Internal problem ID [9613]
Internal file name [OUTPUT/8554_Monday_June_06_2022_03_50_33_AM_46267223/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1285.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x \left (4 x -1\right ) y^{\prime \prime }+\left (\left (4 a +2\right ) x -a \right ) y^{\prime }+a \left (a -1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (4 x^{2}-x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\left (4 x -1\right ) a +2 x \right ) y^{\prime }+\left (a^{2}-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 {a \left (a -1\right ) y}{x \left (4 x -1\right )}-\frac {\left (4 x a -a +2 x \right ) y^{\prime }}{x \left (4 x -1\right )} \\ \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 (4 x a -a +2 x \right ) y^{\prime }}{x \left (4 x -1\right )}+\frac {a \left (a -1\right ) y}{x \left (4 x -1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {4 x a -a +2 x}{x \left (4 x -1\right )}, P_{3}\left (x \right )=\frac {a \left (a -1\right )}{x \left (4 x -1\right )}\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}}}=a \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (4 x -1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (4 x a -a +2 x \right ) y^{\prime }+a \left (a -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^{\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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-1+r +a \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k +r +a \right )+a_{k} \left (2 r +2 k +a \right ) \left (2 r -1+2 k +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+r +a \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -a +1\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -a_{k +1} \left (k +1+r \right ) \left (k +r +a \right )+a_{k} \left (2 r +2 k +a \right ) \left (2 r -1+2 k +a \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 r +2 k +a \right ) \left (2 r -1+2 k +a \right )}{\left (k +1+r \right ) \left (k +r +a \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (2 k +a \right ) \left (-1+2 k +a \right )}{\left (k +1\right ) \left (k +a \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 (-1+2 k +a \right )}{\left (k +1\right ) \left (k +a \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-a +1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (-a +2+2 k \right ) \left (-a +1+2 k \right )}{\left (k +2-a \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-a +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -a +1}, a_{k +1}=\frac {a_{k} \left (-a +2+2 k \right ) \left (-a +1+2 k \right )}{\left (k +2-a \right ) \left (k +1\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 -a +1}\right ), b_{k +1}=\frac {b_{k} \left (2 k +a \right ) \left (-1+2 k +a \right )}{\left (k +1\right ) \left (k +a \right )}, c_{k +1}=\frac {c_{k} \left (-a +2+2 k \right ) \left (-a +1+2 k \right )}{\left (k +2-a \right ) \left (k +1\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 56
dsolve(x*(4*x-1)*diff(diff(y(x),x),x)+((4*a+2)*x-a)*diff(y(x),x)+a*(a-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} 2^{a -1} \left (1+\sqrt {-4 x +1}\right )^{-a +1}+c_{2} x^{-a +1} 2^{-a +1} \left (1+\sqrt {-4 x +1}\right )^{a -1} \]
✓ Solution by Mathematica
Time used: 0.487 (sec). Leaf size: 186
DSolve[(-1 + a)*a*y[x] + (-a + (2 + 4*a)*x)*y'[x] + x*(-1 + 4*x)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {\sqrt [4]{4 x-1} x^{\frac {1}{2}-\frac {a}{2}} \left (1-i \sqrt {4 x-1}\right )^{-i \sqrt {-(a-1)^2}} e^{\sqrt {-(a-1)^2} \arctan \left (\sqrt {4 x-1}\right )} \left (4 \sqrt {-(a-1)^2} c_1 \left (1-i \sqrt {4 x-1}\right )^{i \sqrt {-(a-1)^2}}-c_2 \left (1+i \sqrt {4 x-1}\right )^{i \sqrt {-(a-1)^2}}\right )}{2 \sqrt {-(a-1)^2} \sqrt [4]{1-4 x}} \]