Internal problem ID [9587]
Internal file name [OUTPUT/8528_Monday_June_06_2022_03_38_48_AM_5852290/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1259.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[_Jacobi]
Unable to solve or complete the solution.
\[ \boxed {x \left (x -1\right ) y^{\prime \prime }+\left (\left (a +1\right ) x +b \right ) y^{\prime }-l y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b +x \right ) y^{\prime }-l 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 {l y}{x \left (x -1\right )}-\frac {\left (a x +b +x \right ) y^{\prime }}{x \left (x -1\right )} \\ \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 (a x +b +x \right ) y^{\prime }}{x \left (x -1\right )}-\frac {l y}{x \left (x -1\right )}=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 {a x +b +x}{x \left (x -1\right )}, P_{3}\left (x \right )=-\frac {l}{x \left (x -1\right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-b \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b +x \right ) y^{\prime }-l y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1-r +b \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (-k -r +b \right )+a_{k} \left (a k +a r +k^{2}+2 k r +r^{2}-l \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1-r +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, b +1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (-k -r +b \right )+a_{k} \left (k^{2}+\left (a +2 r \right ) k +a r +r^{2}-l \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (a k +a r +k^{2}+2 k r +r^{2}-l \right )}{\left (k +1+r \right ) \left (-k -r +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (a k +k^{2}-l \right )}{\left (k +1\right ) \left (-k +b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a_{k} \left (a k +k^{2}-l \right )}{\left (k +1\right ) \left (-k +b \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =b +1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (a k +\left (b +1\right ) a +k^{2}+2 k \left (b +1\right )+\left (b +1\right )^{2}-l \right )}{\left (k +2+b \right ) \left (-1-k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =b +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{b +k +1}, a_{k +1}=-\frac {a_{k} \left (a k +\left (b +1\right ) a +k^{2}+2 k \left (b +1\right )+\left (b +1\right )^{2}-l \right )}{\left (k +2+b \right ) \left (-1-k \right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +b +1}\right ), c_{k +1}=-\frac {c_{k} \left (a k +k^{2}-l \right )}{\left (k +1\right ) \left (-k +b \right )}, d_{k +1}=-\frac {d_{k} \left (a k +\left (b +1\right ) a +k^{2}+2 k \left (b +1\right )+\left (b +1\right )^{2}-l \right )}{\left (k +2+b \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 <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 92
dsolve(x*(x-1)*diff(diff(y(x),x),x)+((a+1)*x+b)*diff(y(x),x)-l*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [\frac {a}{2}-\frac {\sqrt {a^{2}+4 l}}{2}, \frac {a}{2}+\frac {\sqrt {a^{2}+4 l}}{2}\right ], \left [-b \right ], x\right )+c_{2} x^{b +1} \operatorname {hypergeom}\left (\left [\frac {a}{2}-\frac {\sqrt {a^{2}+4 l}}{2}+b +1, \frac {a}{2}+\frac {\sqrt {a^{2}+4 l}}{2}+b +1\right ], \left [b +2\right ], x\right ) \]
✓ Solution by Mathematica
Time used: 0.183 (sec). Leaf size: 111
DSolve[-(l*y[x]) + (b + (1 + a)*x)*y'[x] + (-1 + x)*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (a-\sqrt {a^2+4 l}\right ),\frac {1}{2} \left (a+\sqrt {a^2+4 l}\right ),-b,x\right )-(-1)^b c_2 x^{b+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} \left (a+2 b-\sqrt {a^2+4 l}+2\right ),\frac {1}{2} \left (a+2 b+\sqrt {a^2+4 l}+2\right ),b+2,x\right ) \]