Internal problem ID [10384]
Internal file name [OUTPUT/9332_Monday_June_06_2022_01_56_49_PM_58358154/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power
Functions
Problem number: 55.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {\left (x^{2}-1\right ) y^{\prime }+\lambda \left (y^{2}-2 x y+1\right )=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {\lambda \left (-2 y x +y^{2}+1\right )}{x^{2}-1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {2 \lambda y x}{x^{2}-1}-\frac {\lambda \,y^{2}}{x^{2}-1}-\frac {\lambda }{x^{2}-1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {\lambda }{x^{2}-1}\), \(f_1(x)=\frac {2 \lambda x}{x^{2}-1}\) and \(f_2(x)=-\frac {\lambda }{x^{2}-1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {\lambda u}{x^{2}-1}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=\frac {2 \lambda x}{\left (x^{2}-1\right )^{2}}\\ f_1 f_2 &=-\frac {2 \lambda ^{2} x}{\left (x^{2}-1\right )^{2}}\\ f_2^2 f_0 &=-\frac {\lambda ^{3}}{\left (x^{2}-1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {\lambda u^{\prime \prime }\left (x \right )}{x^{2}-1}-\left (-\frac {2 \lambda ^{2} x}{\left (x^{2}-1\right )^{2}}+\frac {2 \lambda x}{\left (x^{2}-1\right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {\lambda ^{3} u \left (x \right )}{\left (x^{2}-1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {LegendreP}\left (\lambda -1, x\right ) c_{1} +\operatorname {LegendreQ}\left (\lambda -1, x\right ) c_{2} \right ) \left (x^{2}-1\right )^{\frac {\lambda }{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \lambda \left (x^{2}-1\right )^{-1+\frac {\lambda }{2}} \left (\operatorname {LegendreQ}\left (\lambda , x\right ) c_{2} +\operatorname {LegendreP}\left (\lambda , x\right ) c_{1} \right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (x^{2}-1\right )^{-1+\frac {\lambda }{2}} \left (\operatorname {LegendreQ}\left (\lambda , x\right ) c_{2} +\operatorname {LegendreP}\left (\lambda , x\right ) c_{1} \right ) \left (x^{2}-1\right ) \left (x^{2}-1\right )^{-\frac {\lambda }{2}}}{\operatorname {LegendreP}\left (\lambda -1, x\right ) c_{1} +\operatorname {LegendreQ}\left (\lambda -1, x\right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\operatorname {LegendreQ}\left (\lambda , x\right )+\operatorname {LegendreP}\left (\lambda , x\right ) c_{3}}{\operatorname {LegendreP}\left (\lambda -1, x\right ) c_{3} +\operatorname {LegendreQ}\left (\lambda -1, x\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\operatorname {LegendreQ}\left (\lambda , x\right )+\operatorname {LegendreP}\left (\lambda , x\right ) c_{3}}{\operatorname {LegendreP}\left (\lambda -1, x\right ) c_{3} +\operatorname {LegendreQ}\left (\lambda -1, x\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\operatorname {LegendreQ}\left (\lambda , x\right )+\operatorname {LegendreP}\left (\lambda , x\right ) c_{3}}{\operatorname {LegendreP}\left (\lambda -1, x\right ) c_{3} +\operatorname {LegendreQ}\left (\lambda -1, x\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) y^{\prime }+\lambda \left (y^{2}-2 x y+1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\lambda \left (y^{2}-2 x y+1\right )}{x^{2}-1} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: <- Abel AIR successful: ODE belongs to the 2F1 3-parameter class`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 280
dsolve((x^2-1)*diff(y(x),x)+lambda*(y(x)^2-2*x*y(x)+1)=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {2 \left (\left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 \lambda } \left (1+x \right ) \left (-1+x \right )^{2} \operatorname {HeunCPrime}\left (0, 2 \lambda -1, 0, 0, \lambda ^{2}-\lambda +\frac {1}{2}, \frac {2}{1+x}\right )+8 \left (-1+x \right )^{2} \operatorname {HeunCPrime}\left (0, -2 \lambda +1, 0, 0, \lambda ^{2}-\lambda +\frac {1}{2}, \frac {2}{1+x}\right ) c_{1} -8 \left (1+x \right )^{2} \left (\left (\left (\lambda -\frac {1}{2}\right ) x -\frac {\lambda }{2}+\frac {1}{2}\right ) c_{1} \left (\frac {1+x}{-1+x}\right )^{-\lambda } \operatorname {hypergeom}\left (\left [1-\lambda , 1-\lambda \right ], \left [-2 \lambda +2\right ], -\frac {2}{-1+x}\right )+\frac {\lambda \left (\frac {1+x}{-1+x}\right )^{\lambda } \left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 \lambda } \operatorname {hypergeom}\left (\left [\lambda , \lambda \right ], \left [2 \lambda \right ], -\frac {2}{-1+x}\right ) \left (-1+x \right )}{16}\right )\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 \lambda } \left (\frac {1+x}{-1+x}\right )^{\lambda }}{\left (8 c_{1} \operatorname {hypergeom}\left (\left [1-\lambda , 1-\lambda \right ], \left [-2 \lambda +2\right ], -\frac {2}{-1+x}\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 \lambda }+\operatorname {hypergeom}\left (\left [\lambda , \lambda \right ], \left [2 \lambda \right ], -\frac {2}{-1+x}\right ) \left (\frac {1+x}{-1+x}\right )^{2 \lambda } \left (-1+x \right )\right ) \left (1+x \right )^{2} \lambda } \]
✓ Solution by Mathematica
Time used: 0.643 (sec). Leaf size: 47
DSolve[(x^2-1)*y'[x]+\[Lambda]*(y[x]^2-2*x*y[x]+1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\operatorname {LegendreQ}(\lambda ,x)+c_1 \operatorname {LegendreP}(\lambda ,x)}{\operatorname {LegendreQ}(\lambda -1,x)+c_1 \operatorname {LegendreP}(\lambda -1,x)} \\ y(x)\to \frac {\operatorname {LegendreP}(\lambda ,x)}{\operatorname {LegendreP}(\lambda -1,x)} \\ \end{align*}