20.8 problem 41

20.8.1 Solving as riccati ode
20.8.2 Maple step by step solution

Internal problem ID [10633]
Internal file name [OUTPUT/9581_Monday_June_06_2022_03_10_54_PM_90496217/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 1.2.8-2. Equations containing arbitrary functions and their derivatives.
Problem number: 41.
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]

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

20.8.1 Solving as riccati ode

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

This is a Riccati ODE. Comparing the ODE to solve \[ y' = f \left (x \right ) y^{2}+g^{\prime }\left (x \right ) y +a f \left (x \right ) {\mathrm e}^{2 g \left (x \right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a f \left (x \right ) {\mathrm e}^{2 g \left (x \right )}\), \(f_1(x)=g^{\prime }\left (x \right )\) 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 &=f \left (x \right ) g^{\prime }\left (x \right )\\ f_2^2 f_0 &=f \left (x \right )^{3} a \,{\mathrm e}^{2 g \left (x \right )} \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 )+f \left (x \right ) g^{\prime }\left (x \right )\right ) u^{\prime }\left (x \right )+f \left (x \right )^{3} a \,{\mathrm e}^{2 g \left (x \right )} u \left (x \right ) &=0 \end {align*}

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

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

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

Summary

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

Verification of solutions

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

20.8.2 Maple step by step solution

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

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 
<- Chini successful`
 

Solution by Maple

Time used: 0.031 (sec). Leaf size: 28

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

\[ y \left (x \right ) = -\tan \left (-\sqrt {a}\, \left (\int f \left (x \right ) {\mathrm e}^{g \left (x \right )}d x \right )+c_{1} \right ) \sqrt {a}\, {\mathrm e}^{g \left (x \right )} \]

Solution by Mathematica

Time used: 0.635 (sec). Leaf size: 41

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

\[ y(x)\to \sqrt {a} e^{g(x)} \tan \left (\sqrt {a} \int _1^xe^{g(K[1])} f(K[1])dK[1]+c_1\right ) \]