Internal problem ID [10987]
Internal file name [OUTPUT/10244_Sunday_December_31_2023_11_13_32_AM_79566971/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 2.1.2-5 Equation of form
\((a x^2+b x+c) y''+f(x)y'+g(x)y=0\)
Problem number: 164.
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.
\[ \boxed {\left (-a^{2}+x^{2}\right ) y^{\prime \prime }+2 b x y^{\prime }+b \left (b -1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-a^{2}+x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+2 b x y^{\prime }+\left (b^{2}-b \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 {b \left (b -1\right ) y}{a^{2}-x^{2}}+\frac {2 b x y^{\prime }}{a^{2}-x^{2}} \\ \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 {2 b x y^{\prime }}{a^{2}-x^{2}}-\frac {b \left (b -1\right ) y}{a^{2}-x^{2}}=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 b x}{a^{2}-x^{2}}, P_{3}\left (x \right )=-\frac {b \left (b -1\right )}{a^{2}-x^{2}}\right ] \\ {} & \circ & \left (-a +x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =a \\ {} & {} & \left (\left (-a +x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}a}}}=b \\ {} & \circ & \left (-a +x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =a \\ {} & {} & \left (\left (-a +x \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}a}}}=0 \\ {} & \circ & x =a \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}=a \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (a^{2}-x^{2}\right )-2 b x y^{\prime }-b \left (b -1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u +a \hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (-2 u a -u^{2}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-2 b a -2 b u \right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-b^{2}+b \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 \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^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a a_{0} r \left (r -1+b \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a a_{k +1} \left (k +1+r \right ) \left (r +k +b \right )-a_{k} \left (r +k +b \right ) \left (r -1+k +b \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 a r \left (r -1+b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -b +1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (r +k +b \right ) \left (\frac {a_{k} \left (r -1+k +b \right )}{2}+a a_{k +1} \left (k +1+r \right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (r -1+k +b \right )}{2 a \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-1+k +b \right )}{2 a \left (k +1\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 +1}=-\frac {a_{k} \left (-1+k +b \right )}{2 a \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =-a +x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (-a +x \right )^{k}, a_{k +1}=-\frac {a_{k} \left (-1+k +b \right )}{2 a \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b +1 \\ {} & {} & a_{k +1}=-\frac {a_{k} k}{2 a \left (k +2-b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b +1 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -b +1}, a_{k +1}=-\frac {a_{k} k}{2 a \left (k +2-b \right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =-a +x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (-a +x \right )^{k -b +1}, a_{k +1}=-\frac {a_{k} k}{2 a \left (k +2-b \right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} \left (-a +x \right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (-a +x \right )^{k -b +1}\right ), c_{k +1}=-\frac {c_{k} \left (-1+k +b \right )}{2 a \left (k +1\right )}, d_{k +1}=-\frac {d_{k} k}{2 a \left (k +2-b \right )}\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 29
dsolve((x^2-a^2)*diff(y(x),x$2)+2*b*x*diff(y(x),x)+b*(b-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \left (a +x \right )^{-b +1}+c_{2} \left (a -x \right )^{-b +1} \]
✓ Solution by Mathematica
Time used: 0.727 (sec). Leaf size: 127
DSolve[(x^2-a^2)*y''[x]+2*b*x*y'[x]+b*(b-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {(x-a)^{\frac {1}{2}-\frac {1}{2} \sqrt {(b-1)^2}} (a+x)^{\frac {1}{2}-\frac {1}{2} \sqrt {(b-1)^2}} \left (x^2-a^2\right )^{-b/2} \left (2 a \sqrt {(b-1)^2} c_1 (x-a)^{\sqrt {(b-1)^2}}-c_2 (a+x)^{\sqrt {(b-1)^2}}\right )}{2 a \sqrt {(b-1)^2}} \]