Internal problem ID [5890]
Internal file name [OUTPUT/5138_Sunday_June_05_2022_03_25_56_PM_22904374/index.tex]
Book: THEORY OF DIFFERENTIAL EQUATIONS IN ENGINEERING AND MECHANICS.
K.T. CHAU, CRC Press. Boca Raton, FL. 2018
Section: Chapter 3. Ordinary Differential Equations. Section 3.6 Summary and Problems. Page
218
Problem number: Problem 3.23.
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.
\[ \boxed {\left (-x^{2}+1\right ) z^{\prime \prime }+\left (1-3 x \right ) z^{\prime }+k z=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d}{d x}z^{\prime }\right )+\left (1-3 x \right ) z^{\prime }+k z=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}z^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}z^{\prime }=-\frac {\left (3 x -1\right ) z^{\prime }}{x^{2}-1}+\frac {k z}{x^{2}-1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} z\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}z^{\prime }+\frac {\left (3 x -1\right ) z^{\prime }}{x^{2}-1}-\frac {k z}{x^{2}-1}=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 {3 x -1}{x^{2}-1}, P_{3}\left (x \right )=-\frac {k}{x^{2}-1}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=2 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}z^{\prime }\right ) \left (x^{2}-1\right )+\left (3 x -1\right ) z^{\prime }-k z=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d}{d u}\frac {d}{d u}z \left (u \right )\right )+\left (3 u -4\right ) \left (\frac {d}{d u}z \left (u \right )\right )-k z \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} z \left (u \right ) \\ {} & {} & z \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 \left (\frac {d}{d u}z \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}z \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}z \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^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}z \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}z \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}z \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r \left (1+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )-a_{k} \left (-k^{2}-2 k r -r^{2}+k -2 k -2 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +r +1\right ) \left (k +2+r \right )-a_{k} \left (-k^{2}+\left (-2 r -2\right ) k -r^{2}+k -2 r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-k^{2}-2 k r -r^{2}+k -2 k -2 r \right )}{2 \left (k +r +1\right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-k^{2}+k +1\right )}{2 k \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [z \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -1}, a_{k +1}=-\frac {a_{k} \left (-k^{2}+k +1\right )}{2 k \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [z=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -1}, a_{k +1}=-\frac {a_{k} \left (-k^{2}+k +1\right )}{2 k \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (-k^{2}+k -2 k \right )}{2 \left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [z \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +1}=-\frac {a_{k} \left (-k^{2}+k -2 k \right )}{2 \left (k +1\right ) \left (k +2\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [z=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +1}=-\frac {a_{k} \left (-k^{2}+k -2 k \right )}{2 \left (k +1\right ) \left (k +2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [z=\left (\moverset {\infty }{\munderset {m =0}{\sum }}a_{m} \left (x +1\right )^{m -1}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}b_{m} \left (x +1\right )^{m}\right ), a_{m +1}=-\frac {a_{m} \left (-m^{2}+k +1\right )}{2 m \left (m +1\right )}, b_{m +1}=-\frac {b_{m} \left (-m^{2}+k -2 m \right )}{2 \left (m +1\right ) \left (m +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 -> hypergeometric -> heuristic approach <- heuristic approach successful -> solution has integrals; searching for one without integrals... -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric solution without integrals succesful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.109 (sec). Leaf size: 99
dsolve((1-x^2)*diff(z(x),x$2)+(1-3*x)*diff(z(x),x)+k*z(x)=0,z(x), singsol=all)
\[ z \left (x \right ) = c_{1} \left (x +1\right )^{-1-\sqrt {k +1}} \operatorname {hypergeom}\left (\left [\sqrt {k +1}, 1+\sqrt {k +1}\right ], \left [1+2 \sqrt {k +1}\right ], \frac {2}{x +1}\right )+c_{2} \left (x +1\right )^{-1+\sqrt {k +1}} \operatorname {hypergeom}\left (\left [-\sqrt {k +1}, 1-\sqrt {k +1}\right ], \left [1-2 \sqrt {k +1}\right ], \frac {2}{x +1}\right ) \]
✓ Solution by Mathematica
Time used: 0.407 (sec). Leaf size: 77
DSolve[(1-x^2)*z''[x]+(1-3*x)*z'[x]+k*z[x]==0,z[x],x,IncludeSingularSolutions -> True]
\[ z(x)\to c_2 G_{2,2}^{2,0}\left (\frac {1-x}{2}| \begin {array}{c} -\sqrt {k+1},\sqrt {k+1} \\ 0,0 \\ \end {array} \right )+c_1 \operatorname {Hypergeometric2F1}\left (1-\sqrt {k+1},\sqrt {k+1}+1,1,\frac {1-x}{2}\right ) \]