19.1 problem 1

Internal problem ID [10593]
Internal file name [OUTPUT/9541_Monday_June_06_2022_03_07_16_PM_10349265/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: 1.
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]

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

19.1.1 Solving as riccati ode

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

This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+f \left (x \right ) y -a^{2}-a f \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a^{2}-a f \left (x \right )\), \(f_1(x)=f \left (x \right )\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=f \left (x \right )\\ f_2^2 f_0 &=-a^{2}-a f \left (x \right ) \end {align*}

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

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

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

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

Summary

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

Verification of solutions

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

19.1.2 Maple step by step solution

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

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: 
   <- Riccati particular case Kamke (b) successful`
 

✓ Solution by Maple

Time used: 0.0 (sec). Leaf size: 52

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

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

✓ Solution by Mathematica

Time used: 0.719 (sec). Leaf size: 166

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

\[ \text {Solve}\left [\int _1^x\frac {\exp \left (-\int _1^{K[2]}(-2 a-f(K[1]))dK[1]\right ) (a+f(K[2])+y(x))}{a-y(x)}dK[2]+\int _1^{y(x)}\left (\frac {\exp \left (-\int _1^x(-2 a-f(K[1]))dK[1]\right )}{(K[3]-a)^2}-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}(-2 a-f(K[1]))dK[1]\right ) (a+f(K[2])+K[3])}{(a-K[3])^2}+\frac {\exp \left (-\int _1^{K[2]}(-2 a-f(K[1]))dK[1]\right )}{a-K[3]}\right )dK[2]\right )dK[3]=c_1,y(x)\right ] \]