Internal problem ID [9711]
Internal file name [OUTPUT/8653_Monday_June_06_2022_04_40_17_AM_57059046/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1384.
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]]
\[ \boxed {y^{\prime \prime }+\frac {\left (-x^{2} \left (a^{2}-1\right )+2 \left (a +3\right ) b x -b^{2}\right ) y}{4 x^{2}}=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\left (-\frac {1}{4} a^{2} x^{2}+\frac {1}{2} a b x -\frac {1}{4} b^{2}+\frac {3}{2} x b +\frac {1}{4} x^{2}\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 &= \sqrt {b^{2}+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 (\sqrt {b^{2}+1}, 2 \sqrt {x}\right )+c_{2} \sqrt {x}\, \operatorname {BesselY}\left (\sqrt {b^{2}+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 (\sqrt {b^{2}+1}, 2 \sqrt {x}\right )+c_{2} \sqrt {x}\, \operatorname {BesselY}\left (\sqrt {b^{2}+1}, 2 \sqrt {x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (\sqrt {b^{2}+1}, 2 \sqrt {x}\right )+c_{2} \sqrt {x}\, \operatorname {BesselY}\left (\sqrt {b^{2}+1}, 2 \sqrt {x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -4 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2} \left (a^{2}-1\right )-2 \left (a +3\right ) b x +b^{2}\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^{2}-2 a b x +b^{2}-6 x b -x^{2}\right ) y}{4 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^{2}-2 a b x +b^{2}-6 x b -x^{2}\right ) y}{4 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^{2}-2 a b x +b^{2}-6 x b -x^{2}}{4 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}}}=-\frac {b^{2}}{4} \\ {} & \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} \\ {} & {} & 4 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (-a^{2} x^{2}+2 a b x -b^{2}+6 x b +x^{2}\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^{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 (b^{2}-4 r^{2}+4 r \right ) x^{r}+\left (-\left (b^{2}-4 r^{2}-4 r \right ) a_{1}+2 a_{0} b \left (a +3\right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k} \left (b^{2}-4 k^{2}-8 k r -4 r^{2}+4 k +4 r \right )+2 a_{k -1} b \left (a +3\right )-a_{k -2} \left (a -1\right ) \left (a +1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -b^{2}+4 r^{2}-4 r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}, \frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -\left (b^{2}-4 r^{2}-4 r \right ) a_{1}+2 a_{0} b \left (a +3\right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {2 a_{0} b \left (a +3\right )}{b^{2}-4 r^{2}-4 r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-b^{2}+4 \left (k +r \right ) \left (k +r -1\right )\right ) a_{k}+2 a_{k -1} b \left (a +3\right )+\left (-a^{2}+1\right ) a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (-b^{2}+4 \left (k +2+r \right ) \left (k +1+r \right )\right ) a_{k +2}+2 a_{k +1} b \left (a +3\right )+\left (-a^{2}+1\right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a^{2}-2 a b a_{k +1}-6 b a_{k +1}-a_{k}}{b^{2}-4 k^{2}-8 k r -4 r^{2}-12 k -12 r -8} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a^{2}-2 a b a_{k +1}-6 b a_{k +1}-a_{k}}{b^{2}-4 k^{2}-8 k \left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )-4 {\left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )}^{2}-12 k -14+6 \sqrt {b^{2}+1}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}}, a_{k +2}=-\frac {a_{k} a^{2}-2 a b a_{k +1}-6 b a_{k +1}-a_{k}}{b^{2}-4 k^{2}-8 k \left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )-4 {\left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )}^{2}-12 k -14+6 \sqrt {b^{2}+1}}, a_{1}=\frac {2 a_{0} b \left (a +3\right )}{b^{2}-4 {\left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )}^{2}-2+2 \sqrt {b^{2}+1}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {a_{k} a^{2}-2 a b a_{k +1}-6 b a_{k +1}-a_{k}}{b^{2}-4 k^{2}-8 k \left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )-4 {\left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )}^{2}-12 k -6 \sqrt {b^{2}+1}-14} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}}, a_{k +2}=-\frac {a_{k} a^{2}-2 a b a_{k +1}-6 b a_{k +1}-a_{k}}{b^{2}-4 k^{2}-8 k \left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )-4 {\left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )}^{2}-12 k -6 \sqrt {b^{2}+1}-14}, a_{1}=\frac {2 a_{0} b \left (a +3\right )}{b^{2}-4 {\left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )}^{2}-2 \sqrt {b^{2}+1}-2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}}\right ), c_{k +2}=-\frac {a^{2} c_{k}-2 a b c_{k +1}-6 b c_{k +1}-c_{k}}{b^{2}-4 k^{2}-8 k \left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )-4 {\left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )}^{2}-12 k -14+6 \sqrt {b^{2}+1}}, c_{1}=\frac {2 c_{0} b \left (a +3\right )}{b^{2}-4 {\left (\frac {1}{2}-\frac {\sqrt {b^{2}+1}}{2}\right )}^{2}-2+2 \sqrt {b^{2}+1}}, d_{k +2}=-\frac {a^{2} d_{k}-2 a b d_{k +1}-6 b d_{k +1}-d_{k}}{b^{2}-4 k^{2}-8 k \left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )-4 {\left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )}^{2}-12 k -6 \sqrt {b^{2}+1}-14}, d_{1}=\frac {2 d_{0} b \left (a +3\right )}{b^{2}-4 {\left (\frac {\sqrt {b^{2}+1}}{2}+\frac {1}{2}\right )}^{2}-2 \sqrt {b^{2}+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: 73
dsolve(diff(diff(y(x),x),x) = -1/4*(-x^2*(a^2-1)+2*(a+3)*b*x-b^2)/x^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {WhittakerM}\left (\frac {b \left (a +3\right )}{2 \sqrt {a^{2}-1}}, \frac {\sqrt {b^{2}+1}}{2}, \sqrt {a^{2}-1}\, x \right )+c_{2} \operatorname {WhittakerW}\left (\frac {b \left (a +3\right )}{2 \sqrt {a^{2}-1}}, \frac {\sqrt {b^{2}+1}}{2}, \sqrt {a^{2}-1}\, x \right ) \]
✓ Solution by Mathematica
Time used: 0.048 (sec). Leaf size: 96
DSolve[y''[x] == -1/4*((-b^2 + 2*(3 + a)*b*x - (-1 + a^2)*x^2)*y[x])/x^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 M_{\frac {(a+3) b}{2 \sqrt {a^2-1}},\frac {\sqrt {b^2+1}}{2}}\left (\sqrt {a^2-1} x\right )+c_2 W_{\frac {(a+3) b}{2 \sqrt {a^2-1}},\frac {\sqrt {b^2+1}}{2}}\left (\sqrt {a^2-1} x\right ) \]