Internal problem ID [9467]
Internal file name [OUTPUT/8407_Monday_June_06_2022_02_58_28_AM_17765466/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1138.
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]]
\[ \boxed {4 y^{\prime \prime } x +4 m y^{\prime }-\left (x -2 m -4 n \right ) y=0} \]
Writing the ode as \begin {align*} x^{2} y^{\prime \prime }+m x y^{\prime }+\left (\frac {1}{2} m x +n 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}-\frac {m}{2}\\ \beta &= 2\\ n &= -m +1\\ \gamma &= {\frac {1}{2}} \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} y = c_{1} x^{\frac {1}{2}-\frac {m}{2}} \operatorname {BesselJ}\left (-m +1, 2 \sqrt {x}\right )+c_{2} x^{\frac {1}{2}-\frac {m}{2}} \operatorname {BesselY}\left (-m +1, 2 \sqrt {x}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}-\frac {m}{2}} \operatorname {BesselJ}\left (-m +1, 2 \sqrt {x}\right )+c_{2} x^{\frac {1}{2}-\frac {m}{2}} \operatorname {BesselY}\left (-m +1, 2 \sqrt {x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{2}-\frac {m}{2}} \operatorname {BesselJ}\left (-m +1, 2 \sqrt {x}\right )+c_{2} x^{\frac {1}{2}-\frac {m}{2}} \operatorname {BesselY}\left (-m +1, 2 \sqrt {x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime } x +4 m y^{\prime }+\left (2 m +4 n -x \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 {m y^{\prime }}{x}-\frac {\left (2 m +4 n -x \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} \\ {} & {} & y^{\prime \prime }+\frac {m y^{\prime }}{x}+\frac {\left (2 m +4 n -x \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 )=\frac {m}{x}, P_{3}\left (x \right )=\frac {2 m +4 n -x}{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}}}=m \\ {} & \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 y^{\prime \prime } x +4 m y^{\prime }+\left (2 m +4 n -x \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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\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 y^{\prime \prime }=\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 +m \right ) x^{-1+r}+\left (4 a_{1} \left (1+r \right ) \left (r +m \right )+2 a_{0} \left (m +2 n \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +r +m \right )+2 a_{k} \left (m +2 n \right )-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 +m \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -m +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) \left (r +m \right )+2 a_{0} \left (m +2 n \right )=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 +m \right )+2 a_{k} m +4 a_{k} n -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 +m \right )+2 m a_{k +1}+4 a_{k +1} n -a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 m a_{k +1}+4 a_{k +1} n -a_{k}}{4 \left (k +2+r \right ) \left (k +1+r +m \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 m a_{k +1}+4 a_{k +1} n -a_{k}}{4 \left (k +2\right ) \left (k +1+m \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 {2 m a_{k +1}+4 a_{k +1} n -a_{k}}{4 \left (k +2\right ) \left (k +1+m \right )}, 4 a_{1} m +2 a_{0} \left (m +2 n \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-m +1 \\ {} & {} & a_{k +2}=-\frac {2 m a_{k +1}+4 a_{k +1} n -a_{k}}{4 \left (k +3-m \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-m +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -m +1}, a_{k +2}=-\frac {2 m a_{k +1}+4 a_{k +1} n -a_{k}}{4 \left (k +3-m \right ) \left (k +2\right )}, 4 a_{1} \left (2-m \right )+2 a_{0} \left (m +2 n \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -m +1}\right ), a_{k +2}=-\frac {2 m a_{k +1}+4 n a_{k +1}-a_{k}}{4 \left (k +2\right ) \left (k +1+m \right )}, 4 a_{1} m +2 a_{0} \left (m +2 n \right )=0, b_{k +2}=-\frac {2 m b_{k +1}+4 n b_{k +1}-b_{k}}{4 \left (k +3-m \right ) \left (k +2\right )}, 4 b_{1} \left (2-m \right )+2 b_{0} \left (m +2 n \right )=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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 26
dsolve(4*x*diff(diff(y(x),x),x)+4*m*diff(y(x),x)-(x-2*m-4*n)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {x}{2}} \left (\operatorname {KummerM}\left (-n , m , x\right ) c_{1} +\operatorname {KummerU}\left (-n , m , x\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.04 (sec). Leaf size: 32
DSolve[(2*m + 4*n - x)*y[x] + 4*m*y'[x] + 4*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x/2} (c_1 \operatorname {HypergeometricU}(-n,m,x)+c_2 L_n^{m-1}(x)) \]