Internal problem ID [10902]
Internal file name [OUTPUT/10159_Sunday_December_31_2023_11_03_03_AM_54718002/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: 79.
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.
\[ \boxed {x y^{\prime \prime }+\left (a b \,x^{2}+b -5\right ) y^{\prime }+2 a^{2} \left (b -2\right ) x^{3} y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a b \,x^{2}+b -5\right ) y^{\prime }+2 a^{2} \left (b -2\right ) x^{3} 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 b \,x^{2}+b -5\right ) y^{\prime }}{x}-2 a^{2} x^{2} \left (b -2\right ) y \\ \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 b \,x^{2}+b -5\right ) y^{\prime }}{x}+2 a^{2} x^{2} \left (b -2\right ) y=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 b \,x^{2}+b -5}{x}, P_{3}\left (x \right )=2 a^{2} x^{2} \left (b -2\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}}}=b -5 \\ {} & \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 b \,x^{2}+b -5\right ) y^{\prime }+2 a^{2} \left (b -2\right ) x^{3} 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^{3}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +3} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -3 \\ {} & {} & x^{3}\cdot y=\moverset {\infty }{\munderset {k =3}{\sum }}a_{k -3} 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 (-6+r +b \right ) x^{-1+r}+a_{1} \left (1+r \right ) \left (-5+r +b \right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (-4+r +b \right )+a_{0} a b r \right ) x^{1+r}+\left (a_{3} \left (3+r \right ) \left (-3+r +b \right )+a_{1} \left (1+r \right ) a b \right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k -5+r +b \right )+a_{k -1} \left (k +r -1\right ) a b +2 a^{2} a_{k -3} \left (b -2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-6+r +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -b +6\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 (-5+r +b \right )=0, a_{2} \left (2+r \right ) \left (-4+r +b \right )+a_{0} a b r =0, a_{3} \left (3+r \right ) \left (-3+r +b \right )+a_{1} \left (1+r \right ) a b =0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=-\frac {a_{0} a b r}{r b +r^{2}+2 b -2 r -8}, a_{3}=0\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 -5+r +b \right )+a_{k -1} \left (k +r -1\right ) a b +2 a^{2} a_{k -3} \left (b -2\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k -2+r +b \right )+a_{k +2} \left (k +2+r \right ) a b +2 a^{2} a_{k} \left (b -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {a \left (2 a b a_{k}+b k a_{k +2}+b r a_{k +2}-4 a a_{k}+2 b a_{k +2}\right )}{\left (k +4+r \right ) \left (k -2+r +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=-\frac {a \left (2 a b a_{k}+b k a_{k +2}-4 a a_{k}+2 b a_{k +2}\right )}{\left (k +4\right ) \left (k -2+b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=-\frac {a \left (2 a b a_{k}+b k a_{k +2}-4 a a_{k}+2 b a_{k +2}\right )}{\left (k +4\right ) \left (k -2+b \right )}, a_{1}=0, a_{2}=0, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b +6 \\ {} & {} & a_{k +4}=-\frac {a \left (2 a b a_{k}+b k a_{k +2}+b \left (-b +6\right ) a_{k +2}-4 a a_{k}+2 b a_{k +2}\right )}{\left (k +10-b \right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b +6 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -b +6}, a_{k +4}=-\frac {a \left (2 a b a_{k}+b k a_{k +2}+b \left (-b +6\right ) a_{k +2}-4 a a_{k}+2 b a_{k +2}\right )}{\left (k +10-b \right ) \left (k +4\right )}, a_{1}=0, a_{2}=-\frac {a_{0} a b \left (-b +6\right )}{\left (-b +6\right ) b +\left (-b +6\right )^{2}+4 b -20}, a_{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k -b +6}\right ), c_{k +4}=-\frac {a \left (2 a b c_{k}+b k c_{k +2}-4 a c_{k}+2 b c_{k +2}\right )}{\left (k +4\right ) \left (k -2+b \right )}, c_{1}=0, c_{2}=0, c_{3}=0, d_{k +4}=-\frac {a \left (2 a b d_{k}+b k d_{k +2}+b \left (-b +6\right ) d_{k +2}-4 a d_{k}+2 b d_{k +2}\right )}{\left (k +10-b \right ) \left (k +4\right )}, d_{1}=0, d_{2}=-\frac {d_{0} a b \left (-b +6\right )}{\left (-b +6\right ) b +\left (-b +6\right )^{2}+4 b -20}, d_{3}=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: 0.187 (sec). Leaf size: 92
dsolve(x*diff(y(x),x$2)+(a*b*x^2+b-5)*diff(y(x),x)+2*a^2*(b-2)*x^3*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-3 \operatorname {KummerU}\left (\frac {b}{2}+1, -2+\frac {b}{2}, \frac {a \left (b -4\right ) x^{2}}{2}\right ) c_{2} b +\left (a \left (b -4\right ) x^{2}+b +4\right ) c_{2} \operatorname {KummerU}\left (\frac {b}{2}, -2+\frac {b}{2}, \frac {a \left (b -4\right ) x^{2}}{2}\right )+2 c_{1} {\mathrm e}^{\frac {a \left (b -4\right ) x^{2}}{2}} \left (a \,x^{2}+1\right )\right ) {\mathrm e}^{-\frac {a \left (b -2\right ) x^{2}}{2}}}{2} \]
✓ Solution by Mathematica
Time used: 3.578 (sec). Leaf size: 67
DSolve[x*y''[x]+(a*b*x^2+b-5)*y'[x]+2*a^2*(b-2)*x^3*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-a x^2} \left (a x^2+1\right ) \left (c_2 \int _1^x\frac {e^{-\frac {1}{2} a (b-4) K[1]^2} K[1]^{5-b}}{\left (a K[1]^2+1\right )^2}dK[1]+c_1\right ) \]