2.20.9 Problem 42

Internal problem ID [13489]
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 : 42
Date solved : Wednesday, December 31, 2025 at 09:51:48 PM
CAS classification : [_Riccati]

2.20.9.1 Solved using first_order_ode_riccati

12.247 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=y^{2}-\frac {f^{\prime \prime }\left (x \right )}{f \left (x \right )} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {f \left (x \right ) y^{2}-f^{\prime \prime }\left (x \right )}{f \left (x \right )} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \textit {the\_rhs} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=-\frac {f^{\prime \prime }\left (x \right )}{f \left (x \right )}\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &=-\frac {f^{\prime \prime }\left (x \right )}{f \left (x \right )} \end{align*}

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )-\frac {f^{\prime \prime }\left (x \right ) u \left (x \right )}{f \left (x \right )} = 0 \]

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} u \left (x \right ) = \left (\int \frac {1}{f \left (x \right )^{2}}d x c_1 +c_2 \right ) f \left (x \right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1}{f \left (x \right )}+\left (\int \frac {1}{f \left (x \right )^{2}}d x c_1 +c_2 \right ) f^{\prime }\left (x \right ) \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= -\frac {\frac {c_1}{f \left (x \right )}+\left (\int \frac {1}{f \left (x \right )^{2}}d x c_1 +c_2 \right ) f^{\prime }\left (x \right )}{\left (\int \frac {1}{f \left (x \right )^{2}}d x c_1 +c_2 \right ) f \left (x \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {1}{f \left (x \right )}+\left (\int \frac {1}{f \left (x \right )^{2}}d x +c_3 \right ) f^{\prime }\left (x \right )}{\left (\int \frac {1}{f \left (x \right )^{2}}d x +c_3 \right ) f \left (x \right )} \]

Summary of solutions found

\begin{align*} y &= -\frac {\frac {1}{f \left (x \right )}+\left (\int \frac {1}{f \left (x \right )^{2}}d x +c_3 \right ) f^{\prime }\left (x \right )}{\left (\int \frac {1}{f \left (x \right )^{2}}d x +c_3 \right ) f \left (x \right )} \\ \end{align*}

2.20.9.2 ✓ Maple. Time used: 0.001 (sec). Leaf size: 44

ode:=diff(y(x),x) = y(x)^2-diff(diff(f(x),x),x)/f(x); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {-\int \frac {1}{f \left (x \right )^{2}}d x f^{\prime }\left (x \right ) f \left (x \right )-c_1 f^{\prime }\left (x \right ) f \left (x \right )-1}{\left (\int \frac {1}{f \left (x \right )^{2}}d x +c_1 \right ) f \left (x \right )^{2}} \]

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 Special 
trying Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = diff(diff(f(x),x),x) 
/f(x)*y(x), y(x) 
      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      <- linear_1 successful 
   <- Riccati to 2nd Order successful
 

2.20.9.3 ✓ Mathematica. Time used: 0.157 (sec). Leaf size: 132

ode=D[y[x],x]==y[x]^2-D[ f[x],{x,2}]/f[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

2.20.9.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
f = Function("f") 
ode = Eq(-y(x)**2 + Derivative(y(x), x) + Derivative(f(x), (x, 2))/f(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -y(x)**2 + Derivative(y(x), x) + Derivative(f(x), (x, 2))/f(x) cannot be solved by the lie group method