2.75 problem 75

Internal problem ID [10404]
Internal file name [OUTPUT/9352_Monday_June_06_2022_02_15_36_PM_48106836/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: 75.
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]

\[ \boxed {\left (x^{n} a +b \,x^{m}+c \right ) y^{\prime }-c y^{2}+b \,x^{m -1} y=a \,x^{-2+n}} \]

2.75.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {c \,y^{2}-b \,x^{m -1} y +a \,x^{-2+n}}{x^{n} a +b \,x^{m}+c} \end {align*}

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

Substituting the above terms back in equation (2) gives \begin {align*} \frac {c u^{\prime \prime }\left (x \right )}{x^{n} a +b \,x^{m}+c}-\left (-\frac {c \left (\frac {x^{n} n a}{x}+\frac {b \,x^{m} m}{x}\right )}{\left (x^{n} a +b \,x^{m}+c \right )^{2}}-\frac {b \,x^{m -1} c}{\left (x^{n} a +b \,x^{m}+c \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {c^{2} a \,x^{-2+n} u \left (x \right )}{\left (x^{n} a +b \,x^{m}+c \right )^{3}} &=0 \end {align*}

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

\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (-x^{n} n a -b \,x^{m} m -b \,x^{m}\right )}{\left (x^{n} a +b \,x^{m}+c \right ) x}+\frac {c a \,x^{-2+n} \textit {\_Y} \left (x \right )}{\left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (-x^{n} n a -b \,x^{m} m -b \,x^{m}\right )}{\left (x^{n} a +b \,x^{m}+c \right ) x}+\frac {c a \,x^{-2+n} \textit {\_Y} \left (x \right )}{\left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (-x^{n} n a -b \,x^{m} m -b \,x^{m}\right )}{\left (x^{n} a +b \,x^{m}+c \right ) x}+\frac {c a \,x^{-2+n} \textit {\_Y} \left (x \right )}{\left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \left (x^{n} a +b \,x^{m}+c \right )}{c \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (-x^{n} n a -b \,x^{m} m -b \,x^{m}\right )}{\left (x^{n} a +b \,x^{m}+c \right ) x}+\frac {c a \,x^{-2+n} \textit {\_Y} \left (x \right )}{\left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {\left (x^{n} a +b \,x^{m}+c \right ) \left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2} \textit {\_Y}^{\prime \prime }\left (x \right )+\left (b \left (m +1\right ) x^{m}+x^{n} n a \right ) x \left (x^{n} a +b \,x^{m}+c \right ) \textit {\_Y}^{\prime }\left (x \right )+a c \,x^{n} \textit {\_Y} \left (x \right )}{x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right )}{c \operatorname {DESol}\left (\left \{\frac {\left (x^{2 m} b^{2}+a^{2} x^{2 n}+2 b \left (x^{n} a +c \right ) x^{m}+2 x^{n} a c +c^{2}\right ) x^{2} \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) b^{2} x \left (m +1\right ) x^{2 m}+x^{2 n} \textit {\_Y}^{\prime }\left (x \right ) a^{2} n x +x \left (\left (a \left (1+m +n \right ) x^{n}+c \left (m +1\right )\right ) b \,x^{m}+x^{n} a c n \right ) \textit {\_Y}^{\prime }\left (x \right )+a c \,x^{n} \textit {\_Y} \left (x \right )}{x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (x^{n} a +b \,x^{m}+c \right ) \left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2} \textit {\_Y}^{\prime \prime }\left (x \right )+\left (b \left (m +1\right ) x^{m}+x^{n} n a \right ) x \left (x^{n} a +b \,x^{m}+c \right ) \textit {\_Y}^{\prime }\left (x \right )+a c \,x^{n} \textit {\_Y} \left (x \right )}{x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right )}{c \operatorname {DESol}\left (\left \{\frac {\left (x^{2 m} b^{2}+a^{2} x^{2 n}+2 b \left (x^{n} a +c \right ) x^{m}+2 x^{n} a c +c^{2}\right ) x^{2} \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) b^{2} x \left (m +1\right ) x^{2 m}+x^{2 n} \textit {\_Y}^{\prime }\left (x \right ) a^{2} n x +x \left (\left (a \left (1+m +n \right ) x^{n}+c \left (m +1\right )\right ) b \,x^{m}+x^{n} a c n \right ) \textit {\_Y}^{\prime }\left (x \right )+a c \,x^{n} \textit {\_Y} \left (x \right )}{x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}

Verification of solutions

\[ y = -\frac {\left (x^{n} a +b \,x^{m}+c \right ) \left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2} \textit {\_Y}^{\prime \prime }\left (x \right )+\left (b \left (m +1\right ) x^{m}+x^{n} n a \right ) x \left (x^{n} a +b \,x^{m}+c \right ) \textit {\_Y}^{\prime }\left (x \right )+a c \,x^{n} \textit {\_Y} \left (x \right )}{x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right )}{c \operatorname {DESol}\left (\left \{\frac {\left (x^{2 m} b^{2}+a^{2} x^{2 n}+2 b \left (x^{n} a +c \right ) x^{m}+2 x^{n} a c +c^{2}\right ) x^{2} \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) b^{2} x \left (m +1\right ) x^{2 m}+x^{2 n} \textit {\_Y}^{\prime }\left (x \right ) a^{2} n x +x \left (\left (a \left (1+m +n \right ) x^{n}+c \left (m +1\right )\right ) b \,x^{m}+x^{n} a c n \right ) \textit {\_Y}^{\prime }\left (x \right )+a c \,x^{n} \textit {\_Y} \left (x \right )}{x^{2} \left (x^{n} a +b \,x^{m}+c \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.

2.75.2 Maple step by step solution

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

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) = -(a*n*x^n+b*m*x^m+x^(m-1)*b*x)*(diff(y(x), x))/(x*(x^n*a+b*x^m+c))-c*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 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
      -> 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 
         -> Mathieu 
            -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      -> 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 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
         -> 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 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
         -> 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)-(c*y(x)^2/(x^n*a+b*x^m+c)+y(x)-x^(m-1)*b*y(x)*x/(x^n*a+b*x^m+c)+x^2*x^(n-2)*a/(x^ 
      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: 
         <- Riccati particular case Kamke (b) successful 
   trying Bernoulli 
   <- Bernoulli successful 
   <- Riccati reduced to Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 1236

dsolve((a*x^n+b*x^m+c)*diff(y(x),x)=c*y(x)^2-b*x^(m-1)*y(x)+a*x^(n-2),y(x), singsol=all)
 

\[ \text {Expression too large to display} \]

✗ Solution by Mathematica

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

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

Not solved