3.16 problem 1016

3.16.1 Solving as second order bessel ode ode

Internal problem ID [9349]
Internal file name [OUTPUT/8285_Monday_June_06_2022_02_38_13_AM_12631337/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1016.
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]]

\[ \boxed {y^{\prime \prime }+\left (a \,x^{2 c}+b \,x^{c -1}\right ) y=0} \]

3.16.1 Solving as second order bessel ode ode

Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\left (x^{2} a \,x^{2 c}+x 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}}\\ \beta &= 2\\ n &= -1\\ \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 \sqrt {x}\right )-c_{2} \sqrt {x}\, \operatorname {BesselY}\left (1, 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 \sqrt {x}\right )-c_{2} \sqrt {x}\, \operatorname {BesselY}\left (1, 2 \sqrt {x}\right ) \\ \end{align*}

Verification of solutions

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

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying an equivalence, under non-integer power transformations, 
   to LODEs admitting Liouvillian solutions. 
   -> 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: 0.468 (sec). Leaf size: 89

dsolve(diff(diff(y(x),x),x)+(a*x^(2*c)+b*x^(c-1))*y(x)=0,y(x), singsol=all)
 

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

Solution by Mathematica

Time used: 0.269 (sec). Leaf size: 225

DSolve[(b*x^(-1 + c) + a*x^(2*c))*y[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 2^{\frac {c}{2 c+2}} x^{-c/2} \left (x^{c+1}\right )^{\frac {c}{2 c+2}} e^{-\frac {\sqrt {a} x^{c+1}}{\sqrt {-(c+1)^2}}} \left (c_1 \operatorname {HypergeometricU}\left (-\frac {(c+1) \left (c b+b+\sqrt {a} c \sqrt {-(c+1)^2}\right )}{2 \sqrt {a} \left (-(c+1)^2\right )^{3/2}},\frac {c}{c+1},\frac {2 \sqrt {a} x^{c+1}}{\sqrt {-(c+1)^2}}\right )+c_2 L_{\frac {(c+1) \left (c b+b+\sqrt {a} c \sqrt {-(c+1)^2}\right )}{2 \sqrt {a} \left (-(c+1)^2\right )^{3/2}}}^{-\frac {1}{c+1}}\left (\frac {2 \sqrt {a} x^{c+1}}{\sqrt {-(c+1)^2}}\right )\right ) \]