Internal problem ID [9461]
Internal file name [OUTPUT/8401_Monday_June_06_2022_02_57_29_AM_41745384/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1132.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[_Laguerre]
Unable to solve or complete the solution.
\[ \boxed {2 x y^{\prime \prime }-\left (2 x -1\right ) y^{\prime }+a y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 x +1\right ) y^{\prime }+a 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 {a y}{2 x}+\frac {\left (2 x -1\right ) y^{\prime }}{2 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 (2 x -1\right ) y^{\prime }}{2 x}+\frac {a y}{2 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 {2 x -1}{2 x}, P_{3}\left (x \right )=\frac {a}{2 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}}}=\frac {1}{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}}}=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} \\ {} & {} & 2 x \left (\frac {d}{d x}y^{\prime }\right )+\left (-2 x +1\right ) y^{\prime }+a 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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) 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} \\ {} & {} & a_{0} r \left (-1+2 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k} \left (-2 k -2 r +a \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +1+r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+a_{k} \left (-2 k -2 r +a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-2 k -2 r +a \right )}{\left (k +1+r \right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-2 k +a \right )}{\left (k +1\right ) \left (2 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 +1}=-\frac {a_{k} \left (-2 k +a \right )}{\left (k +1\right ) \left (2 k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-2 k -1+a \right )}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=-\frac {a_{k} \left (-2 k -1+a \right )}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\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 +\frac {1}{2}}\right ), b_{k +1}=-\frac {b_{k} \left (-2 k +a \right )}{\left (k +1\right ) \left (2 k +1\right )}, c_{k +1}=-\frac {c_{k} \left (-2 k -1+a \right )}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\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: 0.125 (sec). Leaf size: 29
dsolve(2*x*diff(diff(y(x),x),x)-(2*x-1)*diff(y(x),x)+a*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {x}\, \left (\operatorname {KummerM}\left (-\frac {a}{2}+\frac {1}{2}, \frac {3}{2}, x\right ) c_{1} +\operatorname {KummerU}\left (-\frac {a}{2}+\frac {1}{2}, \frac {3}{2}, x\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 44
DSolve[a*y[x] - (-1 + 2*x)*y'[x] + 2*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {x} \left (c_1 \operatorname {HypergeometricU}\left (\frac {1-a}{2},\frac {3}{2},x\right )+c_2 L_{\frac {a-1}{2}}^{\frac {1}{2}}(x)\right ) \]