Internal problem ID [10973]
Internal file name [OUTPUT/10230_Sunday_December_31_2023_11_10_28_AM_77036271/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: 150.
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.
\[ \boxed {\left (-a^{2}+x^{2}\right ) y^{\prime \prime }+b y^{\prime }-6 y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-a^{2}+x^{2}\right ) \left (\frac {d}{d x}y^{\prime }\right )+b y^{\prime }-6 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 {6 y}{a^{2}-x^{2}}+\frac {b 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 {b y^{\prime }}{a^{2}-x^{2}}+\frac {6 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 {b}{a^{2}-x^{2}}, P_{3}\left (x \right )=\frac {6}{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}}}=\frac {b}{2 a} \\ {} & \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 )-b y^{\prime }+6 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 )-b \left (\frac {d}{d u}y \left (u \right )\right )+6 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} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+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} \\ {} & {} & -a_{0} r \left (2 a r -2 a +b \right ) u^{r -1}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 a \left (k +1\right )+2 a r -2 a +b \right )-a_{k} \left (k +r +2\right ) \left (k +r -3\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (2 a r -2 a +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {2 a -b}{2 a}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (a k +a r +\frac {1}{2} b \right ) \left (k +1+r \right ) a_{k +1}-a_{k} \left (k +r +2\right ) \left (k +r -3\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r +2\right ) \left (k +r -3\right )}{\left (2 a k +2 a r +b \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =3 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +2\right ) \left (k -3\right )}{\left (2 a k +b \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {6 a_{0}}{b} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {3 a_{1}}{2 a +b} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {18 a_{0}}{b \left (2 a +b \right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=\frac {4 a_{2}}{3 \left (4 a +b \right )} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=\frac {24 a_{0}}{b \left (2 a +b \right ) \left (4 a +b \right )} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1+\frac {6 u}{b}+\frac {18 u^{2}}{b \left (2 a +b \right )}+\frac {24 u^{3}}{b \left (2 a +b \right ) \left (4 a +b \right )}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =-a +x \\ {} & {} & \left [y=\frac {a_{0} \left (-10 a^{2} b -24 a^{2} x +b^{3}+6 b^{2} x +18 b \,x^{2}+24 x^{3}\right )}{b \left (2 a +b \right ) \left (4 a +b \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {2 a -b}{2 a} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +\frac {2 a -b}{2 a}+2\right ) \left (k +\frac {2 a -b}{2 a}-3\right )}{\left (2 a k +2 a \right ) \left (k +1+\frac {2 a -b}{2 a}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {2 a -b}{2 a} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {2 a -b}{2 a}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {2 a -b}{2 a}+2\right ) \left (k +\frac {2 a -b}{2 a}-3\right )}{\left (2 a k +2 a \right ) \left (k +1+\frac {2 a -b}{2 a}\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 +\frac {2 a -b}{2 a}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {2 a -b}{2 a}+2\right ) \left (k +\frac {2 a -b}{2 a}-3\right )}{\left (2 a k +2 a \right ) \left (k +1+\frac {2 a -b}{2 a}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\frac {c_{0} \left (-10 a^{2} b -24 a^{2} x +b^{3}+6 b^{2} x +18 b \,x^{2}+24 x^{3}\right )}{b \left (2 a +b \right ) \left (4 a +b \right )}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} \left (-a +x \right )^{k +\frac {2 a -b}{2 a}}\right ), d_{k +1}=-\frac {d_{k} \left (k +\frac {2 a -b}{2 a}+2\right ) \left (k +\frac {2 a -b}{2 a}-3\right )}{\left (2 a k +2 a \right ) \left (k +1+\frac {2 a -b}{2 a}\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: 0.015 (sec). Leaf size: 76
dsolve((x^2-a^2)*diff(y(x),x$2)+b*diff(y(x),x)-6*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {c_{1} \left (10 a^{2} b +24 a^{2} x -b^{3}-6 b^{2} x -18 x^{2} b -24 x^{3}\right )}{24}+c_{2} \left (a +x \right ) \left (a -x \right ) \left (b -4 x \right ) \left (\frac {a +x}{a -x}\right )^{\frac {b}{2 a}} \]
✓ Solution by Mathematica
Time used: 13.059 (sec). Leaf size: 1171
DSolve[(x^2-a^2)*y''[x]+b*y'[x]-6*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{\frac {b \text {arctanh}\left (\frac {x}{a}\right )}{2 a}+\frac {\left (b^5-20 a^2 b^3+64 a^4 b+\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}\right ) \text {RootSum}\left [-b^3-6 \text {$\#$1} b^2+10 a^2 b-18 \text {$\#$1}^2 b-24 \text {$\#$1}^3+24 a^2 \text {$\#$1}\&,\log (x-\text {$\#$1})\&\right ]}{2 \left (b^5-20 a^2 b^3+64 a^4 b\right )}} (a+x)^{\frac {1}{2}-\frac {\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}}{4 a b \left (32 a^3-16 b a^2-2 b^2 a+b^3\right )}} (4 x-b)^{\frac {b^5}{2 \left (b^5-20 a^2 b^3+64 a^4 b\right )}-\frac {10 a^2 b^3}{b^5-20 a^2 b^3+64 a^4 b}+\frac {32 a^4 b}{b^5-20 a^2 b^3+64 a^4 b}-\frac {\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}}{2 \left (b^5-20 a^2 b^3+64 a^4 b\right )}} c_2 \int _1^x-\frac {e^{-\frac {\left (b^5-20 a^2 b^3+64 a^4 b+\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}\right ) \text {RootSum}\left [-b^3-6 \text {$\#$1} b^2+10 a^2 b-18 \text {$\#$1}^2 b-24 \text {$\#$1}^3+24 a^2 \text {$\#$1}\&,\log (K[1]-\text {$\#$1})\&\right ]}{b^5-20 a^2 b^3+64 a^4 b}} (K[1]-a)^{\frac {\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}}{2 a (4 a-b) b (2 a+b) (4 a+b)}} (a+K[1])^{\frac {\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}}{2 a b \left (32 a^3-16 b a^2-2 b^2 a+b^3\right )}-1} (4 K[1]-b)^{-\frac {b^5}{b^5-20 a^2 b^3+64 a^4 b}+\frac {20 a^2 b^3}{b^5-20 a^2 b^3+64 a^4 b}-\frac {64 a^4 b}{b^5-20 a^2 b^3+64 a^4 b}+\frac {\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}}{b^5-20 a^2 b^3+64 a^4 b}}}{a-K[1]}dK[1] (x-a)^{\frac {1}{2}-\frac {\sqrt {b^2 \left (64 a^4-20 b^2 a^2+b^4\right )^2}}{4 a (4 a-b) b (2 a+b) (4 a+b)}}+e^{\frac {1}{2} \left (\frac {b \text {arctanh}\left (\frac {x}{a}\right )}{a}+\frac {\left (b^5-20 a^2 b^3+64 a^4 b+\sqrt {\left (b^5-20 a^2 b^3+64 a^4 b\right )^2}\right ) \text {RootSum}\left [-b^3-6 \text {$\#$1} b^2+10 a^2 b-18 \text {$\#$1}^2 b-24 \text {$\#$1}^3+24 a^2 \text {$\#$1}\&,\log (x-\text {$\#$1})\&\right ]}{b^5-20 a^2 b^3+64 a^4 b}\right )} (a+x)^{\frac {1}{4} \left (2-\frac {\sqrt {\left (b^5-20 a^2 b^3+64 a^4 b\right )^2}}{a b \left (32 a^3-16 b a^2-2 b^2 a+b^3\right )}\right )} (4 x-b)^{\frac {b^5-20 a^2 b^3+64 a^4 b-\sqrt {\left (b^5-20 a^2 b^3+64 a^4 b\right )^2}}{2 \left (b^5-20 a^2 b^3+64 a^4 b\right )}} c_1 (x-a)^{\frac {1}{2}-\frac {\sqrt {\left (b^5-20 a^2 b^3+64 a^4 b\right )^2}}{4 a (4 a-b) b (2 a+b) (4 a+b)}} \]