4.17 problem Problem 3.24

Internal problem ID [5891]
Internal file name [OUTPUT/5139_Sunday_June_05_2022_03_25_59_PM_29405420/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 3.24.
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.

\[ \boxed {\left (-x^{2}+1\right ) \eta ^{\prime \prime }-\left (x +1\right ) \eta ^{\prime }+\left (k +1\right ) \eta =0} \]

4.17.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-x^{2}+1\right ) \left (\frac {d}{d x}\eta ^{\prime }\right )+\left (-x -1\right ) \eta ^{\prime }+\left (k +1\right ) \eta =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}\eta ^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\eta ^{\prime }=\frac {\left (k +1\right ) \eta }{x^{2}-1}-\frac {\eta ^{\prime }}{x -1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} \eta \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}\eta ^{\prime }+\frac {\eta ^{\prime }}{x -1}-\frac {\left (k +1\right ) \eta }{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 {1}{x -1}, P_{3}\left (x \right )=-\frac {k +1}{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}}}=0 \\ {} & \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}\eta ^{\prime }\right ) \left (x -1\right ) \left (x^{2}-1\right )+\eta ^{\prime } \left (x^{2}-1\right )-\left (x -1\right ) \left (k +1\right ) \eta =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^{3}-4 u^{2}+4 u \right ) \left (\frac {d}{d u}\frac {d}{d u}\eta \left (u \right )\right )+\left (u^{2}-2 u \right ) \left (\frac {d}{d u}\eta \left (u \right )\right )+\left (-u k +2 k -u +2\right ) \eta \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} \eta \left (u \right ) \\ {} & {} & \eta \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 \eta \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot \eta \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 \eta \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}\eta \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\eta \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}\eta \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}\eta \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}\eta \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}\eta \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} \\ {} & {} & 4 a_{0} r \left (-1+r \right ) u^{-1+r}+\left (4 a_{1} \left (1+r \right ) r +2 a_{0} \left (-2 r^{2}+k +r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +r \right )+2 a_{k} \left (-2 k^{2}-4 k r -2 r^{2}+k +k +r +1\right )-a_{k -1} \left (-\left (k -1\right )^{2}-2 \left (k -1\right ) r -r^{2}+k +1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) r +2 a_{0} \left (-2 r^{2}+k +r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) k^{2}+\left (\left (-8 a_{k}+2 a_{k -1}+8 a_{k +1}\right ) r +2 a_{k}-2 a_{k -1}+4 a_{k +1}\right ) k +\left (-4 a_{k}+a_{k -1}+4 a_{k +1}\right ) r^{2}+\left (2 a_{k}-2 a_{k -1}+4 a_{k +1}\right ) r +\left (2 k +2\right ) a_{k}-k a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )^{2}+\left (\left (-8 a_{k +1}+2 a_{k}+8 a_{k +2}\right ) r +2 a_{k +1}-2 a_{k}+4 a_{k +2}\right ) \left (k +1\right )+\left (-4 a_{k +1}+a_{k}+4 a_{k +2}\right ) r^{2}+\left (2 a_{k +1}-2 a_{k}+4 a_{k +2}\right ) r +\left (2 k +2\right ) a_{k +1}-a_{k} k =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}-2 k r a_{k}+8 k r a_{k +1}-r^{2} a_{k}+4 r^{2} a_{k +1}+a_{k} k -2 k a_{k +1}+6 k a_{k +1}+6 r a_{k +1}+a_{k}}{4 \left (k^{2}+2 k r +r^{2}+3 k +3 r +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}+a_{k} k -2 k a_{k +1}+6 k a_{k +1}+a_{k}}{4 \left (k^{2}+3 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [\eta \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}+a_{k} k -2 k a_{k +1}+6 k a_{k +1}+a_{k}}{4 \left (k^{2}+3 k +2\right )}, 2 a_{0} \left (k +1\right )=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [\eta =\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}+a_{k} k -2 k a_{k +1}+6 k a_{k +1}+a_{k}}{4 \left (k^{2}+3 k +2\right )}, 2 a_{0} \left (k +1\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}+a_{k} k -2 k a_{k +1}-2 k a_{k}+14 k a_{k +1}+10 a_{k +1}}{4 \left (k^{2}+5 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [\eta \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +1}, a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}+a_{k} k -2 k a_{k +1}-2 k a_{k}+14 k a_{k +1}+10 a_{k +1}}{4 \left (k^{2}+5 k +6\right )}, 2 a_{0} k +8 a_{1}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [\eta =\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k +1}, a_{k +2}=\frac {-k^{2} a_{k}+4 k^{2} a_{k +1}+a_{k} k -2 k a_{k +1}-2 k a_{k}+14 k a_{k +1}+10 a_{k +1}}{4 \left (k^{2}+5 k +6\right )}, 2 a_{0} k +8 a_{1}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [\eta =\left (\moverset {\infty }{\munderset {m =0}{\sum }}a_{m} \left (x +1\right )^{m}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}b_{m} \left (x +1\right )^{m +1}\right ), a_{m +2}=\frac {-m^{2} a_{m}+4 m^{2} a_{m +1}+k a_{m}-2 k a_{m +1}+6 m a_{m +1}+a_{m}}{4 \left (m^{2}+3 m +2\right )}, 2 a_{0} \left (k +1\right )=0, b_{m +2}=\frac {-m^{2} b_{m}+4 m^{2} b_{m +1}+k b_{m}-2 k b_{m +1}-2 m b_{m}+14 m b_{m +1}+10 b_{m +1}}{4 \left (m^{2}+5 m +6\right )}, 2 k b_{0}+8 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 
   -> Whittaker 
      -> 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: 0.11 (sec). Leaf size: 95

dsolve((1-x^2)*diff(eta(x),x$2)-(1+x)*diff(eta(x),x)+(k+1)*eta(x)=0,eta(x), singsol=all)
 

\[ \eta \left (x \right ) = c_{1} \left (x +1\right )^{\sqrt {k +1}} \operatorname {hypergeom}\left (\left [-\sqrt {k +1}, 1-\sqrt {k +1}\right ], \left [1-2 \sqrt {k +1}\right ], \frac {2}{x +1}\right )+c_{2} \left (x +1\right )^{-\sqrt {k +1}} \operatorname {hypergeom}\left (\left [\sqrt {k +1}, 1+\sqrt {k +1}\right ], \left [1+2 \sqrt {k +1}\right ], \frac {2}{x +1}\right ) \]

✓ Solution by Mathematica

Time used: 0.283 (sec). Leaf size: 77

DSolve[(1-x^2)*z''[x]-(1+x)*z'[x]+(k+1)*z[x]==0,z[x],x,IncludeSingularSolutions -> True]
 

\[ z(x)\to c_2 G_{2,2}^{2,0}\left (\frac {1-x}{2}| \begin {array}{c} 1-\sqrt {k+1},\sqrt {k+1}+1 \\ 0,0 \\ \end {array} \right )+c_1 \operatorname {Hypergeometric2F1}\left (-\sqrt {k+1},\sqrt {k+1},1,\frac {1-x}{2}\right ) \]