Internal problem ID [10901]
Internal file name [OUTPUT/10158_Sunday_December_31_2023_11_03_03_AM_20057251/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: 78.
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.
\[ \boxed {x y^{\prime \prime }-\left (2 a x +1\right ) y^{\prime }+y b \,x^{3}=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 a x -1\right ) y^{\prime }+y b \,x^{3}=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 }=-b \,x^{2} y+\frac {\left (2 a x +1\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 (2 a x +1\right ) y^{\prime }}{x}+b \,x^{2} 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 {2 a x +1}{x}, P_{3}\left (x \right )=b \,x^{2}\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}}}=-1 \\ {} & \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 (-2 a x -1\right ) y^{\prime }+y b \,x^{3}=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..1 \\ {} & {} & 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 (-2+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-1+r \right )-2 a_{0} a r \right ) x^{r}+\left (a_{2} \left (2+r \right ) r -2 a a_{1} \left (1+r \right )\right ) x^{1+r}+\left (a_{3} \left (3+r \right ) \left (1+r \right )-2 a a_{2} \left (2+r \right )\right ) x^{2+r}+\left (\moverset {\infty }{\munderset {k =3}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )-2 a a_{k} \left (k +r \right )+a_{k -3} b \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\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 (-1+r \right )-2 a_{0} a r =0, a_{2} \left (2+r \right ) r -2 a a_{1} \left (1+r \right )=0, a_{3} \left (3+r \right ) \left (1+r \right )-2 a a_{2} \left (2+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {2 a_{0} a r}{r^{2}-1}, a_{2}=\frac {4 a^{2} a_{0}}{r^{2}+r -2}, a_{3}=\frac {8 a^{3} a_{0}}{r^{3}+3 r^{2}-r -3}\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 -1\right )-2 a a_{k} \left (k +r \right )+a_{k -3} b =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +3 \\ {} & {} & a_{k +4} \left (k +4+r \right ) \left (k +2+r \right )-2 a a_{k +3} \left (k +r +3\right )+a_{k} b =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=\frac {2 a k a_{k +3}+2 a r a_{k +3}+6 a a_{k +3}-a_{k} b}{\left (k +4+r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +4}=\frac {2 a k a_{k +3}+6 a a_{k +3}-a_{k} b}{\left (k +4\right ) \left (k +2\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 {2 a k a_{k +3}+6 a a_{k +3}-a_{k} b}{\left (k +4\right ) \left (k +2\right )}, a_{1}=0, a_{2}=-2 a^{2} a_{0}, a_{3}=-\frac {8 a^{3} a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +4}=\frac {2 a k a_{k +3}+10 a a_{k +3}-a_{k} b}{\left (k +6\right ) \left (k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +4}=\frac {2 a k a_{k +3}+10 a a_{k +3}-a_{k} b}{\left (k +6\right ) \left (k +4\right )}, a_{1}=\frac {4 a a_{0}}{3}, a_{2}=a^{2} a_{0}, a_{3}=\frac {8 a^{3} a_{0}}{15}\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 +2}\right ), c_{k +4}=\frac {2 a k c_{k +3}+6 a c_{k +3}-b c_{k}}{\left (k +4\right ) \left (k +2\right )}, c_{1}=0, c_{2}=-2 a^{2} c_{0}, c_{3}=-\frac {8 a^{3} c_{0}}{3}, d_{k +4}=\frac {2 a k d_{k +3}+10 a d_{k +3}-b d_{k}}{\left (k +6\right ) \left (k +4\right )}, d_{1}=\frac {4 a d_{0}}{3}, d_{2}=a^{2} d_{0}, d_{3}=\frac {8 a^{3} d_{0}}{15}\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: 0.188 (sec). Leaf size: 106
dsolve(x*diff(y(x),x$2)-(2*a*x+1)*diff(y(x),x)+b*x^3*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x^{2} \operatorname {HeunB}\left (2, 0, \frac {a^{2}}{\sqrt {-b}}, -\frac {2 i a}{\left (-b \right )^{\frac {1}{4}}}, i \left (-b \right )^{\frac {1}{4}} x \right ) {\mathrm e}^{a x +\frac {x^{2} \sqrt {-b}}{2}} \left (c_{1} +c_{2} \left (\int \frac {{\mathrm e}^{-x^{2} \sqrt {-b}}}{\operatorname {HeunB}\left (2, 0, \frac {a^{2}}{\sqrt {-b}}, -\frac {2 i a}{\left (-b \right )^{\frac {1}{4}}}, i \left (-b \right )^{\frac {1}{4}} x \right )^{2} x^{3}}d x \right )\right ) \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[x*y''[x]-(2*a*x+1)*y'[x]+b*x^3*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
Not solved