problem 54

Internal problem ID [10383]
Internal ﬁle name [OUTPUT/9331_Monday_June_06_2022_01_52_31_PM_88664424/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: 54.
ODE order: 1.
ODE degree: 1.

\[ \boxed {x^{2} y^{\prime }-\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right ) y^{2}-\left (x^{n} a +b \right ) x y=c \,x^{2}} \]

2.54.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{2 n} \alpha \,y^{2}+x^{n} a x y +x^{n} \beta \,y^{2}+b x y +c \,x^{2}+\gamma \,y^{2}}{x^{2}} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{2 n} \alpha \,y^{2}}{x^{2}}+\frac {x^{n} a y}{x}+\frac {x^{n} \beta \,y^{2}}{x^{2}}+\frac {y b}{x}+c +\frac {\gamma \,y^{2}}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c\), \(f_1(x)=\frac {x^{n} a x +b x}{x^{2}}\) and \(f_2(x)=\frac {\alpha \,x^{2 n}+\beta \,x^{n}+\gamma }{x^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right ) u}{x^{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 {\frac {2 \alpha \,x^{2 n} n}{x}+\frac {\beta \,x^{n} n}{x}}{x^{2}}-\frac {2 \left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )}{x^{3}}\\ f_1 f_2 &=\frac {\left (x^{n} a x +b x \right ) \left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )}{x^{4}}\\ f_2^2 f_0 &=\frac {\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )^{2} c}{x^{4}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} \frac {\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right ) u^{\prime \prime }\left (x \right )}{x^{2}}-\left (\frac {\frac {2 \alpha \,x^{2 n} n}{x}+\frac {\beta \,x^{n} n}{x}}{x^{2}}-\frac {2 \left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )}{x^{3}}+\frac {\left (x^{n} a x +b x \right ) \left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )}{x^{4}}\right ) u^{\prime }\left (x \right )+\frac {\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )^{2} c u \left (x \right )}{x^{4}} &=0 \end {align*}

\[ \text {Expression too large to display} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {c \left (\left (-n +\sqrt {b^{2}-4 c \gamma -2 b +1}\right ) \left (\left (3 \gamma ^{2} \alpha +3 \beta ^{2} \gamma \right ) x^{2 n -\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+\left (6 \alpha \beta \gamma +\beta ^{3}\right ) x^{3 n -\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+\left (3 \alpha ^{2} \gamma +3 \alpha \,\beta ^{2}\right ) x^{4 n -\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+\gamma ^{3} x^{-\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+3 \gamma ^{2} x^{n -\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}} \beta +3 x^{5 n -\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}} \beta \,\alpha ^{2}+x^{6 n -\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}} \alpha ^{3}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {-\sqrt {b^{2}-4 c \gamma -2 b +1}\, \sqrt {a^{2}-4 \alpha c}+\sqrt {a^{2}-4 \alpha c}\, n +\left (b +n -1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}\right ], \left [1-\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{n}\right ], \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right )+\operatorname {hypergeom}\left (\left [\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}\, \sqrt {a^{2}-4 \alpha c}+\sqrt {a^{2}-4 \alpha c}\, n +\left (b +n -1\right ) a -2 \beta c}{2 \sqrt {a^{2}-4 \alpha c}\, n}\right ], \left [1+\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{n}\right ], \frac {\sqrt {a^{2}-4 \alpha c}\, x^{n}}{n}\right ) \left (n +\sqrt {b^{2}-4 c \gamma -2 b +1}\right ) c_{2} \left (\left (3 \gamma ^{2} \alpha +3 \beta ^{2} \gamma \right ) x^{2 n +\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+\left (6 \alpha \beta \gamma +\beta ^{3}\right ) x^{3 n +\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+\left (3 \alpha ^{2} \gamma +3 \alpha \,\beta ^{2}\right ) x^{4 n +\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+\gamma ^{3} x^{\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}}+3 \gamma ^{2} x^{n +\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}} \beta +3 x^{5 n +\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}} \beta \,\alpha ^{2}+x^{6 n +\frac {\sqrt {b^{2}-4 c \gamma -2 b +1}}{2}} \alpha ^{3}\right )\right ) {\mathrm e}^{-\frac {x^{n} \left (-a +\sqrt {a^{2}-4 \alpha c}\right )}{2 n}} x^{-\frac {3}{2}+\frac {b}{2}}}{\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right )^{2}} \] 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

2.54.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right ) y^{2}-\left (x^{n} a +b \right ) x y=c \,x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (\alpha \,x^{2 n}+\beta \,x^{n}+\gamma \right ) y^{2}+\left (x^{n} a +b \right ) x y+c \,x^{2}}{x^{2}} \end {array} \]

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_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (x^(n-1)*x^(2*n-2)*a*alpha*x^3+x^(n-1)*x^(n-2)*a*beta*x^3+x^(2*n-2)*al 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
         -> 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 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
                  <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
         <- hypergeometric successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful`

2.54 problem 54

2.54.1 Solving as riccati ode

2.54.2 Maple step by step solution