Internal problem ID [10940]
Internal file name [OUTPUT/10197_Sunday_December_31_2023_11_05_15_AM_29391360/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 2.1.2-4 Equation of form
\(x^2 y''+f(x)y'+g(x)y=0\)
Problem number: 117.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_bessel_ode"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }-\left (a^{2} x^{4}+a \left (2 b -1\right ) x^{2}+b \left (1+b \right )\right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\left (-a^{2} x^{4}-2 a \,x^{2} b +a \,x^{2}-b^{2}-b \right ) y = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end {align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following \begin {align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end {align*}
With the standard solution \begin {align*} y&=x^{\alpha } \left (c_{1} \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_{2} \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end {align*}
Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives \begin {align*} \alpha &= {\frac {1}{2}}\\ \beta &= 2\\ n &= -1-2 b\\ \gamma &= {\frac {1}{2}} \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (-1-2 b , 2 \sqrt {x}\right )+c_{2} \sqrt {x}\, \operatorname {BesselY}\left (-1-2 b , 2 \sqrt {x}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (-1-2 b , 2 \sqrt {x}\right )+c_{2} \sqrt {x}\, \operatorname {BesselY}\left (-1-2 b , 2 \sqrt {x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (-1-2 b , 2 \sqrt {x}\right )+c_{2} \sqrt {x}\, \operatorname {BesselY}\left (-1-2 b , 2 \sqrt {x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-a^{2} x^{4}-2 x^{2} \left (b -\frac {1}{2}\right ) a -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 {\left (a^{2} x^{4}+2 a \,x^{2} b -a \,x^{2}+b^{2}+b \right ) y}{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 {\left (a^{2} x^{4}+2 a \,x^{2} b -a \,x^{2}+b^{2}+b \right ) y}{x^{2}}=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 )=0, P_{3}\left (x \right )=-\frac {a^{2} x^{4}+2 a \,x^{2} b -a \,x^{2}+b^{2}+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}}}=0 \\ {} & \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}}}=-b^{2}-b \\ {} & \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^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-a^{2} x^{4}-2 a \,x^{2} b +a \,x^{2}-b^{2}-b \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..4 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} \left (r +b \right ) \left (-r +1+b \right ) x^{r}-a_{1} \left (r +1+b \right ) \left (-r +b \right ) x^{1+r}+\left (-a_{2} \left (r +2+b \right ) \left (-r -1+b \right )-a_{0} a \left (2 b -1\right )\right ) x^{2+r}+\left (-a_{3} \left (r +3+b \right ) \left (-r -2+b \right )-a_{1} a \left (2 b -1\right )\right ) x^{3+r}+\left (\moverset {\infty }{\munderset {k =4}{\sum }}\left (-a_{k} \left (r +k +b \right ) \left (-r +1-k +b \right )-a_{k -2} a \left (2 b -1\right )-a_{k -4} a^{2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (r +b \right ) \left (-r +1+b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-b , 1+b \right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-a_{1} \left (r +1+b \right ) \left (-r +b \right )=0, -a_{2} \left (r +2+b \right ) \left (-r -1+b \right )-a_{0} a \left (2 b -1\right )=0, -a_{3} \left (r +3+b \right ) \left (-r -2+b \right )-a_{1} a \left (2 b -1\right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=-\frac {a_{0} a \left (2 b -1\right )}{b^{2}-r^{2}+b -3 r -2}, a_{3}=0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k} \left (r +k +b \right ) \left (-r +1-k +b \right )-a \left (a a_{k -4}+2 b a_{k -2}-a_{k -2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +4 \\ {} & {} & -a_{k +4} \left (r +k +4+b \right ) \left (-r -3-k +b \right )-a \left (a_{k} a +2 b a_{k +2}-a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +4}=-\frac {a \left (a_{k} a +2 b a_{k +2}-a_{k +2}\right )}{\left (r +k +4+b \right ) \left (-r -3-k +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b \\ {} & {} & a_{k +4}=-\frac {a \left (a_{k} a +2 b a_{k +2}-a_{k +2}\right )}{\left (k +4\right ) \left (2 b -3-k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -b}, a_{k +4}=-\frac {a \left (a_{k} a +2 b a_{k +2}-a_{k +2}\right )}{\left (k +4\right ) \left (2 b -3-k \right )}, a_{1}=0, a_{2}=-\frac {a_{0} a \left (2 b -1\right )}{4 b -2}, a_{3}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+b \\ {} & {} & a_{k +4}=-\frac {a \left (a_{k} a +2 b a_{k +2}-a_{k +2}\right )}{\left (5+2 b +k \right ) \left (-k -4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+b \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1+b}, a_{k +4}=-\frac {a \left (a_{k} a +2 b a_{k +2}-a_{k +2}\right )}{\left (5+2 b +k \right ) \left (-k -4\right )}, a_{1}=0, a_{2}=-\frac {a_{0} a \left (2 b -1\right )}{b^{2}-\left (1+b \right )^{2}-2 b -5}, a_{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k -b}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +1+b}\right ), c_{k +4}=-\frac {a \left (a c_{k}+2 b c_{k +2}-c_{k +2}\right )}{\left (k +4\right ) \left (2 b -3-k \right )}, c_{1}=0, c_{2}=-\frac {c_{0} a \left (2 b -1\right )}{4 b -2}, c_{3}=0, d_{k +4}=-\frac {a \left (a d_{k}+2 b d_{k +2}-d_{k +2}\right )}{\left (5+2 b +k \right ) \left (-k -4\right )}, d_{1}=0, d_{2}=-\frac {d_{0} a \left (2 b -1\right )}{b^{2}-\left (1+b \right )^{2}-2 b -5}, d_{3}=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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 38
dsolve(x^2*diff(y(x),x$2)-(a^2*x^4+a*(2*b-1)*x^2+b*(b+1))*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x^{-b} {\mathrm e}^{-\frac {a \,x^{2}}{2}} \left (c_{2} \Gamma \left (b +\frac {1}{2}\right )-c_{2} \Gamma \left (b +\frac {1}{2}, -a \,x^{2}\right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.45 (sec). Leaf size: 66
DSolve[x^2*y''[x]-(a^2*x^4+a*(2*b-1)*x^2+b*(b+1))*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} e^{-\frac {a x^2}{2}} x^{-b} \left (a c_2 x^{2 b+3} \left (-a x^2\right )^{-b-\frac {3}{2}} \Gamma \left (b+\frac {1}{2},-a x^2\right )+2 c_1\right ) \]