19.6 problem 6

19.6.1 Solving as riccati ode
19.6.2 Maple step by step solution

Internal problem ID [10598]
Internal file name [OUTPUT/9546_Monday_June_06_2022_03_07_28_PM_86558589/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. subsection 1.2.8-1. Equations containing arbitrary functions (but not containing their derivatives).
Problem number: 6.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "riccati"

Maple gives the following as the ode type

[_Riccati]

\[ \boxed {y^{\prime }+\left (1+n \right ) x^{n} y^{2}-x^{1+n} f \left (x \right ) y=-f \left (x \right )} \]

19.6.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -x^{n} y^{2} n -x^{n} y^{2}+x^{1+n} f \left (x \right ) y -f \left (x \right ) \end {align*}

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

Substituting the above terms back in equation (2) gives \begin {align*} \left (-x^{n} n -x^{n}\right ) u^{\prime \prime }\left (x \right )-\left (-\frac {x^{n} n^{2}}{x}-\frac {x^{n} n}{x}+x^{1+n} f \left (x \right ) \left (-x^{n} n -x^{n}\right )\right ) u^{\prime }\left (x \right )-\left (-x^{n} n -x^{n}\right )^{2} f \left (x \right ) u \left (x \right ) &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = x^{1+n} \left (\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}d x \right ) c_{2} +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = c_{2} x^{-1-n} {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x}+x^{n} \left (1+n \right ) \left (\left (\int {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x} x^{-2 n -2}d x \right ) c_{2} +c_{1} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (c_{2} x^{-1-n} {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x}+x^{n} \left (1+n \right ) \left (\left (\int {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x} x^{-2 n -2}d x \right ) c_{2} +c_{1} \right )\right ) x^{-1-n}}{\left (-x^{n} n -x^{n}\right ) \left (\left (\int x^{-2 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}d x \right ) c_{2} +c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {\left (1+n \right ) \left (\int x^{-2 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}d x +c_{3} \right ) x^{-1-n}+x^{-3 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}}{\left (1+n \right ) \left (\int {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x} x^{-2 n -2}d x +c_{3} \right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (1+n \right ) \left (\int x^{-2 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}d x +c_{3} \right ) x^{-1-n}+x^{-3 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}}{\left (1+n \right ) \left (\int {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x} x^{-2 n -2}d x +c_{3} \right )} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (1+n \right ) \left (\int x^{-2 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}d x +c_{3} \right ) x^{-1-n}+x^{-3 n -2} {\mathrm e}^{\int \left (x^{1+n} f \left (x \right )+\frac {n}{x}\right )d x}}{\left (1+n \right ) \left (\int {\mathrm e}^{\int \frac {x^{n +2} f \left (x \right )+n}{x}d x} x^{-2 n -2}d x +c_{3} \right )} \] Verified OK.

19.6.2 Maple step by step solution

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

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)*f(x)*x+n)*(diff(y(x), x))/x-x^n*(n+1)*f(x)*y(x), y(x)`      * 
      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 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
      <- unable to find a useful change of variables 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         trying 2nd order, integrating factor of the form mu(x,y) 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
         -> Trying changes of variables to rationalize or make the ODE simpler 
         <- unable to find a useful change of variables 
            trying a symmetry of the form [xi=0, eta=F(x)] 
         trying to convert to an ODE of Bessel type 
   -> Trying a change of variables to reduce to Bernoulli 
   -> Calling odsolve with the ODE`, diff(y(x), x)-((-x^n*n-x^n)*y(x)^2+y(x)+x^(n+1)*f(x)*y(x)*x-f(x)*x^2)/x, y(x), explicit`      * 
      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 inverse_Riccati 
      trying 1st order ODE linearizable_by_differentiation 
   -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
   -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
   -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
trying inverse_Riccati 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
`, `-> Computing symmetries using: way = 4 
`, `-> Computing symmetries using: way = 2
 

Solution by Maple

Time used: 0.015 (sec). Leaf size: 169

dsolve(diff(y(x),x)=-(n+1)*x^n*y(x)^2+x^(n+1)*f(x)*y(x)-f(x),y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {x^{-n -1} \left (x^{n +1} {\mathrm e}^{\int \frac {x^{n +1} f \left (x \right ) x -2 n -2}{x}d x}+\left (\int x^{n} {\mathrm e}^{\int x^{n +1} f \left (x \right )d x +\left (-2 n -2\right ) \left (\int \frac {1}{x}d x \right )}d x \right ) n +\int x^{n} {\mathrm e}^{\int x^{n +1} f \left (x \right )d x +\left (-2 n -2\right ) \left (\int \frac {1}{x}d x \right )}d x -c_{1} \right )}{\left (\int x^{n} {\mathrm e}^{\int x^{n +1} f \left (x \right )d x +\left (-2 n -2\right ) \left (\int \frac {1}{x}d x \right )}d x \right ) n +\int x^{n} {\mathrm e}^{\int x^{n +1} f \left (x \right )d x +\left (-2 n -2\right ) \left (\int \frac {1}{x}d x \right )}d x -c_{1}} \]

Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[y'[x]==-(n+1)*x^n*y[x]^2+x^(n+1)*f[x]*y[x]-f[x],y[x],x,IncludeSingularSolutions -> True]
 

Not solved