t^{1+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} \\ \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 }}2 t^{n +r} 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}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-t^{n +r} a_{n} \left (n +r \right )+a_{n} t^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 t^{r} a_{0} r \left (-1+r \right )-t^{r} a_{0} r +a_{0} t^{r} = 0 \] Or \[ \left (2 t^{r} r \left (-1+r \right )-t^{r} r +t^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}-3 r +1\right ) t^{r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ 2 r^{2}-3 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}-3 r +1\right ) t^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) 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^{1+n}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n +\frac {1}{2}} \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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 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 \[ a_{n} = -\frac {a_{n -1}}{2 n +2 r -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -\frac {a_{n -1}}{2 n +1}\tag {5} \] 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 \[ a_{1}=-\frac {1}{1+2 r} \] Which for the root \(r = 1\) becomes \[ a_{1}=-{\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{4 r^{2}+8 r +3} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{15}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{8 r^{3}+36 r^{2}+46 r +15} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {1}{105}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{15}\) |
| \(a_{3}\) | \(-\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {1}{105}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1}{945}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{15}\) |
| \(a_{3}\) | \(-\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {1}{105}}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{945}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {1}{10395}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{15}\) |
| \(a_{3}\) | \(-\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {1}{105}}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{945}\) |
| \(a_{5}\) | \(-\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(-{\frac {1}{10395}}\) |
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}}{15}-\frac {t^{3}}{105}+\frac {t^{4}}{945}-\frac {t^{5}}{10395}+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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 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} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1}}{2 n +2 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{n} = -\frac {b_{n -1}}{2 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ b_{1}=-\frac {1}{1+2 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{2}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {1}{8 r^{3}+36 r^{2}+46 r +15} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{3}=-{\frac {1}{48}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(-\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {1}{48}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{4}={\frac {1}{384}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(-\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {1}{48}}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{384}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{5}=-{\frac {1}{3840}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{1+2 r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(-\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-{\frac {1}{48}}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{384}\) |
| \(b_{5}\) | \(-\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(-{\frac {1}{3840}}\) |
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 ) \\ &= \sqrt {t}\, \left (1-\frac {t}{2}+\frac {t^{2}}{8}-\frac {t^{3}}{48}+\frac {t^{4}}{384}-\frac {t^{5}}{3840}+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 \left (1-\frac {t}{3}+\frac {t^{2}}{15}-\frac {t^{3}}{105}+\frac {t^{4}}{945}-\frac {t^{5}}{10395}+O\left (t^{6}\right )\right ) + c_{2} \sqrt {t}\, \left (1-\frac {t}{2}+\frac {t^{2}}{8}-\frac {t^{3}}{48}+\frac {t^{4}}{384}-\frac {t^{5}}{3840}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t \left (1-\frac {t}{3}+\frac {t^{2}}{15}-\frac {t^{3}}{105}+\frac {t^{4}}{945}-\frac {t^{5}}{10395}+O\left (t^{6}\right )\right )+c_{2} \sqrt {t}\, \left (1-\frac {t}{2}+\frac {t^{2}}{8}-\frac {t^{3}}{48}+\frac {t^{4}}{384}-\frac {t^{5}}{3840}+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 {t}{3}+\frac {t^{2}}{15}-\frac {t^{3}}{105}+\frac {t^{4}}{945}-\frac {t^{5}}{10395}+O\left (t^{6}\right )\right )+c_{2} \sqrt {t}\, \left (1-\frac {t}{2}+\frac {t^{2}}{8}-\frac {t^{3}}{48}+\frac {t^{4}}{384}-\frac {t^{5}}{3840}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t \left (1-\frac {t}{3}+\frac {t^{2}}{15}-\frac {t^{3}}{105}+\frac {t^{4}}{945}-\frac {t^{5}}{10395}+O\left (t^{6}\right )\right )+c_{2} \sqrt {t}\, \left (1-\frac {t}{2}+\frac {t^{2}}{8}-\frac {t^{3}}{48}+\frac {t^{4}}{384}-\frac {t^{5}}{3840}+O\left (t^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 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}{2 t^{2}}-\frac {\left (t -1\right ) y^{\prime }}{2 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 }}{2 t}+\frac {y}{2 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}{2 t}, P_{3}\left (t \right )=\frac {1}{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}}}=-\frac {1}{2} \\ {} & \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}}}=\frac {1}{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} \\ {} & {} & 2 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+2 r \right ) \left (-1+r \right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 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+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r -1\right ) \left (\left (k +r -\frac {1}{2}\right ) a_{k}+\frac {a_{k -1}}{2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (k +r \right ) \left (\left (k +\frac {1}{2}+r \right ) a_{k +1}+\frac {a_{k}}{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 k +1+2 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 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}}{2 k +3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{2 k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +\frac {1}{2}}, a_{k +1}=-\frac {a_{k}}{2 k +2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{1+k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} t^{k +\frac {1}{2}}\right ), a_{1+k}=-\frac {a_{k}}{2 k +3}, b_{1+k}=-\frac {b_{k}}{2 k +2}\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 <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker 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: 45
Order:=6; dsolve(2*t^2*diff(y(t),t$2)+(t^2-t)*diff(y(t),t)+y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} \sqrt {t}\, \left (1-\frac {1}{2} t +\frac {1}{8} t^{2}-\frac {1}{48} t^{3}+\frac {1}{384} t^{4}-\frac {1}{3840} t^{5}+\operatorname {O}\left (t^{6}\right )\right )+c_{2} t \left (1-\frac {1}{3} t +\frac {1}{15} t^{2}-\frac {1}{105} t^{3}+\frac {1}{945} t^{4}-\frac {1}{10395} t^{5}+\operatorname {O}\left (t^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 86
AsymptoticDSolveValue[2*t^2*y''[t]+(t^2-t)*y'[t]+y[t]==0,y[t],{t,0,5}]
\[ y(t)\to c_1 t \left (-\frac {t^5}{10395}+\frac {t^4}{945}-\frac {t^3}{105}+\frac {t^2}{15}-\frac {t}{3}+1\right )+c_2 \sqrt {t} \left (-\frac {t^5}{3840}+\frac {t^4}{384}-\frac {t^3}{48}+\frac {t^2}{8}-\frac {t}{2}+1\right ) \]