Internal problem ID [9463]
Internal file name [OUTPUT/8403_Monday_June_06_2022_02_57_51_AM_3382579/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1134.
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]]
\[ \boxed {4 x y^{\prime \prime }-\left (x +a \right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+\left (-\frac {1}{4} a x -\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 &= -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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x \left (\frac {d}{d x}y^{\prime }\right )+\left (-x -a \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 (x +a \right ) y}{4 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 {\left (x +a \right ) y}{4 x}=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 {x +a}{4 x}\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}}}=0 \\ {} & \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 \left (\frac {d}{d x}y^{\prime }\right )+\left (-x -a \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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (-1+r \right ) x^{-1+r}+\left (4 a_{1} \left (1+r \right ) r -a a_{0}\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )-a a_{k}-a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) r -a a_{0}=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )-a a_{k}-a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 a_{k +2} \left (k +2+r \right ) \left (k +1+r \right )-a a_{k +1}-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a a_{k +1}+a_{k}}{4 \left (k +2+r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {a a_{k +1}+a_{k}}{4 \left (k +2\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {a a_{k +1}+a_{k}}{4 \left (k +2\right ) \left (k +1\right )}, -a a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {a a_{k +1}+a_{k}}{4 \left (k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=\frac {a a_{k +1}+a_{k}}{4 \left (k +3\right ) \left (k +2\right )}, -a a_{0}+8 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k +1}\right ), b_{k +2}=\frac {a b_{k +1}+b_{k}}{4 \left (k +2\right ) \left (k +1\right )}, -a b_{0}=0, c_{k +2}=\frac {a c_{k +1}+c_{k}}{4 \left (k +3\right ) \left (k +2\right )}, -a c_{0}+8 c_{1}=0\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: 0.031 (sec). Leaf size: 21
dsolve(4*x*diff(diff(y(x),x),x)-(x+a)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {WhittakerM}\left (-\frac {a}{4}, \frac {1}{2}, x\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {a}{4}, \frac {1}{2}, x\right ) \]
✓ Solution by Mathematica
Time used: 0.132 (sec). Leaf size: 44
DSolve[(-a - x)*y[x] + 4*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{4} e^{-x/2} x \left (c_2 \operatorname {Hypergeometric1F1}\left (\frac {a}{4}+1,2,x\right )+c_1 \operatorname {HypergeometricU}\left (\frac {a}{4}+1,2,x\right )\right ) \]