Internal problem ID [9666]
Internal file name [OUTPUT/8608_Monday_June_06_2022_04_25_27_AM_52839156/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1339.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (a \left (b +2\right ) x^{2}+\left (c -d +1\right ) x \right ) y^{\prime }}{\left (x a +1\right ) x^{2}}+\frac {\left (a b x -c d \right ) y}{\left (x a +1\right ) x^{2}}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{3}+x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (a \left (b +2\right ) x +c -d +1\right ) y^{\prime }+\left (a b x -c d \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 (a b x -c d \right ) y}{\left (x a +1\right ) x^{2}}-\frac {\left (a b x +2 x a +c -d +1\right ) y^{\prime }}{x \left (x a +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 b x +2 x a +c -d +1\right ) y^{\prime }}{x \left (x a +1\right )}+\frac {\left (a b x -c d \right ) y}{\left (x a +1\right ) 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 {a b x +2 x a +c -d +1}{\left (x a +1\right ) x}, P_{3}\left (x \right )=\frac {a b x -c d}{\left (x a +1\right ) x^{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}}}=c -d +1 \\ {} & \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}}}=-c d \\ {} & \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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x a +1\right ) x^{2}+x \left (a b x +2 x a +c -d +1\right ) y^{\prime }+\left (a b x -c d \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 =1..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 =2..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} \\ {} & {} & -a_{0} \left (d -r \right ) \left (c +r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (d -k -r \right ) \left (c +k +r \right )+a a_{k -1} \left (k +r \right ) \left (k -1+b +r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (d -r \right ) \left (c +r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{d , -c \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k} \left (d -k -r \right ) \left (c +k +r \right )+a a_{k -1} \left (k +r \right ) \left (k -1+b +r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -a_{k +1} \left (d -k -1-r \right ) \left (c +k +1+r \right )+a a_{k} \left (k +r +1\right ) \left (k +b +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a a_{k} \left (k +r +1\right ) \left (k +b +r \right )}{\left (d -k -1-r \right ) \left (c +k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =d \\ {} & {} & a_{k +1}=\frac {a a_{k} \left (k +d +1\right ) \left (k +b +d \right )}{\left (-k -1\right ) \left (c +k +1+d \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =d \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +d}, a_{k +1}=\frac {a a_{k} \left (k +d +1\right ) \left (k +b +d \right )}{\left (-k -1\right ) \left (c +k +1+d \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-c \\ {} & {} & a_{k +1}=\frac {a a_{k} \left (k -c +1\right ) \left (k +b -c \right )}{\left (d -k -1+c \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-c \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -c}, a_{k +1}=\frac {a a_{k} \left (k -c +1\right ) \left (k +b -c \right )}{\left (d -k -1+c \right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +d}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k -c}\right ), e_{k +1}=\frac {a e_{k} \left (k +d +1\right ) \left (k +b +d \right )}{\left (-k -1\right ) \left (c +k +1+d \right )}, f_{k +1}=\frac {a f_{k} \left (k -c +1\right ) \left (k +b -c \right )}{\left (d -k -1+c \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: 0.157 (sec). Leaf size: 76
dsolve(diff(diff(y(x),x),x) = -(a*(b+2)*x^2+(c-d+1)*x)/(a*x+1)/x^2*diff(y(x),x)-(a*b*x-c*d)/(a*x+1)/x^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (x^{-c} \operatorname {hypergeom}\left (\left [-d , 1-b -d \right ], \left [1-d -c \right ], -a x \right ) c_{2} +x^{d} \operatorname {hypergeom}\left (\left [c , 1-b +c \right ], \left [1+d +c \right ], -a x \right ) c_{1} \right ) \left (a x +1\right )^{-b +c -d} \]
✓ Solution by Mathematica
Time used: 0.268 (sec). Leaf size: 66
DSolve[y''[x] == -(((-(c*d) + a*b*x)*y[x])/(x^2*(1 + a*x))) - (((1 + c - d)*x + a*(2 + b)*x^2)*y'[x])/(x^2*(1 + a*x)),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 a^{-c} x^{-c} \operatorname {Hypergeometric2F1}(1-c,b-c,-c-d+1,-a x)+c_2 a^d x^d \operatorname {Hypergeometric2F1}(d+1,b+d,c+d+1,-a x) \]