Internal problem ID [10908]
Internal file name [OUTPUT/10165_Sunday_December_31_2023_11_03_13_AM_53294476/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: 85.
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.
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{2}+b x +2\right ) y^{\prime }+\left (c \,x^{2}+d x +b \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b x +2\right ) y^{\prime }+\left (c \,x^{2}+d x +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 (c \,x^{2}+d x +b \right ) y}{x}-\frac {\left (a \,x^{2}+b x +2\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 +2\right ) y^{\prime }}{x}+\frac {\left (c \,x^{2}+d x +b \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 +2}{x}, P_{3}\left (x \right )=\frac {c \,x^{2}+d x +b}{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}}}=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}=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 +2\right ) y^{\prime }+\left (c \,x^{2}+d x +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..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 \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2+r \right )+a_{0} b \left (1+r \right )\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (3+r \right )+a_{1} b \left (2+r \right )+a_{0} \left (a r +d \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +2+r \right )+a_{k} b \left (k +1+r \right )+a_{k -1} \left (a \left (k -1\right )+a r +d \right )+a_{k -2} c \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 \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\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 (2+r \right )+a_{0} b \left (1+r \right )=0, a_{2} \left (2+r \right ) \left (3+r \right )+a_{1} b \left (2+r \right )+a_{0} \left (a r +d \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0} b}{2+r}, a_{2}=-\frac {a_{0} \left (a r -b^{2}+d \right )}{r^{2}+5 r +6}\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 +2+r \right )+a_{k} b \left (k +1+r \right )+a_{k -1} \left (\left (k +r -1\right ) a +d \right )+a_{k -2} c =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +4+r \right )+a_{k +2} b \left (k +3+r \right )+a_{k +1} \left (\left (k +1+r \right ) a +d \right )+a_{k} c =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 +1}+3 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +3+r \right ) \left (k +4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+2 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +3}=-\frac {a k a_{k +1}+b k a_{k +2}+2 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +2\right ) \left (k +3\right )}, a_{1}=-a_{0} b , a_{2}=-\frac {a_{0} \left (-b^{2}-a +d \right )}{2}\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 +1}+3 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +3\right ) \left (k +4\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 +1}+3 b a_{k +2}+a_{k} c +d a_{k +1}}{\left (k +3\right ) \left (k +4\right )}, a_{1}=-\frac {a_{0} b}{2}, a_{2}=-\frac {a_{0} \left (-b^{2}+d \right )}{6}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k}\right ), e_{k +3}=-\frac {a k e_{k +1}+b k e_{k +2}+2 b e_{k +2}+c e_{k}+d e_{k +1}}{\left (k +2\right ) \left (k +3\right )}, e_{1}=-e_{0} b , e_{2}=-\frac {e_{0} \left (-b^{2}-a +d \right )}{2}, f_{k +3}=-\frac {a k f_{k +1}+b k f_{k +2}+a f_{k +1}+3 b f_{k +2}+c f_{k}+d f_{k +1}}{\left (k +3\right ) \left (k +4\right )}, f_{1}=-\frac {f_{0} b}{2}, f_{2}=-\frac {f_{0} \left (-b^{2}+d \right )}{6}\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 <- hyper3 successful: indirect Equivalence to 0F1 under \`\`^ @ Moebius\`\` is resolved <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.046 (sec). Leaf size: 143
dsolve(x*diff(y(x),x$2)+(a*x^2+b*x+2)*diff(y(x),x)+(c*x^2+d*x+b)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-\frac {x \left (a^{2} x +2 a b -2 c \right )}{2 a}} \left (\operatorname {hypergeom}\left (\left [\frac {3 a^{3}-d \,a^{2}+a b c -c^{2}}{2 a^{3}}\right ], \left [\frac {3}{2}\right ], \frac {\left (a^{2} x +a b -2 c \right )^{2}}{2 a^{3}}\right ) \left (a^{2} x +a b -2 c \right ) c_{2} +c_{1} \operatorname {hypergeom}\left (\left [\frac {2 a^{3}-d \,a^{2}+a b c -c^{2}}{2 a^{3}}\right ], \left [\frac {1}{2}\right ], \frac {\left (a^{2} x +a b -2 c \right )^{2}}{2 a^{3}}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.155 (sec). Leaf size: 134
DSolve[x*y''[x]+(a*x^2+b*x+2)*y'[x]+(c*x^2+d*x+b)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {e^{-\frac {1}{2} x \left (-\frac {2 c}{a}+a x+2 b\right )} \left (c_2 \operatorname {Hypergeometric1F1}\left (-\frac {-2 a^3+d a^2-b c a+c^2}{2 a^3},\frac {1}{2},\frac {\left (x a^2+b a-2 c\right )^2}{2 a^3}\right )+c_1 \operatorname {HermiteH}\left (\frac {-2 a^3+d a^2-b c a+c^2}{a^3},\frac {x a^2+b a-2 c}{\sqrt {2} a^{3/2}}\right )\right )}{x} \]