Internal problem ID [11014]
Internal file name [OUTPUT/10271_Sunday_December_31_2023_11_33_37_AM_1689924/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: 191.
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.
\[ \boxed {x^{2} \left (a x +b \right ) y^{\prime \prime }+\left (c \,x^{2}+\left (a \lambda +2 b \right ) x +b \lambda \right ) y^{\prime }+\lambda \left (c -2 a \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (a x +b \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (c \,x^{2}+\left (a \lambda +2 b \right ) x +b \lambda \right ) y^{\prime }+\lambda \left (c -2 a \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 {\lambda \left (2 a -c \right ) y}{x^{2} \left (a x +b \right )}-\frac {\left (a x \lambda +c \,x^{2}+b \lambda +2 b x \right ) y^{\prime }}{x^{2} \left (a x +b \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 \lambda +c \,x^{2}+b \lambda +2 b x \right ) y^{\prime }}{x^{2} \left (a x +b \right )}-\frac {\lambda \left (2 a -c \right ) y}{x^{2} \left (a x +b \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 \lambda +c \,x^{2}+b \lambda +2 b x}{x^{2} \left (a x +b \right )}, P_{3}\left (x \right )=-\frac {\lambda \left (2 a -c \right )}{x^{2} \left (a x +b \right )}\right ] \\ {} & \circ & \left (x +\frac {b}{a}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {b}{a} \\ {} & {} & \left (\left (x +\frac {b}{a}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {b}{a}}}}=\frac {\left (\frac {b^{2} c}{a^{2}}-\frac {2 b^{2}}{a}\right ) a}{b^{2}} \\ {} & \circ & \left (x +\frac {b}{a}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {b}{a} \\ {} & {} & \left (\left (x +\frac {b}{a}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {b}{a}}}}=0 \\ {} & \circ & x =-\frac {b}{a}\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}=-\frac {b}{a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (a x +b \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (a x \lambda +c \,x^{2}+b \lambda +2 b x \right ) y^{\prime }-\lambda \left (2 a -c \right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {b}{a}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (a \,u^{3}-2 u^{2} b +\frac {u \,b^{2}}{a}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (a \lambda u +c \,u^{2}-\frac {2 c u b}{a}+\frac {b^{2} c}{a^{2}}+2 b u -\frac {2 b^{2}}{a}\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-2 a \lambda +c \lambda \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..2 \\ {} & {} & 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..3 \\ {} & {} & 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} \\ {} & {} & \frac {a_{0} r \,b^{2} \left (a r -3 a +c \right ) u^{r -1}}{a^{2}}+\left (\frac {a_{1} \left (1+r \right ) b^{2} \left (a r -2 a +c \right )}{a^{2}}+\frac {a_{0} \left (a r -2 a +c \right ) \left (a \lambda -2 b r \right )}{a}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (\frac {a_{k +1} \left (k +1+r \right ) b^{2} \left (a \left (k +1\right )+a r -3 a +c \right )}{a^{2}}+\frac {a_{k} \left (a k +a r -2 a +c \right ) \left (a \lambda -2 b k -2 b r \right )}{a}+a_{k -1} \left (k +r -1\right ) \left (a \left (k -1\right )+a r -a +c \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \frac {r \,b^{2} \left (a r -3 a +c \right )}{a^{2}}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3 a -c}{a}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \frac {a_{1} \left (1+r \right ) b^{2} \left (a r -2 a +c \right )}{a^{2}}+\frac {a_{0} \left (a r -2 a +c \right ) \left (a \lambda -2 b r \right )}{a}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {\left (\left (k a_{k -1}+\lambda a_{k}+r a_{k -1}-a_{k -1}\right ) a^{2}-2 a_{k} \left (k +r \right ) b a +a_{k +1} \left (k +1+r \right ) b^{2}\right ) \left (\left (k +r -2\right ) a +c \right )}{a^{2}}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {\left (\left (\left (k +1\right ) a_{k}+\lambda a_{k +1}+a_{k} r -a_{k}\right ) a^{2}-2 a_{k +1} \left (k +1+r \right ) b a +a_{k +2} \left (k +2+r \right ) b^{2}\right ) \left (\left (k +r -1\right ) a +c \right )}{a^{2}}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+a r a_{k}-2 b k a_{k +1}-2 b r a_{k +1}-2 b a_{k +1}\right )}{\left (k +2+r \right ) b^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}-2 b k a_{k +1}-2 b a_{k +1}\right )}{\left (k +2\right ) b^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}-2 b k a_{k +1}-2 b a_{k +1}\right )}{\left (k +2\right ) b^{2}}, \frac {a_{1} b^{2} \left (c -2 a \right )}{a^{2}}+a_{0} \left (c -2 a \right ) \lambda =0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {b}{a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {b}{a}\right )^{k}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}-2 b k a_{k +1}-2 b a_{k +1}\right )}{\left (k +2\right ) b^{2}}, \frac {a_{1} b^{2} \left (c -2 a \right )}{a^{2}}+a_{0} \left (c -2 a \right ) \lambda =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3 a -c}{a} \\ {} & {} & a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+\left (3 a -c \right ) a_{k}-2 b k a_{k +1}-\frac {2 b \left (3 a -c \right ) a_{k +1}}{a}-2 b a_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3 a -c}{a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {3 a -c}{a}}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+\left (3 a -c \right ) a_{k}-2 b k a_{k +1}-\frac {2 b \left (3 a -c \right ) a_{k +1}}{a}-2 b a_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}}, \frac {a_{1} \left (1+\frac {3 a -c}{a}\right ) b^{2}}{a}+a_{0} \left (a \lambda -\frac {2 b \left (3 a -c \right )}{a}\right )=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {b}{a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {b}{a}\right )^{k +\frac {3 a -c}{a}}, a_{k +2}=-\frac {a \left (a k a_{k}+a \lambda a_{k +1}+\left (3 a -c \right ) a_{k}-2 b k a_{k +1}-\frac {2 b \left (3 a -c \right ) a_{k +1}}{a}-2 b a_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}}, \frac {a_{1} \left (1+\frac {3 a -c}{a}\right ) b^{2}}{a}+a_{0} \left (a \lambda -\frac {2 b \left (3 a -c \right )}{a}\right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (x +\frac {b}{a}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +\frac {b}{a}\right )^{k +\frac {3 a -c}{a}}\right ), d_{k +2}=-\frac {a \left (a k d_{k}+a \lambda d_{k +1}-2 b k d_{k +1}-2 b d_{k +1}\right )}{\left (k +2\right ) b^{2}}, \frac {d_{1} b^{2} \left (c -2 a \right )}{a^{2}}+d_{0} \left (c -2 a \right ) \lambda =0, e_{k +2}=-\frac {a \left (a k e_{k}+a \lambda e_{k +1}+\left (3 a -c \right ) e_{k}-2 b k e_{k +1}-\frac {2 b \left (3 a -c \right ) e_{k +1}}{a}-2 b e_{k +1}\right )}{\left (k +2+\frac {3 a -c}{a}\right ) b^{2}}, \frac {e_{1} \left (1+\frac {3 a -c}{a}\right ) b^{2}}{a}+e_{0} \left (a \lambda -\frac {2 b \left (3 a -c \right )}{a}\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 -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a = 0, e <> 0, c <> 0 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.469 (sec). Leaf size: 169
dsolve(x^2*(a*x+b)*diff(y(x),x$2)+(c*x^2+(2*b+a*lambda)*x+b*lambda)*diff(y(x),x)+lambda*(c-2*a)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (a x +b \right )^{\frac {3 a -c}{a}} \left (c_{1} x^{\frac {-3 a +c}{a}} \operatorname {HeunC}\left (\frac {\lambda a}{b}, 1-\frac {c}{a}, 3-\frac {c}{a}, 0, -\frac {\lambda a}{b}+\frac {c \lambda }{2 b}+\frac {5}{2}-\frac {2 c}{a}+\frac {c^{2}}{2 a^{2}}, -\frac {b}{a x}\right ) x^{2}+c_{2} \operatorname {HeunC}\left (\frac {\lambda a}{b}, \frac {c}{a}-1, 3-\frac {c}{a}, 0, -\frac {\lambda a}{b}+\frac {c \lambda }{2 b}+\frac {5}{2}-\frac {2 c}{a}+\frac {c^{2}}{2 a^{2}}, -\frac {b}{a x}\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 1.48 (sec). Leaf size: 55
DSolve[x^2*(a*x+b)*y''[x]+(c*x^2+(2*b+a*\[Lambda])*x+b*\[Lambda])*y'[x]+\[Lambda]*(c-2*a)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{\frac {\lambda }{x}} \left (c_2 \int _1^x\frac {e^{-\frac {\lambda }{K[1]}} (b+a K[1])^{2-\frac {c}{a}}}{K[1]^2}dK[1]+c_1\right ) \]