Internal problem ID [10629]
Internal file name [OUTPUT/9577_Monday_June_06_2022_03_10_45_PM_76170662/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: 37.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-g \left (x \right ) \left (y-f \left (x \right )\right )^{2}=f^{\prime }\left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f \left (x \right )^{2} g \left (x \right )-2 f \left (x \right ) g \left (x \right ) y +g \left (x \right ) y^{2}+f^{\prime }\left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = f \left (x \right )^{2} g \left (x \right )-2 f \left (x \right ) g \left (x \right ) y +g \left (x \right ) y^{2}+f^{\prime }\left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=f \left (x \right )^{2} g \left (x \right )+f^{\prime }\left (x \right )\), \(f_1(x)=-2 f \left (x \right ) g \left (x \right )\) and \(f_2(x)=g \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{g \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' &=g^{\prime }\left (x \right )\\ f_1 f_2 &=-2 f \left (x \right ) g \left (x \right )^{2}\\ f_2^2 f_0 &=g \left (x \right )^{2} \left (f \left (x \right )^{2} g \left (x \right )+f^{\prime }\left (x \right )\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} g \left (x \right ) u^{\prime \prime }\left (x \right )-\left (g^{\prime }\left (x \right )-2 f \left (x \right ) g \left (x \right )^{2}\right ) u^{\prime }\left (x \right )+g \left (x \right )^{2} \left (f \left (x \right )^{2} g \left (x \right )+f^{\prime }\left (x \right )\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (c_{1} +\int g \left (x \right )d x \right ) {\mathrm e}^{-\left (\int f \left (x \right ) g \left (x \right )d x \right )} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = g \left (x \right ) {\mathrm e}^{-\left (\int f \left (x \right ) g \left (x \right )d x \right )} c_{2} \left (1-f \left (x \right ) \left (\int g \left (x \right )d x \right )-c_{1} f \left (x \right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {1-f \left (x \right ) \left (\int g \left (x \right )d x \right )-c_{1} f \left (x \right )}{c_{1} +\int g \left (x \right )d x} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-1+f \left (x \right ) \left (\int g \left (x \right )d x \right )+c_{3} f \left (x \right )}{c_{3} +\int g \left (x \right )d x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-1+f \left (x \right ) \left (\int g \left (x \right )d x \right )+c_{3} f \left (x \right )}{c_{3} +\int g \left (x \right )d x} \\ \end{align*}
Verification of solutions
\[ y = \frac {-1+f \left (x \right ) \left (\int g \left (x \right )d x \right )+c_{3} f \left (x \right )}{c_{3} +\int g \left (x \right )d x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-g \left (x \right ) \left (y-f \left (x \right )\right )^{2}=f^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=g \left (x \right ) \left (y-f \left (x \right )\right )^{2}+f^{\prime }\left (x \right ) \end {array} \]
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 (d) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 17
dsolve(diff(y(x),x)=g(x)*(y(x)-f(x))^2+diff(f(x),x),y(x), singsol=all)
\[ y \left (x \right ) = f \left (x \right )+\frac {1}{c_{1} -\left (\int g \left (x \right )d x \right )} \]
✓ Solution by Mathematica
Time used: 0.35 (sec). Leaf size: 31
DSolve[y'[x]==g[x]*(y[x]-f[x])^2+f'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to f(x)+\frac {1}{-\int _1^xg(K[2])dK[2]+c_1} \\ y(x)\to f(x) \\ \end{align*}