19.13 problem 13

Internal problem ID [10605]
Internal file name [OUTPUT/9553_Monday_June_06_2022_03_08_39_PM_4505664/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: 13.
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]

\[ \boxed {y^{\prime }-f \left (x \right ) y^{2}+a \,{\mathrm e}^{\lambda x} f \left (x \right ) y=a \lambda \,{\mathrm e}^{\lambda x}} \]

19.13.1 Solving as riccati ode

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

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

Substituting the above terms back in equation (2) gives \begin {align*} f \left (x \right ) u^{\prime \prime }\left (x \right )-\left (f^{\prime }\left (x \right )-{\mathrm e}^{\lambda x} f \left (x \right )^{2} a \right ) u^{\prime }\left (x \right )+\lambda \,{\mathrm e}^{\lambda x} f \left (x \right )^{2} a u \left (x \right ) &=0 \end {align*}

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

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

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

Summary

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

Verification of solutions

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

19.13.2 Maple step by step solution

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

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) = -(exp(lambda*x)*f(x)^2*a-(diff(f(x), x)))*(diff(y(x), x))/f(x)-exp(lam 
      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 
         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 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 
      <- 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 
            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 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 
         <- 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)-(f(x)*y(x)^2+y(x)-a*exp(lambda*x)*f(x)*y(x)*x+x^2*a*lambda*exp(lambda*x))/x, y(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: 
         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 
`, `-> Computing symmetries using: way = 6`
 

✗ Solution by Maple

dsolve(diff(y(x),x)=f(x)*y(x)^2-a*exp(lambda*x)*f(x)*y(x)+a*lambda*exp(lambda*x),y(x), singsol=all)
 

\[ \text {No solution found} \]

✓ Solution by Mathematica

Time used: 48.456 (sec). Leaf size: 207

DSolve[y'[x]==f[x]*y[x]^2-a*Exp[\[Lambda]*x]*f[x]*y[x]+a*\[Lambda]*Exp[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {a \exp \left (\int _1^{e^{x \lambda }}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]+2 \lambda x\right ) \left (\int _1^{e^{x \lambda }}\frac {\exp \left (-\int _1^{K[2]}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]\right )}{K[2]^2}dK[2]+c_1\right )}{\exp \left (\int _1^{e^{x \lambda }}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]+\lambda x\right ) \int _1^{e^{x \lambda }}\frac {\exp \left (-\int _1^{K[2]}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]\right )}{K[2]^2}dK[2]+c_1 \exp \left (\int _1^{e^{x \lambda }}-\frac {a f\left (\frac {\log (K[1])}{\lambda }\right )}{\lambda }dK[1]+\lambda x\right )+1} \\ y(x)\to a e^{\lambda x} \\ \end{align*}