Internal problem ID [11020]
Internal file name [OUTPUT/10277_Sunday_December_31_2023_11_39_47_AM_81148635/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 2.1.2-6 Equation of form
\((a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 197.
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.
\[ \boxed {\left (a \,x^{3}+x^{2} b +c x \right ) y^{\prime \prime }+\left (-2 a \,x^{2}-\left (b +1\right ) x +k \right ) y^{\prime }+2 \left (a x +1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (a \,x^{2}+b x +c \right )+\left (-2 a \,x^{2}+\left (-b -1\right ) x +k \right ) y^{\prime }+\left (2 a x +2\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 {2 \left (a x +1\right ) y}{x \left (a \,x^{2}+b x +c \right )}+\frac {\left (2 a \,x^{2}+b x -k +x \right ) y^{\prime }}{x \left (a \,x^{2}+b x +c \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 (2 a \,x^{2}+b x -k +x \right ) y^{\prime }}{x \left (a \,x^{2}+b x +c \right )}+\frac {2 \left (a x +1\right ) y}{x \left (a \,x^{2}+b x +c \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 {2 a \,x^{2}+b x -k +x}{x \left (a \,x^{2}+b x +c \right )}, P_{3}\left (x \right )=\frac {2 \left (a x +1\right )}{x \left (a \,x^{2}+b x +c \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 {k}{c} \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (a \,x^{2}+b x +c \right )+\left (-2 a \,x^{2}-b x +k -x \right ) y^{\prime }+\left (2 a x +2\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..1 \\ {} & {} & 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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & 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} r \left (c r -c +k \right ) x^{r -1}+\left (a_{1} \left (1+r \right ) \left (c r +k \right )+a_{0} \left (r -2\right ) \left (b r -1\right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (c \left (k +1\right )+c r -c +k \right )+a_{k} \left (k +r -2\right ) \left (b k +b r -1\right )+a a_{k -1} \left (k +r -2\right ) \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (c r -c +k \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {c -k}{c}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (c r +k \right )+a_{0} \left (r -2\right ) \left (b r -1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (\left (k +r \right ) c +k \right ) \left (k +1+r \right ) a_{k +1}+\left (-1+\left (k +r \right ) b \right ) a_{k} \left (k +r -2\right )+a a_{k -1} \left (k +r -2\right ) \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1+r \right ) c +k \right ) \left (k +2+r \right ) a_{k +2}+\left (-1+\left (k +1+r \right ) b \right ) a_{k +1} \left (k +r -1\right )+a a_{k} \left (k +r -1\right ) \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}+2 a k r a_{k}+a \,r^{2} a_{k}+b \,k^{2} a_{k +1}+2 b k r a_{k +1}+b \,r^{2} a_{k +1}-3 a k a_{k}-3 a r a_{k}+2 a a_{k}-b a_{k +1}-k a_{k +1}-r a_{k +1}+a_{k +1}}{\left (c k +c r +c +k \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}+b \,k^{2} a_{k +1}-3 a k a_{k}+2 a a_{k}-b a_{k +1}-k a_{k +1}+a_{k +1}}{\left (c k +c +k \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 +2}=-\frac {a \,k^{2} a_{k}+b \,k^{2} a_{k +1}-3 a k a_{k}+2 a a_{k}-b a_{k +1}-k a_{k +1}+a_{k +1}}{\left (c k +c +k \right ) \left (k +2\right )}, k a_{1}+2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {c -k}{c} \\ {} & {} & a_{k +2}=-\frac {a \,k^{2} a_{k}+\frac {2 a k \left (c -k \right ) a_{k}}{c}+\frac {a \left (c -k \right )^{2} a_{k}}{c^{2}}+b \,k^{2} a_{k +1}+\frac {2 b k \left (c -k \right ) a_{k +1}}{c}+\frac {b \left (c -k \right )^{2} a_{k +1}}{c^{2}}-3 a k a_{k}-\frac {3 a \left (c -k \right ) a_{k}}{c}+2 a a_{k}-b a_{k +1}-k a_{k +1}-\frac {\left (c -k \right ) a_{k +1}}{c}+a_{k +1}}{\left (c k +2 c \right ) \left (k +2+\frac {c -k}{c}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {c -k}{c} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {c -k}{c}}, a_{k +2}=-\frac {a \,k^{2} a_{k}+\frac {2 a k \left (c -k \right ) a_{k}}{c}+\frac {a \left (c -k \right )^{2} a_{k}}{c^{2}}+b \,k^{2} a_{k +1}+\frac {2 b k \left (c -k \right ) a_{k +1}}{c}+\frac {b \left (c -k \right )^{2} a_{k +1}}{c^{2}}-3 a k a_{k}-\frac {3 a \left (c -k \right ) a_{k}}{c}+2 a a_{k}-b a_{k +1}-k a_{k +1}-\frac {\left (c -k \right ) a_{k +1}}{c}+a_{k +1}}{\left (c k +2 c \right ) \left (k +2+\frac {c -k}{c}\right )}, a_{1} \left (1+\frac {c -k}{c}\right ) c +a_{0} \left (\frac {c -k}{c}-2\right ) \left (\frac {b \left (c -k \right )}{c}-1\right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {m =0}{\sum }}d_{m} x^{m}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}e_{m} x^{m +\frac {c -k}{c}}\right ), d_{m +2}=-\frac {a \,m^{2} d_{m}+b \,m^{2} d_{m +1}-3 a m d_{m}+2 a d_{m}-b d_{m +1}-m d_{m +1}+d_{m +1}}{\left (c m +c +k \right ) \left (m +2\right )}, k d_{1}+2 d_{0}=0, e_{m +2}=-\frac {a \,m^{2} e_{m}+\frac {2 a m \left (c -k \right ) e_{m}}{c}+\frac {a \left (c -k \right )^{2} e_{m}}{c^{2}}+b \,m^{2} e_{m +1}+\frac {2 b m \left (c -k \right ) e_{m +1}}{c}+\frac {b \left (c -k \right )^{2} e_{m +1}}{c^{2}}-3 a m e_{m}-\frac {3 a \left (c -k \right ) e_{m}}{c}+2 a e_{m}-b e_{m +1}-m e_{m +1}-\frac {\left (c -k \right ) e_{m +1}}{c}+e_{m +1}}{\left (c m +2 c \right ) \left (m +2+\frac {c -k}{c}\right )}, e_{1} \left (1+\frac {c -k}{c}\right ) c +e_{0} \left (\frac {c -k}{c}-2\right ) \left (\frac {b \left (c -k \right )}{c}-1\right )=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 -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.437 (sec). Leaf size: 2278
dsolve((a*x^3+b*x^2+c*x)*diff(y(x),x$2)+(-2*a*x^2-(b+1)*x+k)*diff(y(x),x)+2*(a*x+1)*y(x)=0,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 10.126 (sec). Leaf size: 186
DSolve[(a*x^3+b*x^2+c*x)*y''[x]+(-2*a*x^2-(b+1)*x+k)*y'[x]+2*(a*x+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {\left (-k x (a x+2)-(b-1) x^2+c (k-2 x)+k^2\right ) \left (c_2 \int _1^x\frac {\exp \left (\frac {(2 c+b k) \arctan \left (\frac {b+2 a K[1]}{\sqrt {4 a c-b^2}}\right )}{c \sqrt {4 a c-b^2}}\right ) K[1]^{-\frac {k}{c}} (c+K[1] (b+a K[1]))^{\frac {k}{2 c}+1}}{\left (k^2-K[1] (a K[1]+2) k-(b-1) K[1]^2+c (k-2 K[1])\right )^2}dK[1]+c_1\right )}{a k+b-c (k-2)-k^2+2 k-1} \]