2.30 problem 30

Internal problem ID [10359]
Internal file name [OUTPUT/9307_Monday_June_06_2022_01_50_57_PM_28947109/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: 30.
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]

\[ \boxed {y^{\prime }-a \,x^{n} y^{2}-b \,x^{m} y=x^{m} b c -a \,c^{2} x^{n}} \]

2.30.1 Solving as riccati ode

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

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

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

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

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

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

Summary

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

Verification of solutions

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

2.30.2 Maple step by step solution

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

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*x^m*b+n)*(diff(y(x), x))/x+a*x^n*c*(a*x^n*c-b*x^m)*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 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Kummer 
            -> 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)-(a*x^n*y(x)^2+y(x)+b*x^m*y(x)*x+x^2*(x^m*b*c-a*c^2*x^n))/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)] 
   <- symmetry pattern of the form [0, F(x)*G(y)] successful 
   <- Riccati with symmetry pattern of the form [0,F(x)*G(y)] successful`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 98

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

\[ \frac {a \left (c +y \left (x \right )\right ) \left (\int _{}^{x}{\mathrm e}^{-\frac {2 \left (-\frac {b \left (n +1\right ) \textit {\_a}^{m}}{2}+a \,\textit {\_a}^{n} c \left (1+m \right )\right ) \textit {\_a}}{\left (1+m \right ) \left (n +1\right )}} \textit {\_a}^{n}d \textit {\_a} \right )+c_{1} y \left (x \right )+c_{1} c +{\mathrm e}^{-\frac {2 \left (-\frac {b \left (n +1\right ) x^{m}}{2}+a \,x^{n} c \left (1+m \right )\right ) x}{\left (1+m \right ) \left (n +1\right )}}}{c +y \left (x \right )} = 0 \]

✓ Solution by Mathematica

Time used: 3.436 (sec). Leaf size: 286

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

\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {e^{\frac {b x^{m+1}}{m+1}-\frac {2 a c x^{n+1}}{n+1}}}{a b (m-n) (c+K[2])^2}-\int _1^x\left (-\frac {\exp \left (\frac {b K[1]^{m+1}}{m+1}-\frac {2 a c K[1]^{n+1}}{n+1}\right ) K[1]^n}{b (m-n) (c+K[2])}-\frac {\exp \left (\frac {b K[1]^{m+1}}{m+1}-\frac {2 a c K[1]^{n+1}}{n+1}\right ) \left (-b K[1]^m+a c K[1]^n-a K[2] K[1]^n\right )}{a b (m-n) (c+K[2])^2}\right )dK[1]\right )dK[2]+\int _1^x\frac {\exp \left (\frac {b K[1]^{m+1}}{m+1}-\frac {2 a c K[1]^{n+1}}{n+1}\right ) \left (-b K[1]^m+a c K[1]^n-a y(x) K[1]^n\right )}{a b (m-n) (c+y(x))}dK[1]=c_1,y(x)\right ] \]