Internal problem ID [9498]
Internal file name [OUTPUT/8438_Monday_June_06_2022_03_04_30_AM_46133983/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1169.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+2 y^{\prime } x +\left (x a -b^{2}\right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+2 y^{\prime } x +\left (x a -b^{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 \sqrt {a}\\ n &= \sqrt {4 b^{2}+1}\\ \gamma &= {\frac {1}{2}} \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = \frac {c_{1} \operatorname {BesselJ}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \operatorname {BesselJ}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \operatorname {BesselJ}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}}+\frac {c_{2} \operatorname {BesselY}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+2 y^{\prime } x +\left (x a -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 {2 y^{\prime }}{x}-\frac {\left (x a -b^{2}\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 {2 y^{\prime }}{x}+\frac {\left (x a -b^{2}\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 {2}{x}, P_{3}\left (x \right )=\frac {x a -b^{2}}{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}}}=2 \\ {} & \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} \\ {} & \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 )+2 y^{\prime } x +\left (x a -b^{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..1 \\ {} & {} & 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 (b^{2}-r^{2}-r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (b^{2}-k^{2}-2 k r -r^{2}-k -r \right )+a_{k -1} a \right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -b^{2}+r^{2}+r =0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}, -\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r +1\right ) k -b^{2}+r^{2}+r \right ) a_{k}+a_{k -1} a =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (2 r +1\right ) \left (k +1\right )-b^{2}+r^{2}+r \right ) a_{k +1}+a_{k} a =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} a}{b^{2}-k^{2}-2 k r -r^{2}-3 k -3 r -2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} a}{b^{2}-k^{2}-2 k \left (-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}\right )-{\left (-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}\right )}^{2}-3 k -\frac {1}{2}+\frac {3 \sqrt {4 b^{2}+1}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}}, a_{k +1}=\frac {a_{k} a}{b^{2}-k^{2}-2 k \left (-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}\right )-{\left (-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}\right )}^{2}-3 k -\frac {1}{2}+\frac {3 \sqrt {4 b^{2}+1}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2} \\ {} & {} & a_{k +1}=\frac {a_{k} a}{b^{2}-k^{2}-2 k \left (-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right )-{\left (-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right )}^{2}-3 k -\frac {1}{2}-\frac {3 \sqrt {4 b^{2}+1}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}}, a_{k +1}=\frac {a_{k} a}{b^{2}-k^{2}-2 k \left (-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right )-{\left (-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right )}^{2}-3 k -\frac {1}{2}-\frac {3 \sqrt {4 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 {4 b^{2}+1}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k -\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}}\right ), c_{k +1}=\frac {c_{k} a}{b^{2}-k^{2}-2 k \left (-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}\right )-{\left (-\frac {1}{2}-\frac {\sqrt {4 b^{2}+1}}{2}\right )}^{2}-3 k -\frac {1}{2}+\frac {3 \sqrt {4 b^{2}+1}}{2}}, d_{k +1}=\frac {d_{k} a}{b^{2}-k^{2}-2 k \left (-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right )-{\left (-\frac {1}{2}+\frac {\sqrt {4 b^{2}+1}}{2}\right )}^{2}-3 k -\frac {1}{2}-\frac {3 \sqrt {4 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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 49
dsolve(x^2*diff(diff(y(x),x),x)+2*x*diff(y(x),x)+(a*x-b^2)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} \operatorname {BesselJ}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (\sqrt {4 b^{2}+1}, 2 \sqrt {a}\, \sqrt {x}\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.103 (sec). Leaf size: 103
DSolve[(-b^2 + a*x)*y[x] + 2*x*y'[x] + x^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1 \operatorname {Gamma}\left (1-\sqrt {4 b^2+1}\right ) \operatorname {BesselJ}\left (-\sqrt {4 b^2+1},2 \sqrt {a} \sqrt {x}\right )+c_2 \operatorname {Gamma}\left (\sqrt {4 b^2+1}+1\right ) \operatorname {BesselJ}\left (\sqrt {4 b^2+1},2 \sqrt {a} \sqrt {x}\right )}{\sqrt {a} \sqrt {x}} \]