Internal problem ID [9593]
Internal file name [OUTPUT/8534_Monday_June_06_2022_03_40_47_AM_38326904/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1265.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {\left (x -1\right ) \left (x -2\right ) y^{\prime \prime }-\left (2 x -3\right ) y^{\prime }+y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x^{2}-3 x +2\right )+\left (-2 x +3\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {y}{x^{2}-3 x +2}+\frac {\left (2 x -3\right ) y^{\prime }}{x^{2}-3 x +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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (2 x -3\right ) y^{\prime }}{x^{2}-3 x +2}+\frac {y}{x^{2}-3 x +2}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x -3}{x^{2}-3 x +2}, P_{3}\left (x \right )=\frac {1}{x^{2}-3 x +2}\right ] \\ {} & \circ & \left (x -1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =1 \\ {} & {} & \left (\left (x -1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}1}}}=-1 \\ {} & \circ & \left (x -1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =1 \\ {} & {} & \left (\left (x -1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}1}}}=0 \\ {} & \circ & x =1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x^{2}-3 x +2\right )+\left (-2 x +3\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-2 u +1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-2+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )+a_{k} \left (k^{2}+2 k r +r^{2}-3 k -3 r +1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r -3\right ) k +r^{2}-3 r +1\right ) a_{k}-a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (k^{2}+2 k r +r^{2}-3 k -3 r +1\right ) a_{k}}{\left (k +1+r \right ) \left (k +r -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {\left (k^{2}-3 k +1\right ) a_{k}}{\left (k +1\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {\left (k^{2}-3 k +1\right ) a_{k}}{\left (k +1\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=\frac {\left (k^{2}+k -1\right ) a_{k}}{\left (k +3\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2}, a_{k +1}=\frac {\left (k^{2}+k -1\right ) a_{k}}{\left (k +3\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x -1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x -1\right )^{k +2}, a_{k +1}=\frac {\left (k^{2}+k -1\right ) a_{k}}{\left (k +3\right ) \left (k +1\right )}\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 <- heuristic approach successful -> solution has integrals; searching for one without integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric solution without integrals succesful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.234 (sec). Leaf size: 94
dsolve((x-1)*(x-2)*diff(diff(y(x),x),x)-(2*x-3)*diff(y(x),x)+y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x -2\right )^{2} \left (c_{1} \operatorname {hypergeom}\left (\left [\frac {1}{2}-\frac {\sqrt {5}}{2}, \frac {5}{2}-\frac {\sqrt {5}}{2}\right ], \left [-\sqrt {5}+1\right ], \frac {1}{x -1}\right ) \left (x -1\right )^{\frac {\sqrt {5}}{2}}+c_{2} \operatorname {hypergeom}\left (\left [\frac {1}{2}+\frac {\sqrt {5}}{2}, \frac {5}{2}+\frac {\sqrt {5}}{2}\right ], \left [\sqrt {5}+1\right ], \frac {1}{x -1}\right ) \left (x -1\right )^{-\frac {\sqrt {5}}{2}}\right )}{\sqrt {x -1}} \]
✓ Solution by Mathematica
Time used: 0.072 (sec). Leaf size: 57
DSolve[y[x] - (-3 + 2*x)*y'[x] + (-2 + x)*(-1 + x)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \left (x^2-3 x+2\right ) \left (c_1 P_{\frac {1}{2} \left (-1+\sqrt {5}\right )}^2(2 x-3)+c_2 Q_{\frac {1}{2} \left (-1+\sqrt {5}\right )}^2(2 x-3)\right ) \]