Internal problem ID [11019]
Internal file name [OUTPUT/10276_Sunday_December_31_2023_11_39_46_AM_38210555/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-6 Equation of form
\((a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 196.
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.
\[ \boxed {\left (a \,x^{3}+x^{2} b +c x \right ) y^{\prime \prime }+\left (\alpha \,x^{2}+\beta x +2 c \right ) y^{\prime }-\left (x \alpha +2 b -\beta \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (a \,x^{2}+b x +c \right )+\left (\alpha \,x^{2}+\beta x +2 c \right ) y^{\prime }+\left (-x \alpha -2 b +\beta \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 (x \alpha +2 b -\beta \right ) y}{x \left (a \,x^{2}+b x +c \right )}-\frac {\left (\alpha \,x^{2}+\beta x +2 c \right ) y^{\prime }}{x \left (a \,x^{2}+b x +c \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 (\alpha \,x^{2}+\beta x +2 c \right ) y^{\prime }}{x \left (a \,x^{2}+b x +c \right )}-\frac {\left (x \alpha +2 b -\beta \right ) y}{x \left (a \,x^{2}+b x +c \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 {\alpha \,x^{2}+\beta x +2 c}{x \left (a \,x^{2}+b x +c \right )}, P_{3}\left (x \right )=-\frac {x \alpha +2 b -\beta }{x \left (a \,x^{2}+b x +c \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}}}=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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (a \,x^{2}+b x +c \right )+\left (\alpha \,x^{2}+\beta x +2 c \right ) y^{\prime }+\left (-x \alpha -2 b +\beta \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} \\ {} & {} & c a_{0} r \left (r +1\right ) x^{r -1}+\left (c a_{1} \left (r +1\right ) \left (2+r \right )+a_{0} \left (r +1\right ) \left (b r -2 b +\beta \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (c a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+a_{k} \left (k +r +1\right ) \left (b k +b r -2 b +\beta \right )+a_{k -1} \left (k +r -2\right ) \left (a \left (k -1\right )+a r +\alpha \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & c r \left (r +1\right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & c a_{1} \left (r +1\right ) \left (2+r \right )+a_{0} \left (r +1\right ) \left (b r -2 b +\beta \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & c a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )+\left (k +r +1\right ) \left (\left (k +r -2\right ) b +\beta \right ) a_{k}+\left (\left (k +r -1\right ) a +\alpha \right ) \left (k +r -2\right ) a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & c a_{k +2} \left (k +2+r \right ) \left (k +3+r \right )+\left (k +2+r \right ) \left (\left (k +r -1\right ) b +\beta \right ) a_{k +1}+\left (\left (k +r \right ) a +\alpha \right ) \left (k +r -1\right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}+2 a k r a_{k}+a \,r^{2} a_{k}+b \,k^{2} a_{k +1}+2 b k r a_{k +1}+b \,r^{2} a_{k +1}-a k a_{k}-a r a_{k}+\alpha k a_{k}+\alpha r a_{k}+b k a_{k +1}+b r a_{k +1}+\beta k a_{k +1}+\beta r a_{k +1}-a_{k} \alpha -2 b a_{k +1}+2 \beta a_{k +1}}{c \left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}+b \,k^{2} a_{k +1}-3 a k a_{k}+\alpha k a_{k}-b k a_{k +1}+\beta k a_{k +1}+2 a a_{k}-2 a_{k} \alpha -2 b a_{k +1}+\beta a_{k +1}}{c \left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=-\frac {a \,k^{2} a_{k}+b \,k^{2} a_{k +1}-3 a k a_{k}+\alpha k a_{k}-b k a_{k +1}+\beta k a_{k +1}+2 a a_{k}-2 a_{k} \alpha -2 b a_{k +1}+\beta a_{k +1}}{c \left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}+b \,k^{2} a_{k +1}-a k a_{k}+\alpha k a_{k}+b k a_{k +1}+\beta k a_{k +1}-a_{k} \alpha -2 b a_{k +1}+2 \beta a_{k +1}}{c \left (k +2\right ) \left (k +3\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 {a \,k^{2} a_{k}+b \,k^{2} a_{k +1}-a k a_{k}+\alpha k a_{k}+b k a_{k +1}+\beta k a_{k +1}-a_{k} \alpha -2 b a_{k +1}+2 \beta a_{k +1}}{c \left (k +2\right ) \left (k +3\right )}, 2 c a_{1}+a_{0} \left (\beta -2 b \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k}\right ), d_{k +2}=-\frac {a \,k^{2} d_{k}+b \,k^{2} d_{k +1}-3 a k d_{k}+\alpha k d_{k}-b k d_{k +1}+\beta k d_{k +1}+2 a d_{k}-2 \alpha d_{k}-2 b d_{k +1}+\beta d_{k +1}}{c \left (k +1\right ) \left (k +2\right )}, 0=0, e_{k +2}=-\frac {a \,k^{2} e_{k}+b \,k^{2} e_{k +1}-a k e_{k}+\alpha k e_{k}+b k e_{k +1}+\beta k e_{k +1}-\alpha e_{k}-2 b e_{k +1}+2 \beta e_{k +1}}{c \left (k +2\right ) \left (k +3\right )}, 2 c e_{1}+e_{0} \left (\beta -2 b \right )=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> 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 <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 1.062 (sec). Leaf size: 1505
dsolve((a*x^3+b*x^2+c*x)*diff(y(x),x$2)+(alpha*x^2+beta*x+2*c)*diff(y(x),x)-(alpha*x+2*b-beta)*y(x)=0,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 6.092 (sec). Leaf size: 224
DSolve[(a*x^3+b*x^2+c*x)*y''[x]+(\[Alpha]*x^2+\[Beta]*x+2*c)*y'[x]-(\[Alpha]*x+2*b-\[Beta])*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\left (b (2 a x-3 \beta -2 \alpha x)-a \alpha x^2-2 a \beta x+2 b^2+\beta ^2+\alpha c+\alpha ^2 x^2+2 \alpha \beta x\right ) \left (c_2 \int _1^x\frac {\exp \left (\frac {(b \alpha +2 a (b-\beta )) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{a \sqrt {4 a c-b^2}}\right ) (c+K[1] (b+a K[1]))^{1-\frac {\alpha }{2 a}}}{\left (2 b^2-3 \beta b+2 (a-\alpha ) K[1] b+\beta ^2+\alpha ^2 K[1]^2-a \alpha K[1]^2+c \alpha -2 a \beta K[1]+2 \alpha \beta K[1]\right )^2}dK[1]+c_1\right )}{x \left (2 b^2+\beta ^2-3 \beta b+\alpha c\right )} \]