29.22 problem 131

29.22.1 Solving as second order bessel ode ode
29.22.2 Maple step by step solution

Internal problem ID [10954]
Internal file name [OUTPUT/10211_Sunday_December_31_2023_11_08_37_AM_37366069/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: 131.
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]]

\[ \boxed {x^{2} y^{\prime \prime }+\lambda x y^{\prime }+\left (a \,x^{2}+b x +c \right ) y=0} \]

29.22.1 Solving as second order bessel ode ode

Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\lambda x y^{\prime }+\left (a \,x^{2}+b x +c \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}-\frac {\lambda }{2}\\ \beta &= 2\\ n &= \sqrt {\lambda ^{2}-4 c -2 \lambda +1}\\ \gamma &= {\frac {1}{2}} \end {align*}

Substituting all the above into (4) gives the solution as \begin {align*} y = c_{1} x^{\frac {1}{2}-\frac {\lambda }{2}} \operatorname {BesselJ}\left (\sqrt {\lambda ^{2}-4 c -2 \lambda +1}, 2 \sqrt {x}\right )+c_{2} x^{\frac {1}{2}-\frac {\lambda }{2}} \operatorname {BesselY}\left (\sqrt {\lambda ^{2}-4 c -2 \lambda +1}, 2 \sqrt {x}\right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}-\frac {\lambda }{2}} \operatorname {BesselJ}\left (\sqrt {\lambda ^{2}-4 c -2 \lambda +1}, 2 \sqrt {x}\right )+c_{2} x^{\frac {1}{2}-\frac {\lambda }{2}} \operatorname {BesselY}\left (\sqrt {\lambda ^{2}-4 c -2 \lambda +1}, 2 \sqrt {x}\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} x^{\frac {1}{2}-\frac {\lambda }{2}} \operatorname {BesselJ}\left (\sqrt {\lambda ^{2}-4 c -2 \lambda +1}, 2 \sqrt {x}\right )+c_{2} x^{\frac {1}{2}-\frac {\lambda }{2}} \operatorname {BesselY}\left (\sqrt {\lambda ^{2}-4 c -2 \lambda +1}, 2 \sqrt {x}\right ) \] Verified OK.

29.22.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\lambda x y^{\prime }+\left (a \,x^{2}+b x +c \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 \,x^{2}+b x +c \right ) y}{x^{2}}-\frac {\lambda y^{\prime }}{x} \\ \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 {\lambda y^{\prime }}{x}+\frac {\left (a \,x^{2}+b x +c \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 )=\frac {\lambda }{x}, P_{3}\left (x \right )=\frac {a \,x^{2}+b x +c}{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}}}=\lambda \\ {} & \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}}}=c \\ {} & \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 )+\lambda x y^{\prime }+\left (a \,x^{2}+b x +c \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..2 \\ {} & {} & 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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) 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 (\lambda r +r^{2}+c -r \right ) x^{r}+\left (\left (\lambda r +r^{2}+c +\lambda +r \right ) a_{1}+a_{0} b \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+\lambda k +2 k r +\lambda r +r^{2}+c -k -r \right )+a_{k -1} b +a_{k -2} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \lambda r +r^{2}+c -r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}, \frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (\lambda r +r^{2}+c +\lambda +r \right ) a_{1}+a_{0} b =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} b}{\lambda r +r^{2}+c +\lambda +r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r +\lambda -1\right ) k +r^{2}+\left (-1+\lambda \right ) r +c \right ) a_{k}+a_{k -2} a +a_{k -1} b =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+\left (2 r +\lambda -1\right ) \left (k +2\right )+r^{2}+\left (-1+\lambda \right ) r +c \right ) a_{k +2}+a_{k} a +a_{k +1} b =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+\lambda k +2 k r +\lambda r +r^{2}+c +3 k +2 \lambda +3 r +2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+\lambda k +2 k \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+\lambda \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +3 k +\frac {\lambda }{2}+\frac {7}{2}-\frac {3 \sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+\lambda k +2 k \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+\lambda \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +3 k +\frac {\lambda }{2}+\frac {7}{2}-\frac {3 \sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, a_{1}=-\frac {a_{0} b}{\lambda \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +\frac {\lambda }{2}+\frac {1}{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+\lambda k +2 k \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+\lambda \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +3 k +\frac {\lambda }{2}+\frac {7}{2}+\frac {3 \sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, a_{k +2}=-\frac {a_{k} a +a_{k +1} b}{k^{2}+\lambda k +2 k \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+\lambda \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +3 k +\frac {\lambda }{2}+\frac {7}{2}+\frac {3 \sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, a_{1}=-\frac {a_{0} b}{\lambda \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +\frac {\lambda }{2}+\frac {1}{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k +\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}\right ), d_{k +2}=-\frac {a d_{k}+b d_{k +1}}{k^{2}+\lambda k +2 k \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+\lambda \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +3 k +\frac {\lambda }{2}+\frac {7}{2}-\frac {3 \sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, d_{1}=-\frac {d_{0} b}{\lambda \left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +\frac {\lambda }{2}+\frac {1}{2}-\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, e_{k +2}=-\frac {a e_{k}+b e_{k +1}}{k^{2}+\lambda k +2 k \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+\lambda \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +3 k +\frac {\lambda }{2}+\frac {7}{2}+\frac {3 \sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}, e_{1}=-\frac {e_{0} b}{\lambda \left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )+{\left (\frac {1}{2}-\frac {\lambda }{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}\right )}^{2}+c +\frac {\lambda }{2}+\frac {1}{2}+\frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}}\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 
   -> Whittaker 
      -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
   <- Whittaker successful 
<- special function solution successful`
 

Solution by Maple

Time used: 0.11 (sec). Leaf size: 75

dsolve(x^2*diff(y(x),x$2)+lambda*x*diff(y(x),x)+(a*x^2+b*x+c)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = x^{-\frac {\lambda }{2}} \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 \sqrt {a}}, \frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}, 2 i \sqrt {a}\, x \right ) c_{2} +\operatorname {WhittakerM}\left (-\frac {i b}{2 \sqrt {a}}, \frac {\sqrt {\lambda ^{2}-4 c -2 \lambda +1}}{2}, 2 i \sqrt {a}\, x \right ) c_{1} \right ) \]

Solution by Mathematica

Time used: 0.224 (sec). Leaf size: 159

DSolve[x^2*y''[x]+\[Lambda]*x*y'[x]+(a*x^2+b*x+c)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-i \sqrt {a} x} x^{\frac {1}{2} \left (\sqrt {(\lambda -1)^2-4 c}-\lambda +1\right )} \left (c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {i b}{\sqrt {a}}+\sqrt {(\lambda -1)^2-4 c}+1\right ),\sqrt {(\lambda -1)^2-4 c}+1,2 i \sqrt {a} x\right )+c_2 L_{\frac {1}{2} \left (-\frac {i b}{\sqrt {a}}-\sqrt {(\lambda -1)^2-4 c}-1\right )}^{\sqrt {(\lambda -1)^2-4 c}}\left (2 i \sqrt {a} x\right )\right ) \]