Internal problem ID [10601]
Internal file name [OUTPUT/9549_Monday_June_06_2022_03_07_35_PM_92126063/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: 9.
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]
\[ \boxed {y^{\prime }-f \left (x \right ) y^{2}-g \left (x \right ) y=-a^{2} f \left (x \right )-a g \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f \left (x \right ) y^{2}+g \left (x \right ) y -a^{2} f \left (x \right )-a g \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = f \left (x \right ) y^{2}+g \left (x \right ) y -a^{2} f \left (x \right )-a g \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a^{2} f \left (x \right )-a g \left (x \right )\), \(f_1(x)=g \left (x \right )\) 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 &=g \left (x \right ) f \left (x \right )\\ f_2^2 f_0 &=f \left (x \right )^{2} \left (-a^{2} f \left (x \right )-a g \left (x \right )\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 )+g \left (x \right ) f \left (x \right )\right ) u^{\prime }\left (x \right )+f \left (x \right )^{2} \left (-a^{2} f \left (x \right )-a g \left (x \right )\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -\left (\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{1} \right ) {\mathrm e}^{-a \left (\int f \left (x \right )d x \right )} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = c_{2} f \left (x \right ) \left (-{\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x -a \left (\int f \left (x \right )d x \right )}+a \,{\mathrm e}^{-a \left (\int f \left (x \right )d x \right )} \left (\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{1} \right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (-{\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x -a \left (\int f \left (x \right )d x \right )}+a \,{\mathrm e}^{-a \left (\int f \left (x \right )d x \right )} \left (\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{1} \right )\right ) {\mathrm e}^{\int a f \left (x \right )d x}}{\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-{\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x}+a c_{3} +\left (\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x \right ) a}{\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-{\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x}+a c_{3} +\left (\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x \right ) a}{\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {-{\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x}+a c_{3} +\left (\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x \right ) a}{\int {\mathrm e}^{\int \left (2 a f \left (x \right )+g \left (x \right )\right )d x} f \left (x \right )d x +c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-f \left (x \right ) y^{2}-g \left (x \right ) y=-a^{2} f \left (x \right )-a g \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=f \left (x \right ) y^{2}+g \left (x \right ) y-a^{2} f \left (x \right )-a g \left (x \right ) \end {array} \]
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 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: 67
dsolve(diff(y(x),x)=f(x)*y(x)^2+g(x)*y(x)-a^2*f(x)-a*g(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-a \left (\int {\mathrm e}^{\int g \left (x \right )d x +2 a \left (\int f \left (x \right )d x \right )} f \left (x \right )d x \right )+c_{1} a +{\mathrm e}^{\int g \left (x \right )d x +2 a \left (\int f \left (x \right )d x \right )}}{-\left (\int {\mathrm e}^{\int g \left (x \right )d x +2 a \left (\int f \left (x \right )d x \right )} f \left (x \right )d x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 1.122 (sec). Leaf size: 201
DSolve[y'[x]==f[x]*y[x]^2+g[x]*y[x]-a^2*f[x]-a*g[x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x-\frac {\exp \left (-\int _1^{K[2]}(-2 a f(K[1])-g(K[1]))dK[1]\right ) (a f(K[2])+y(x) f(K[2])+g(K[2]))}{a-y(x)}dK[2]+\int _1^{y(x)}\left (-\int _1^x\left (-\frac {\exp \left (-\int _1^{K[2]}(-2 a f(K[1])-g(K[1]))dK[1]\right ) f(K[2])}{a-K[3]}-\frac {\exp \left (-\int _1^{K[2]}(-2 a f(K[1])-g(K[1]))dK[1]\right ) (a f(K[2])+K[3] f(K[2])+g(K[2]))}{(a-K[3])^2}\right )dK[2]-\frac {\exp \left (-\int _1^x(-2 a f(K[1])-g(K[1]))dK[1]\right )}{(K[3]-a)^2}\right )dK[3]=c_1,y(x)\right ] \]