Internal problem ID [1794]
Internal file name [OUTPUT/1795_Sunday_June_05_2022_02_31_38_AM_10633998/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: 2.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Irregular singular point"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {t \left (t -2\right )^{2} y^{\prime \prime }+t y^{\prime }+y=0} \] With the expansion point for the power series method at \(t = 2\).
The ode does not have its expansion point at \(t = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let \[ x = t -2 \] The ode is converted to be in terms of the new independent variable \(x\). This results in \[ x^{2} \left (2+x \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (2+x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right ) = 0 \] With its expansion point and initial conditions now at \(x = 0\). The transformed ODE is now solved. The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{3}+2 x^{2}\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+\left (2+x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right ) = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} \frac {d^{2}}{d x^{2}}y \left (x \right )+p(x) \frac {d}{d x}y \left (x \right ) + q(x) y \left (x \right ) &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {1}{x^{2}}\\ q(x) &= \frac {1}{x^{2} \left (2+x \right )}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-2, \infty ]\)
Irregular singular points : \([0]\)
Since \(x = 0\) is not an ordinary point, then we will now check if it is a regular singular point. Unable to solve since \(x = 0\) is not regular singular point. Terminating. Unable to solve the transformed ode. Terminating.
Verification of solutions N/A
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t \left (t -2\right )^{2} 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^{\prime }}{\left (t -2\right )^{2}}-\frac {y}{t \left (t -2\right )^{2}} \\ \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 }}{\left (t -2\right )^{2}}+\frac {y}{t \left (t -2\right )^{2}}=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}{\left (t -2\right )^{2}}, P_{3}\left (t \right )=\frac {1}{t \left (t -2\right )^{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}}}=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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & t_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & t \left (t -2\right )^{2} y^{\prime \prime }+t 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 \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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & t^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) t^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & t^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) t^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (-1+r \right ) t^{-1+r}+\left (4 a_{1} \left (1+r \right ) r -a_{0} \left (4 r^{2}-5 r -1\right )\right ) t^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )-a_{k} \left (4 k^{2}+8 k r +4 r^{2}-5 k -5 r -1\right )+a_{k -1} \left (k +r -1\right ) \left (k -2+r \right )\right ) t^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 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 must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) r -a_{0} \left (4 r^{2}-5 r -1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) k^{2}+\left (\left (-8 a_{k}+2 a_{k -1}+8 a_{k +1}\right ) r +5 a_{k}-3 a_{k -1}+4 a_{k +1}\right ) k +\left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r^{2}+\left (5 a_{k}-3 a_{k -1}+4 a_{k +1}\right ) r +a_{k}+2 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )^{2}+\left (\left (-8 a_{k +1}+2 a_{k}+8 a_{k +2}\right ) r +5 a_{k +1}-3 a_{k}+4 a_{k +2}\right ) \left (k +1\right )+\left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r^{2}+\left (5 a_{k +1}-3 a_{k}+4 a_{k +2}\right ) r +a_{k +1}+2 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+2 k r a_{k}-8 k r a_{k +1}+r^{2} a_{k}-4 r^{2} a_{k +1}-k a_{k}-3 k a_{k +1}-r a_{k}-3 r a_{k +1}+2 a_{k +1}}{4 \left (k^{2}+2 k r +r^{2}+3 k +3 r +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k a_{k}-3 k a_{k +1}+2 a_{k +1}}{4 \left (k^{2}+3 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}-k a_{k}-3 k a_{k +1}+2 a_{k +1}}{4 \left (k^{2}+3 k +2\right )}, a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k a_{k}-11 k a_{k +1}-5 a_{k +1}}{4 \left (k^{2}+5 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k +1}, a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{k +1}+k a_{k}-11 k a_{k +1}-5 a_{k +1}}{4 \left (k^{2}+5 k +6\right )}, 8 a_{1}+2 a_{0}=0\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^{1+k}\right ), a_{k +2}=-\frac {k^{2} a_{k}-4 k^{2} a_{1+k}-k a_{k}-3 k a_{1+k}+2 a_{1+k}}{4 \left (k^{2}+3 k +2\right )}, a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}-4 k^{2} b_{1+k}+k b_{k}-11 k b_{1+k}-5 b_{1+k}}{4 \left (k^{2}+5 k +6\right )}, 8 b_{1}+2 b_{0}=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 -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a = 0, e <> 0, c <> 0 `
✗ Solution by Maple
Order:=6; dsolve(t*(t-2)^2*diff(y(t),t$2)+t*diff(y(t),t)+y(t)=0,y(t),type='series',t=2);
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 0.039 (sec). Leaf size: 112
AsymptoticDSolveValue[t*(t-2)^2*y''[t]+t*y'[t]+y[t]==0,y[t],{t,2,5}]
\[ y(t)\to c_2 e^{\frac {1}{t-2}} \left (\frac {247853}{240} (t-2)^5+\frac {4069}{24} (t-2)^4+\frac {199}{6} (t-2)^3+8 (t-2)^2+\frac {5 (t-2)}{2}+1\right ) (t-2)^2+c_1 \left (-\frac {641}{480} (t-2)^5+\frac {25}{48} (t-2)^4-\frac {7}{24} (t-2)^3+\frac {1}{4} (t-2)^2+\frac {2-t}{2}+1\right ) \]