Internal problem ID [10930]
Internal file name [OUTPUT/10187_Sunday_December_31_2023_11_03_45_AM_79548268/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: 107.
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.
\[ \boxed {\left (a x +b \right ) y^{\prime \prime }+s \left (c x +d \right ) y^{\prime }-s^{2} \left (\left (a +c \right ) x +b +d \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a x +b \right ) \left (\frac {d}{d x}y^{\prime }\right )+s \left (c x +d \right ) y^{\prime }-s^{2} \left (\left (a +c \right ) x +b +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 {s^{2} \left (a x +c x +b +d \right ) y}{a x +b}-\frac {s \left (c x +d \right ) y^{\prime }}{a x +b} \\ \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 {s \left (c x +d \right ) y^{\prime }}{a x +b}-\frac {s^{2} \left (a x +c x +b +d \right ) y}{a x +b}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {b}{a}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {s \left (c x +d \right )}{a x +b}, P_{3}\left (x \right )=-\frac {s^{2} \left (a x +c x +b +d \right )}{a x +b}\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 {s \left (-\frac {c b}{a}+d \right )}{a} \\ {} & \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}=-\frac {b}{a}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-\frac {b}{a} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (a x +b \right ) \left (\frac {d}{d x}y^{\prime }\right )+s \left (c x +d \right ) y^{\prime }-s^{2} \left (a x +c x +b +d \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 \\ {} & {} & a u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (s c u -\frac {c s b}{a}+d s \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-s^{2} a u -s^{2} c u +\frac {s^{2} b c}{a}-d \,s^{2}\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 y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \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..1 \\ {} & {} & 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 \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & \frac {a_{0} r \left (a^{2} r +d s a -s b c -a^{2}\right ) u^{-1+r}}{a}+\left (\frac {a_{1} \left (1+r \right ) \left (a^{2} r +d s a -s b c \right )}{a}+\frac {a_{0} s \left (a c r -d s a +s b c \right )}{a}\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (\frac {a_{k +1} \left (k +1+r \right ) \left (a^{2} \left (k +1\right )+a^{2} r +d s a -s b c -a^{2}\right )}{a}+\frac {a_{k} s \left (a c k +a c r -d s a +s b c \right )}{a}-s^{2} a_{k -1} \left (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 \left (a^{2} r +d s a -s b c -a^{2}\right )}{a}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {-d s a +s b c +a^{2}}{a^{2}}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \frac {a_{1} \left (1+r \right ) \left (a^{2} r +d s a -s b c \right )}{a}+\frac {a_{0} s \left (a c r -d s a +s b c \right )}{a}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \frac {\left (-s^{2} a_{k -1}+a_{k +1} \left (k +1+r \right ) \left (k +r \right )\right ) a^{2}+\left (\left (-c a_{k -1}-d a_{k}\right ) s +d \left (k +1+r \right ) a_{k +1}+c a_{k} \left (k +r \right )\right ) s a -b s \left (-a_{k} s +a_{k +1} \left (k +1+r \right )\right ) c}{a}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {\left (-s^{2} a_{k}+a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )\right ) a^{2}+\left (\left (-c a_{k}-d a_{k +1}\right ) s +d \left (k +2+r \right ) a_{k +2}+c a_{k +1} \left (k +1+r \right )\right ) s a -b s \left (-a_{k +1} s +a_{k +2} \left (k +2+r \right )\right ) c}{a}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {s \left (a^{2} s a_{k}-a c k a_{k +1}-a c r a_{k +1}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2+r \right ) \left (a^{2} k +a^{2} r +d s a -s b c +a^{2}\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {s \left (a^{2} s a_{k}-a c k a_{k +1}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2\right ) \left (a^{2} k +d s a -s b c +a^{2}\right )} \\ \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 {s \left (a^{2} s a_{k}-a c k a_{k +1}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2\right ) \left (a^{2} k +d s a -s b c +a^{2}\right )}, \frac {a_{1} \left (d s a -s b c \right )}{a}+\frac {a_{0} s \left (-d s a +s b c \right )}{a}=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 {s \left (a^{2} s a_{k}-a c k a_{k +1}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2\right ) \left (a^{2} k +d s a -s b c +a^{2}\right )}, \frac {a_{1} \left (d s a -s b c \right )}{a}+\frac {a_{0} s \left (-d s a +s b c \right )}{a}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {-d s a +s b c +a^{2}}{a^{2}} \\ {} & {} & a_{k +2}=\frac {s \left (a^{2} s a_{k}-a c k a_{k +1}-\frac {c \left (-d s a +s b c +a^{2}\right ) a_{k +1}}{a}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) \left (a^{2} k +2 a^{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {-d s a +s b c +a^{2}}{a^{2}} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {-d s a +s b c +a^{2}}{a^{2}}}, a_{k +2}=\frac {s \left (a^{2} s a_{k}-a c k a_{k +1}-\frac {c \left (-d s a +s b c +a^{2}\right ) a_{k +1}}{a}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) \left (a^{2} k +2 a^{2}\right )}, a_{1} \left (1+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) a +\frac {a_{0} s \left (\frac {c \left (-d s a +s b c +a^{2}\right )}{a}-d s a +s b c \right )}{a}=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 {-d s a +s b c +a^{2}}{a^{2}}}, a_{k +2}=\frac {s \left (a^{2} s a_{k}-a c k a_{k +1}-\frac {c \left (-d s a +s b c +a^{2}\right ) a_{k +1}}{a}+a c s a_{k}+a d s a_{k +1}-b c s a_{k +1}-a c a_{k +1}\right )}{\left (k +2+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) \left (a^{2} k +2 a^{2}\right )}, a_{1} \left (1+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) a +\frac {a_{0} s \left (\frac {c \left (-d s a +s b c +a^{2}\right )}{a}-d s a +s b c \right )}{a}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} \left (x +\frac {b}{a}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} \left (x +\frac {b}{a}\right )^{k +\frac {-d s a +s b c +a^{2}}{a^{2}}}\right ), e_{k +2}=\frac {s \left (a^{2} s e_{k}-a c k e_{k +1}+a c s e_{k}+a d s e_{k +1}-b c s e_{k +1}-a c e_{k +1}\right )}{\left (k +2\right ) \left (a^{2} k +d s a -s b c +a^{2}\right )}, \frac {e_{1} \left (d s a -s b c \right )}{a}+\frac {e_{0} s \left (-d s a +s b c \right )}{a}=0, f_{k +2}=\frac {s \left (a^{2} s f_{k}-a c k f_{k +1}-\frac {c \left (-d s a +s b c +a^{2}\right ) f_{k +1}}{a}+a c s f_{k}+a d s f_{k +1}-b c s f_{k +1}-a c f_{k +1}\right )}{\left (k +2+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) \left (a^{2} k +2 a^{2}\right )}, f_{1} \left (1+\frac {-d s a +s b c +a^{2}}{a^{2}}\right ) a +\frac {f_{0} s \left (\frac {c \left (-d s a +s b c +a^{2}\right )}{a}-d s a +s b c \right )}{a}=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 <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form could result into a too large expression - returning special function form of solution, free of uncomputed <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 166
dsolve((a*x+b)*diff(y(x),x$2)+s*(c*x+d)*diff(y(x),x)-s^2*((a+c)*x+b+d)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (\left (-c_{1} +c_{2} \right ) a^{2}+a d s c_{1} -b c s c_{1} \right ) \Gamma \left (\frac {-d s a +b c s +a^{2}}{a^{2}}, \frac {s \left (2 a +c \right ) \left (a x +b \right )}{a^{2}}\right )+\Gamma \left (\frac {-d s a +b c s +2 a^{2}}{a^{2}}\right ) c_{1} a^{2}\right ) \left (a x +b \right )^{\frac {-d s a +b c s +a^{2}}{a^{2}}} \left (\frac {s \left (2 a +c \right ) \left (a x +b \right )}{a^{2}}\right )^{\frac {d s a -b c s -a^{2}}{a^{2}}} {\mathrm e}^{\frac {s \left (a^{2} x +2 a b +b c \right )}{a^{2}}}}{a^{2}} \]
✓ Solution by Mathematica
Time used: 1.269 (sec). Leaf size: 122
DSolve[(a*x+b)*y''[x]+s*(c*x+d)*y'[x]-s^2*((a+c)*x+b+d)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 e^{s x}-\frac {c_2 e^{s \left (\frac {b (2 a+c)}{a^2}+x\right )} (a x+b)^{\frac {s (b c-a d)}{a^2}+1} \left (\frac {s (2 a+c) (a x+b)}{a^2}\right )^{\frac {s (a d-b c)}{a^2}-1} \Gamma \left (\frac {a^2-d s a+b c s}{a^2},\frac {(2 a+c) s (b+a x)}{a^2}\right )}{a} \]