Internal problem ID [9603]
Internal file name [OUTPUT/8544_Monday_June_06_2022_03_49_04_AM_58836134/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1275.
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }+4 y^{\prime } x +\left (-x^{2}+2 \left (1-m +2 l \right ) x -m^{2}+1\right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+y^{\prime } x +\left (-\frac {1}{4} x^{2}+l x -\frac {1}{2} m x +\frac {1}{2} x -\frac {1}{4} m^{2}+\frac {1}{4}\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 &= 0\\ \beta &= 2\\ n &= \sqrt {m^{2}-1}\\ \gamma &= {\frac {1}{2}} \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = c_{1} \operatorname {BesselJ}\left (\sqrt {m^{2}-1}, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (\sqrt {m^{2}-1}, 2 \sqrt {x}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \operatorname {BesselJ}\left (\sqrt {m^{2}-1}, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (\sqrt {m^{2}-1}, 2 \sqrt {x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \operatorname {BesselJ}\left (\sqrt {m^{2}-1}, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (\sqrt {m^{2}-1}, 2 \sqrt {x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }+4 y^{\prime } x +\left (-x^{2}+\left (4 l -2 m +2\right ) x -m^{2}+1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (4 l x -m^{2}-2 m x -x^{2}+2 x +1\right ) y}{4 x^{2}}-\frac {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} \\ {} & {} & y^{\prime \prime }+\frac {y^{\prime }}{x}+\frac {\left (4 l x -m^{2}-2 m x -x^{2}+2 x +1\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 )=\frac {1}{x}, P_{3}\left (x \right )=\frac {4 l x -m^{2}-2 m x -x^{2}+2 x +1}{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}}}=1 \\ {} & \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 {m^{2}}{4}+\frac {1}{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} y^{\prime \prime }+4 y^{\prime } x +\left (4 l x -m^{2}-2 m x -x^{2}+2 x +1\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 y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\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 (-m^{2}+4 r^{2}+1\right ) x^{r}+\left (a_{1} \left (-m^{2}+4 r^{2}+8 r +5\right )+2 a_{0} \left (1-m +2 l \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (4 k^{2}+8 k r -m^{2}+4 r^{2}+1\right )+2 a_{k -1} \left (1-m +2 l \right )-a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -m^{2}+4 r^{2}+1=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {\sqrt {m^{2}-1}}{2}, \frac {\sqrt {m^{2}-1}}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (-m^{2}+4 r^{2}+8 r +5\right )+2 a_{0} \left (1-m +2 l \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=\frac {2 a_{0} \left (1-m +2 l \right )}{m^{2}-4 r^{2}-8 r -5} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (4 k^{2}+8 k r -m^{2}+4 r^{2}+1\right )+\left (4 l -2 m +2\right ) a_{k -1}-a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (4 \left (k +2\right )^{2}+8 \left (k +2\right ) r -m^{2}+4 r^{2}+1\right )+\left (4 l -2 m +2\right ) a_{k +1}-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 l a_{k +1}-2 m a_{k +1}-a_{k}+2 a_{k +1}}{4 k^{2}+8 k r -m^{2}+4 r^{2}+16 k +16 r +17} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {\sqrt {m^{2}-1}}{2} \\ {} & {} & a_{k +2}=-\frac {4 l a_{k +1}-2 m a_{k +1}-a_{k}+2 a_{k +1}}{4 k^{2}-4 k \sqrt {m^{2}-1}+16+16 k -8 \sqrt {m^{2}-1}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {\sqrt {m^{2}-1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {\sqrt {m^{2}-1}}{2}}, a_{k +2}=-\frac {4 l a_{k +1}-2 m a_{k +1}-a_{k}+2 a_{k +1}}{4 k^{2}-4 k \sqrt {m^{2}-1}+16+16 k -8 \sqrt {m^{2}-1}}, a_{1}=\frac {2 a_{0} \left (1-m +2 l \right )}{-4+4 \sqrt {m^{2}-1}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {\sqrt {m^{2}-1}}{2} \\ {} & {} & a_{k +2}=-\frac {4 l a_{k +1}-2 m a_{k +1}-a_{k}+2 a_{k +1}}{4 k^{2}+4 k \sqrt {m^{2}-1}+16+16 k +8 \sqrt {m^{2}-1}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {\sqrt {m^{2}-1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {\sqrt {m^{2}-1}}{2}}, a_{k +2}=-\frac {4 l a_{k +1}-2 m a_{k +1}-a_{k}+2 a_{k +1}}{4 k^{2}+4 k \sqrt {m^{2}-1}+16+16 k +8 \sqrt {m^{2}-1}}, a_{1}=\frac {2 a_{0} \left (1-m +2 l \right )}{-4-4 \sqrt {m^{2}-1}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {\sqrt {m^{2}-1}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {\sqrt {m^{2}-1}}{2}}\right ), a_{k +2}=-\frac {4 l a_{k +1}-2 m a_{k +1}-a_{k}+2 a_{k +1}}{4 k^{2}-4 k \sqrt {m^{2}-1}+16+16 k -8 \sqrt {m^{2}-1}}, a_{1}=\frac {2 a_{0} \left (1-m +2 l \right )}{-4+4 \sqrt {m^{2}-1}}, b_{k +2}=-\frac {4 l b_{k +1}-2 m b_{k +1}-b_{k}+2 b_{k +1}}{4 k^{2}+4 k \sqrt {m^{2}-1}+16+16 k +8 \sqrt {m^{2}-1}}, b_{1}=\frac {2 b_{0} \left (1-m +2 l \right )}{-4-4 \sqrt {m^{2}-1}}\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: 0.156 (sec). Leaf size: 53
dsolve(4*x^2*diff(diff(y(x),x),x)+4*x*diff(y(x),x)+(-x^2+2*(1-m+2*l)*x-m^2+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{2} \operatorname {WhittakerW}\left (l -\frac {m}{2}+\frac {1}{2}, \frac {\sqrt {m +1}\, \sqrt {m -1}}{2}, x\right )+c_{1} \operatorname {WhittakerM}\left (l -\frac {m}{2}+\frac {1}{2}, \frac {\sqrt {m +1}\, \sqrt {m -1}}{2}, x\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.064 (sec). Leaf size: 97
DSolve[(1 - m^2 + 2*(1 + 2*l - m)*x - x^2)*y[x] + 4*x*y'[x] + 4*x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x/2} x^{\frac {\sqrt {m^2-1}}{2}} \left (c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (-2 l+m+\sqrt {m^2-1}\right ),\sqrt {m^2-1}+1,x\right )+c_2 L_{l-\frac {m}{2}-\frac {\sqrt {m^2-1}}{2}}^{\sqrt {m^2-1}}(x)\right ) \]