Internal problem ID [1105]
Internal file name [OUTPUT/1106_Sunday_June_05_2022_02_02_43_AM_82361730/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.1 Homogeneous linear equations. Page
203
Problem number: 22.
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.
\[ \boxed {\left (2 x +1\right ) y^{\prime \prime }-2 \left (2 x^{2}-1\right ) y^{\prime }-4 \left (x +1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-4 x^{2}+2\right ) y^{\prime }+\left (-4 x -4\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 {4 \left (x +1\right ) y}{2 x +1}+\frac {2 \left (2 x^{2}-1\right ) y^{\prime }}{2 x +1} \\ \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 \left (2 x^{2}-1\right ) y^{\prime }}{2 x +1}-\frac {4 \left (x +1\right ) y}{2 x +1}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {1}{2}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 \left (2 x^{2}-1\right )}{2 x +1}, P_{3}\left (x \right )=-\frac {4 \left (x +1\right )}{2 x +1}\right ] \\ {} & \circ & \left (x +\frac {1}{2}\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {1}{2} \\ {} & {} & \left (\left (x +\frac {1}{2}\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {1}{2}}}}=\frac {1}{2} \\ {} & \circ & \left (x +\frac {1}{2}\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-\frac {1}{2} \\ {} & {} & \left (\left (x +\frac {1}{2}\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-\frac {1}{2}}}}=0 \\ {} & \circ & x =-\frac {1}{2}\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=-\frac {1}{2}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-\frac {1}{2} \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (2 x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (-4 x^{2}+2\right ) y^{\prime }+\left (-4 x -4\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -\frac {1}{2}\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & 2 u \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (-4 u^{2}+4 u +1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-4 u -2\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & u \cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+2 r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (1+2 r \right )+2 a_{0} \left (-1+2 r \right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+2 a_{k} \left (2 k +2 r -1\right )-4 a_{k -1} \left (k +r \right )\right ) u^{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 must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (1+2 r \right )+2 a_{0} \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +\frac {1}{2}+r \right ) \left (k +1+r \right ) a_{k +1}+\left (4 a_{k}-4 a_{k -1}\right ) k +\left (4 a_{k}-4 a_{k -1}\right ) r -2 a_{k}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (k +\frac {3}{2}+r \right ) \left (k +2+r \right ) a_{k +2}+\left (4 a_{k +1}-4 a_{k}\right ) \left (k +1\right )+\left (4 a_{k +1}-4 a_{k}\right ) r -2 a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+2 r a_{k}-2 r a_{k +1}+2 a_{k}-a_{k +1}\right )}{\left (2 k +3+2 r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+2 a_{k}-a_{k +1}\right )}{\left (2 k +3\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+2 a_{k}-a_{k +1}\right )}{\left (2 k +3\right ) \left (k +2\right )}, a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {1}{2}\right )^{k}, a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+2 a_{k}-a_{k +1}\right )}{\left (2 k +3\right ) \left (k +2\right )}, a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+3 a_{k}-2 a_{k +1}\right )}{\left (2 k +4\right ) \left (k +\frac {5}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {1}{2}}, a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+3 a_{k}-2 a_{k +1}\right )}{\left (2 k +4\right ) \left (k +\frac {5}{2}\right )}, 3 a_{1}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {1}{2}\right )^{k +\frac {1}{2}}, a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+3 a_{k}-2 a_{k +1}\right )}{\left (2 k +4\right ) \left (k +\frac {5}{2}\right )}, 3 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +\frac {1}{2}\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +\frac {1}{2}\right )^{k +\frac {1}{2}}\right ), a_{k +2}=\frac {2 \left (2 k a_{k}-2 k a_{k +1}+2 a_{k}-a_{k +1}\right )}{\left (2 k +3\right ) \left (k +2\right )}, a_{1}-2 a_{0}=0, b_{k +2}=\frac {2 \left (2 k b_{k}-2 k b_{k +1}+3 b_{k}-2 b_{k +1}\right )}{\left (2 k +4\right ) \left (k +\frac {5}{2}\right )}, 3 b_{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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 `
✓ Solution by Maple
Time used: 0.187 (sec). Leaf size: 32
dsolve((2*x+1)*diff(y(x),x$2)-2*(2*x^2-1)*diff(y(x),x)-4*(x+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} \operatorname {HeunB}\left (-\frac {1}{2}, -2, -\frac {1}{2}, 3, x +\frac {1}{2}\right )+c_{2} \operatorname {HeunB}\left (\frac {1}{2}, -2, -\frac {1}{2}, 3, x +\frac {1}{2}\right ) \sqrt {4 x +2} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[(2*x+1)*y''[x]-2*(2*x^2-1)*y'[x]-4*(x+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
Not solved