problem 64

Internal problem ID [10393]
Internal ﬁle name [OUTPUT/9341_Monday_June_06_2022_02_12_57_PM_55541898/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: 64.
ODE order: 1.
ODE degree: 1.

\[ \boxed {\left (c_{2} x^{2}+b_{2} x +a_{2} \right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (b_{1} x +a_{1} \right ) y=-a_{0}} \]

2.64.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {c_{2} \lambda \,x^{2} y^{2}+y^{2} b_{2} \lambda x +y^{2} a_{2} \lambda +y b_{1} x +a_{1} y +a_{0}}{c_{2} x^{2}+b_{2} x +a_{2}} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {y^{2} c_{2} \lambda \,x^{2}}{c_{2} x^{2}+b_{2} x +a_{2}}-\frac {y^{2} b_{2} \lambda x}{c_{2} x^{2}+b_{2} x +a_{2}}-\frac {y^{2} a_{2} \lambda }{c_{2} x^{2}+b_{2} x +a_{2}}-\frac {y b_{1} x}{c_{2} x^{2}+b_{2} x +a_{2}}-\frac {a_{1} y}{c_{2} x^{2}+b_{2} x +a_{2}}-\frac {a_{0}}{c_{2} x^{2}+b_{2} x +a_{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {a_{0}}{c_{2} x^{2}+b_{2} x +a_{2}}\), \(f_1(x)=-\frac {b_{1} x +a_{1}}{c_{2} x^{2}+b_{2} x +a_{2}}\) and \(f_2(x)=-\frac {c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda }{c_{2} x^{2}+b_{2} x +a_{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {\left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right ) u}{c_{2} x^{2}+b_{2} x +a_{2}}} \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 c_{2} \lambda x +\lambda b_{2}}{c_{2} x^{2}+b_{2} x +a_{2}}+\frac {\left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right ) \left (2 c_{2} x +b_{2} \right )}{\left (c_{2} x^{2}+b_{2} x +a_{2} \right )^{2}}\\ f_1 f_2 &=\frac {\left (b_{1} x +a_{1} \right ) \left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right )}{\left (c_{2} x^{2}+b_{2} x +a_{2} \right )^{2}}\\ f_2^2 f_0 &=-\frac {\left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right )^{2} a_{0}}{\left (c_{2} x^{2}+b_{2} x +a_{2} \right )^{3}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -\frac {\left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right ) u^{\prime \prime }\left (x \right )}{c_{2} x^{2}+b_{2} x +a_{2}}-\left (-\frac {2 c_{2} \lambda x +\lambda b_{2}}{c_{2} x^{2}+b_{2} x +a_{2}}+\frac {\left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right ) \left (2 c_{2} x +b_{2} \right )}{\left (c_{2} x^{2}+b_{2} x +a_{2} \right )^{2}}+\frac {\left (b_{1} x +a_{1} \right ) \left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right )}{\left (c_{2} x^{2}+b_{2} x +a_{2} \right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {\left (c_{2} \lambda \,x^{2}+\lambda x b_{2} +a_{2} \lambda \right )^{2} a_{0} u \left (x \right )}{\left (c_{2} x^{2}+b_{2} x +a_{2} \right )^{3}} &=0 \end {align*}

\[ u \left (x \right ) = c_{3} \operatorname {hypergeom}\left (\left [\frac {-c_{2} +b_{1} +\sqrt {c_{2}^{2}+\left (-4 a_{0} \lambda -2 b_{1} \right ) c_{2} +b_{1}^{2}}}{2 c_{2}}, -\frac {c_{2} -b_{1} +\sqrt {c_{2}^{2}+\left (-4 a_{0} \lambda -2 b_{1} \right ) c_{2} +b_{1}^{2}}}{2 c_{2}}\right ], \left [\frac {b_{1} \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, c_{2} -2 c_{2} a_{1} +b_{1} b_{2}}{2 c_{2}^{2} \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}}\right ], \frac {-2 \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, x \,c_{2}^{2}-\sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, b_{2} c_{2} +4 c_{2} a_{2} -b_{2}^{2}}{8 c_{2} a_{2} -2 b_{2}^{2}}\right )+c_{4} {\left (2 \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, x \,c_{2}^{2}+\sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, b_{2} c_{2} -4 c_{2} a_{2} +b_{2}^{2}\right )}^{\frac {c_{2} \left (c_{2} -\frac {b_{1}}{2}\right ) \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}+c_{2} a_{1} -\frac {b_{1} b_{2}}{2}}{\sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, c_{2}^{2}}} \operatorname {hypergeom}\left (\left [\frac {c_{2} \left (c_{2} +\sqrt {c_{2}^{2}+\left (-4 a_{0} \lambda -2 b_{1} \right ) c_{2} +b_{1}^{2}}\right ) \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}+2 c_{2} a_{1} -b_{1} b_{2}}{2 \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, c_{2}^{2}}, \frac {c_{2} \left (c_{2} -\sqrt {c_{2}^{2}+\left (-4 a_{0} \lambda -2 b_{1} \right ) c_{2} +b_{1}^{2}}\right ) \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}+2 c_{2} a_{1} -b_{1} b_{2}}{2 \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, c_{2}^{2}}\right ], \left [\frac {4 c_{2}^{2} \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}-b_{1} \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, c_{2} +2 c_{2} a_{1} -b_{1} b_{2}}{2 c_{2}^{2} \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}}\right ], \frac {-2 \sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, x \,c_{2}^{2}-\sqrt {\frac {-4 c_{2} a_{2} +b_{2}^{2}}{c_{2}^{2}}}\, b_{2} c_{2} +4 c_{2} a_{2} -b_{2}^{2}}{8 c_{2} a_{2} -2 b_{2}^{2}}\right ) \] The above shows that \[ \text {Expression too large to display} \] Using the above in (1) gives the solution \[ \text {Expression too large to display} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

Summary

The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}

Veriﬁcation of solutions

\[ \text {Expression too large to display} \] Warning, solution could not be veriﬁed

2.64.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (c_{2} x^{2}+b_{2} x +a_{2} \right ) \left (y^{\prime }+\lambda y^{2}\right )+\left (b_{1} x +a_{1} \right ) y=-a_{0} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2} c_{2} \lambda \,x^{2}+y^{2} b_{2} \lambda x +y^{2} a_{2} \lambda +y b_{1} x +y a_{1} +a_{0}}{c_{2} x^{2}+b_{2} x +a_{2}} \end {array} \]

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: 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(b__1*x+a__1)*(diff(y(x), x))/(c__2*x^2+b__2*x+a__2)-a__0*lambda*y(x) 
      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 
         <- hypergeometric successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful`

2.64 problem 64

2.64.1 Solving as riccati ode

2.64.2 Maple step by step solution