Internal problem ID [11042]
Internal file name [OUTPUT/10299_Wednesday_January_24_2024_10_06_52_PM_19035987/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 2.1.2-7 Equation of form
\((a_4 x^4+a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 219.
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.
\[ \boxed {x^{2} \left (x^{2}+a \right ) y^{\prime \prime }+\left (b \,x^{2}+c \right ) x y^{\prime }+y d=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (b \,x^{2}+c \right ) x y^{\prime }+y d =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 {d y}{x^{2} \left (x^{2}+a \right )}-\frac {\left (b \,x^{2}+c \right ) y^{\prime }}{x \left (x^{2}+a \right )} \\ \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 (b \,x^{2}+c \right ) y^{\prime }}{x \left (x^{2}+a \right )}+\frac {d y}{x^{2} \left (x^{2}+a \right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {b \,x^{2}+c}{x \left (x^{2}+a \right )}, P_{3}\left (x \right )=\frac {d}{x^{2} \left (x^{2}+a \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}}}=\frac {c}{a} \\ {} & \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}}}=\frac {d}{a} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+a \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (b \,x^{2}+c \right ) x y^{\prime }+y d =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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & 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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (a \,r^{2}-a r +c r +d \right ) x^{r}+a_{1} \left (a \,r^{2}+a r +c r +c +d \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (a \,k^{2}+2 a k r +a \,r^{2}-a k -a r +c k +c r +d \right )+a_{k -2} \left (k -2+r \right ) \left (k -3+r +b \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & a \,r^{2}-a r +c r +d =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}, \frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (a \,r^{2}+a r +c r +c +d \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k -2} \left (k -2+r \right ) \left (k -3+r +b \right )+a_{k} \left (a \,k^{2}+\left (2 a r -a +c \right ) k +a \,r^{2}+\left (-a +c \right ) r +d \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k} \left (k +r \right ) \left (k +r -1+b \right )+a_{k +2} \left (a \left (k +2\right )^{2}+\left (2 a r -a +c \right ) \left (k +2\right )+a \,r^{2}+\left (-a +c \right ) r +d \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k +r \right ) \left (k +r -1+b \right )}{a \,k^{2}+2 a k r +a \,r^{2}+3 a k +3 a r +c k +c r +2 a +2 c +d} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right ) \left (k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}-1+b \right )}{a \,k^{2}-k \left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )+\frac {\left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {c}{2}-\frac {3 \sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2}+c k -\frac {c \left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )}{2 a}+d} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}}, a_{k +2}=-\frac {a_{k} \left (k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right ) \left (k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}-1+b \right )}{a \,k^{2}-k \left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )+\frac {\left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {c}{2}-\frac {3 \sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2}+c k -\frac {c \left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )}{2 a}+d}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a} \\ {} & {} & a_{k +2}=-\frac {a_{k} \left (k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right ) \left (k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}-1+b \right )}{a \,k^{2}+k \left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )+\frac {\left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {c}{2}+\frac {3 \sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2}+c k +\frac {c \left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )}{2 a}+d} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}}, a_{k +2}=-\frac {a_{k} \left (k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right ) \left (k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}-1+b \right )}{a \,k^{2}+k \left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )+\frac {\left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {c}{2}+\frac {3 \sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2}+c k +\frac {c \left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )}{2 a}+d}, a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}}\right ), e_{k +2}=-\frac {e_{k} \left (k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right ) \left (k -\frac {-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}-1+b \right )}{a \,k^{2}-k \left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )+\frac {\left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {c}{2}-\frac {3 \sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2}+c k -\frac {c \left (-a +c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )}{2 a}+d}, e_{1}=0, f_{k +2}=-\frac {f_{k} \left (k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}\right ) \left (k +\frac {a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2 a}-1+b \right )}{a \,k^{2}+k \left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )+\frac {\left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )^{2}}{4 a}+3 a k +\frac {7 a}{2}+\frac {c}{2}+\frac {3 \sqrt {a^{2}-2 a c -4 a d +c^{2}}}{2}+c k +\frac {c \left (a -c +\sqrt {a^{2}-2 a c -4 a d +c^{2}}\right )}{2 a}+d}, f_{1}=0\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 <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 286
dsolve(x^2*(x^2+a)*diff(y(x),x$2)+(b*x^2+c)*x*diff(y(x),x)+d*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (x^{2}+a \right )^{\frac {\left (-b +2\right ) a +c}{2 a}} \left (c_{2} x^{-\frac {-a +c +\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}}{2 a}} \operatorname {hypergeom}\left (\left [-\frac {-3 a -c +\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}}{4 a}, \frac {-\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}+\left (-2 b +5\right ) a +c}{4 a}\right ], \left [1-\frac {\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}}{2 a}\right ], -\frac {x^{2}}{a}\right )+c_{1} x^{\frac {a -c +\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}}{2 a}} \operatorname {hypergeom}\left (\left [\frac {3 a +c +\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}}{4 a}, \frac {\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}+\left (-2 b +5\right ) a +c}{4 a}\right ], \left [1+\frac {\sqrt {a^{2}+\left (-2 c -4 d \right ) a +c^{2}}}{2 a}\right ], -\frac {x^{2}}{a}\right )\right ) \]
✓ Solution by Mathematica
Time used: 2.385 (sec). Leaf size: 336
DSolve[x^2*(x^2+a)*y''[x]+(b*x^2+c)*x*y'[x]+d*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to a^{-\frac {\sqrt {a^2-2 a (c+2 d)+c^2}+a-c}{4 a}} x^{-\frac {\sqrt {a^2-2 a (c+2 d)+c^2}-a+c}{2 a}} \left (c_2 x^{\frac {\sqrt {a^2-2 a (c+2 d)+c^2}}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {-2 b a+a+c-\sqrt {a^2-2 (c+2 d) a+c^2}}{4 a},\frac {a-c+\sqrt {a^2-2 (c+2 d) a+c^2}}{4 a},\frac {\sqrt {a^2-2 (c+2 d) a+c^2}}{2 a}+1,-\frac {x^2}{a}\right )+c_1 a^{\frac {\sqrt {a^2-2 a (c+2 d)+c^2}}{2 a}} \operatorname {Hypergeometric2F1}\left (-\frac {-a+c+\sqrt {a^2-2 (c+2 d) a+c^2}}{4 a},-\frac {-2 b a+a+c+\sqrt {a^2-2 (c+2 d) a+c^2}}{4 a},1-\frac {\sqrt {a^2-2 (c+2 d) a+c^2}}{2 a},-\frac {x^2}{a}\right )\right ) \]