Internal problem ID [10907]
Internal file name [OUTPUT/10164_Sunday_December_31_2023_11_03_13_AM_15472296/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 2.1.2-3 Equation of form
\((a x + b)y''+f(x)y'+g(x)y=0\)
Problem number: 84.
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.
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (A \,x^{2}+B x +\operatorname {C0} \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (A \,x^{2}+B x +\mathit {C0} \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 \,x^{2}+B x +\mathit {C0} \right ) y}{x}-\frac {\left (a \,x^{2}+b x +c \right ) y^{\prime }}{x} \\ \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^{2}+b x +c \right ) y^{\prime }}{x}+\frac {\left (A \,x^{2}+B x +\mathit {C0} \right ) y}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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^{2}+b x +c}{x}, P_{3}\left (x \right )=\frac {A \,x^{2}+B x +\mathit {C0}}{x}\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 \\ {} & \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}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (A \,x^{2}+B x +\mathit {C0} \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..2 \\ {} & {} & 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 \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r +c \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} \left (b r +\mathit {C0} \right )\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (1+r +c \right )+a_{1} \left (b r +\mathit {C0} +b \right )+a_{0} \left (a r +B \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +c \right )+a_{k} \left (b k +b r +\mathit {C0} \right )+a_{k -1} \left (a \left (k -1\right )+a r +B \right )+A a_{k -2}\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 +c \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -c +1\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} \left (b r +\mathit {C0} \right )=0, a_{2} \left (2+r \right ) \left (1+r +c \right )+a_{1} \left (b r +\mathit {C0} +b \right )+a_{0} \left (a r +B \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0} \left (b r +\mathit {C0} \right )}{r c +r^{2}+c +r}, a_{2}=-\frac {a_{0} \left (a \,r^{2} c +a \,r^{3}-b^{2} r^{2}+B r c +B \,r^{2}-2 b r \mathit {C0} +a r c +a \,r^{2}-b^{2} r +B c +B r -\mathit {C0}^{2}-\mathit {C0} b \right )}{r^{2} c^{2}+2 r^{3} c +r^{4}+3 r \,c^{2}+7 r^{2} c +4 r^{3}+2 c^{2}+7 r c +5 r^{2}+2 c +2 r}\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 +c \right )+\left (\left (k +r \right ) b +\mathit {C0} \right ) a_{k}+\left (\left (k +r -1\right ) a +B \right ) a_{k -1}+A a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +2+r +c \right )+\left (\left (k +2+r \right ) b +\mathit {C0} \right ) a_{k +2}+\left (\left (k +1+r \right ) a +B \right ) a_{k +1}+A a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+a r a_{k +1}+b k a_{k +2}+b r a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +3+r \right ) \left (k +2+r +c \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +3\right ) \left (k +2+c \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +3\right ) \left (k +2+c \right )}, a_{1}=-\frac {a_{0} \mathit {C0}}{c}, a_{2}=-\frac {a_{0} \left (B c -\mathit {C0}^{2}-\mathit {C0} b \right )}{2 c^{2}+2 c}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-c +1 \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+a \left (-c +1\right ) a_{k +1}+b k a_{k +2}+b \left (-c +1\right ) a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +4-c \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-c +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -c +1}, a_{k +3}=-\frac {a k a_{k +1}+a \left (-c +1\right ) a_{k +1}+b k a_{k +2}+b \left (-c +1\right ) a_{k +2}+A a_{k}+B a_{k +1}+\mathit {C0} a_{k +2}+a a_{k +1}+2 b a_{k +2}}{\left (k +4-c \right ) \left (k +3\right )}, a_{1}=-\frac {a_{0} \left (b \left (-c +1\right )+\mathit {C0} \right )}{\left (-c +1\right ) c +\left (-c +1\right )^{2}+1}, a_{2}=-\frac {a_{0} \left (a \left (-c +1\right )^{2} c +a \left (-c +1\right )^{3}-b^{2} \left (-c +1\right )^{2}+B \left (-c +1\right ) c +B \left (-c +1\right )^{2}-2 b \left (-c +1\right ) \mathit {C0} +a \left (-c +1\right ) c +a \left (-c +1\right )^{2}-b^{2} \left (-c +1\right )+B c +B \left (-c +1\right )-\mathit {C0}^{2}-\mathit {C0} b \right )}{\left (-c +1\right )^{2} c^{2}+2 \left (-c +1\right )^{3} c +\left (-c +1\right )^{4}+3 \left (-c +1\right ) c^{2}+7 \left (-c +1\right )^{2} c +4 \left (-c +1\right )^{3}+2 c^{2}+7 \left (-c +1\right ) c +5 \left (-c +1\right )^{2}+2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -c +1}\right ), d_{k +3}=-\frac {a k d_{k +1}+b k d_{k +2}+A d_{k}+B d_{k +1}+\mathit {C0} d_{k +2}+a d_{k +1}+2 b d_{k +2}}{\left (k +3\right ) \left (k +2+c \right )}, d_{1}=-\frac {d_{0} \mathit {C0}}{c}, d_{2}=-\frac {d_{0} \left (B c -\mathit {C0}^{2}-\mathit {C0} b \right )}{2 c^{2}+2 c}, e_{k +3}=-\frac {a k e_{k +1}+a \left (-c +1\right ) e_{k +1}+b k e_{k +2}+b \left (-c +1\right ) e_{k +2}+A e_{k}+B e_{k +1}+\mathit {C0} e_{k +2}+a e_{k +1}+2 b e_{k +2}}{\left (k +4-c \right ) \left (k +3\right )}, e_{1}=-\frac {e_{0} \left (b \left (-c +1\right )+\mathit {C0} \right )}{\left (-c +1\right ) c +\left (-c +1\right )^{2}+1}, e_{2}=-\frac {e_{0} \left (a \left (-c +1\right )^{2} c +a \left (-c +1\right )^{3}-b^{2} \left (-c +1\right )^{2}+B \left (-c +1\right ) c +B \left (-c +1\right )^{2}-2 b \left (-c +1\right ) \mathit {C0} +a \left (-c +1\right ) c +a \left (-c +1\right )^{2}-b^{2} \left (-c +1\right )+B c +B \left (-c +1\right )-\mathit {C0}^{2}-\mathit {C0} b \right )}{\left (-c +1\right )^{2} c^{2}+2 \left (-c +1\right )^{3} c +\left (-c +1\right )^{4}+3 \left (-c +1\right ) c^{2}+7 \left (-c +1\right )^{2} c +4 \left (-c +1\right )^{3}+2 c^{2}+7 \left (-c +1\right ) c +5 \left (-c +1\right )^{2}+2}\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 -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.281 (sec). Leaf size: 186
dsolve(x*diff(y(x),x$2)+(a*x^2+b*x+c)*diff(y(x),x)+(A*x^2+B*x+C0)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {x \left (-a^{2} x -2 a b +2 A \right )}{2 a}} \left (x^{-c +1} \operatorname {HeunB}\left (-c +1, -\frac {\sqrt {2}\, \left (-a b +2 A \right )}{a^{\frac {3}{2}}}, \frac {\left (-c -1\right ) a^{3}+2 B \,a^{2}-2 A a b +2 A^{2}}{a^{3}}, \frac {\left (b c -2 \operatorname {C0} \right ) \sqrt {2}}{\sqrt {a}}, \frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) c_{2} +\operatorname {HeunB}\left (c -1, -\frac {\sqrt {2}\, \left (-a b +2 A \right )}{a^{\frac {3}{2}}}, \frac {\left (-c -1\right ) a^{3}+2 B \,a^{2}-2 A a b +2 A^{2}}{a^{3}}, \frac {\left (b c -2 \operatorname {C0} \right ) \sqrt {2}}{\sqrt {a}}, \frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) c_{1} \right ) \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[x*y''[x]+(a*x^2+b*x+c)*y'[x]+(A*x^2+B*x+C0)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
Not solved