Internal problem ID [9662]
Internal file name [OUTPUT/8604_Monday_June_06_2022_04_24_36_AM_68695018/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1335.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (3 x -1\right ) y^{\prime }}{2 x \left (x -1\right )}+\frac {\left (x a +b \right ) y}{4 x \left (x -1\right )^{2}}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 \left (\frac {d}{d x}y^{\prime }\right ) x \left (x -1\right )^{2}+\left (6 x^{2}-8 x +2\right ) y^{\prime }+\left (x a +b \right ) 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 {\left (3 x -1\right ) y^{\prime }}{2 x \left (x -1\right )}-\frac {\left (x a +b \right ) y}{4 x \left (x -1\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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (3 x -1\right ) y^{\prime }}{2 x \left (x -1\right )}+\frac {\left (x a +b \right ) y}{4 x \left (x -1\right )^{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 {3 x -1}{2 x \left (x -1\right )}, P_{3}\left (x \right )=\frac {x a +b}{4 x \left (x -1\right )^{2}}\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}}}=\frac {1}{2} \\ {} & \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} \\ {} & {} & 4 \left (\frac {d}{d x}y^{\prime }\right ) x \left (x -1\right )^{2}+2 \left (3 x -1\right ) \left (x -1\right ) y^{\prime }+\left (x a +b \right ) 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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & 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..3 \\ {} & {} & 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} \\ {} & {} & 2 a_{0} r \left (-1+2 r \right ) x^{-1+r}+\left (2 a_{1} \left (1+r \right ) \left (1+2 r \right )+a_{0} \left (-8 r^{2}+b \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (2 a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k} \left (-8 k^{2}-16 k r -8 r^{2}+b \right )+a_{k -1} \left (4 \left (k -1\right )^{2}+8 \left (k -1\right ) r +4 r^{2}+a +2 k -2+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 2 r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 2 a_{1} \left (1+r \right ) \left (1+2 r \right )+a_{0} \left (-8 r^{2}+b \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+2 \left (4 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r -3 a_{k -1}+3 a_{k +1}\right ) k +4 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+6 \left (-a_{k -1}+a_{k +1}\right ) r +\left (a +2\right ) a_{k -1}+a_{k} b +2 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) \left (k +1\right )^{2}+2 \left (4 \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r -3 a_{k}+3 a_{k +2}\right ) \left (k +1\right )+4 \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r^{2}+6 \left (-a_{k}+a_{k +2}\right ) r +\left (a +2\right ) a_{k}+a_{k +1} b +2 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+8 k r a_{k}-16 k r a_{k +1}+4 r^{2} a_{k}-8 r^{2} a_{k +1}+a_{k} a +a_{k +1} b +2 k a_{k}-16 k a_{k +1}+2 r a_{k}-16 r a_{k +1}-8 a_{k +1}}{2 \left (2 k^{2}+4 k r +2 r^{2}+7 k +7 r +6\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +2 k a_{k}-16 k a_{k +1}-8 a_{k +1}}{2 \left (2 k^{2}+7 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +2 k a_{k}-16 k a_{k +1}-8 a_{k +1}}{2 \left (2 k^{2}+7 k +6\right )}, b a_{0}+2 a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +6 k a_{k}-24 k a_{k +1}+2 a_{k}-18 a_{k +1}}{2 \left (2 k^{2}+9 k +10\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+a_{k} a +a_{k +1} b +6 k a_{k}-24 k a_{k +1}+2 a_{k}-18 a_{k +1}}{2 \left (2 k^{2}+9 k +10\right )}, 6 a_{1}+a_{0} \left (b -2\right )=0\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 +\frac {1}{2}}\right ), c_{k +2}=-\frac {4 k^{2} c_{k}-8 k^{2} c_{k +1}+a c_{k}+b c_{k +1}+2 k c_{k}-16 k c_{k +1}-8 c_{k +1}}{2 \left (2 k^{2}+7 k +6\right )}, b c_{0}+2 c_{1}=0, d_{k +2}=-\frac {4 k^{2} d_{k}-8 k^{2} d_{k +1}+a d_{k}+b d_{k +1}+6 k d_{k}-24 k d_{k +1}+2 d_{k}-18 d_{k +1}}{2 \left (2 k^{2}+9 k +10\right )}, 6 d_{1}+d_{0} \left (b -2\right )=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 <- Legendre successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 57
dsolve(diff(diff(y(x),x),x) = -1/2/x*(3*x-1)/(x-1)*diff(y(x),x)-1/4*(a*x+b)/x/(x-1)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {-4 a +1}}{2}-\frac {1}{2}, \sqrt {-a -b}, \sqrt {x}\right )+c_{2} \operatorname {LegendreQ}\left (\frac {\sqrt {-4 a +1}}{2}-\frac {1}{2}, \sqrt {-a -b}, \sqrt {x}\right ) \]
✓ Solution by Mathematica
Time used: 0.834 (sec). Leaf size: 510
DSolve[y''[x] == -1/4*((b + a*x)*y[x])/((-1 + x)^2*x) - ((-1 + 3*x)*y'[x])/(2*(-1 + x)*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {(x-1)^{\frac {2 a \sqrt {-4 \sqrt {(4 a-1) (a+b)}-8 a-4 b+1}+2 b \left (\sqrt {-4 \sqrt {(4 a-1) (a+b)}-8 a-4 b+1}+2\right )-\sqrt {(4 a-1) (a+b)} \sqrt {-4 \sqrt {(4 a-1) (a+b)}-8 a-4 b+1}+1}{8 b+2}} \left (c_1 \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (\sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}+1\right ),\frac {8 \sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1} a+4 b \left (\sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}+1\right )-4 \sqrt {(4 a-1) (a+b)} \sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}-\sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}+1}{16 b+4},\frac {1}{2},x\right )+i c_2 \sqrt {x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (\sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}+3\right ),\frac {8 \sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1} a+4 b \left (\sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}+3\right )-4 \sqrt {(4 a-1) (a+b)} \sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}-\sqrt {-8 a-4 b-4 \sqrt {(4 a-1) (a+b)}+1}+3}{16 b+4},\frac {3}{2},x\right )\right )}{\sqrt {1-x}} \]