2.15.3 Problem 12

Internal problem ID [13429]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.7-2. Equations containing arccosine.
Problem number : 12
Date solved : Wednesday, December 31, 2025 at 09:27:23 PM
CAS classification : [_Riccati]

2.15.3.1 Solved using first_order_ode_riccati

22.839 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=-\left (k +1\right ) x^{k} y^{2}+\lambda \arccos \left (x \right )^{n} \left (x^{k +1} y-1\right ) \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -x^{k} y^{2} k +\arccos \left (x \right )^{n} x^{k +1} \lambda y-x^{k} y^{2}-\arccos \left (x \right )^{n} \lambda \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 \(f_0(x)=-\arccos \left (x \right )^{n} \lambda \), \(f_1(x)=\lambda x \,x^{k} \arccos \left (x \right )^{n}\) and \(f_2(x)=-x^{k} k -x^{k}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u \left (-x^{k} k -x^{k}\right )} \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' &=-\frac {x^{k} k^{2}}{x}-\frac {x^{k} k}{x}\\ f_1 f_2 &=\lambda x \,x^{k} \arccos \left (x \right )^{n} \left (-x^{k} k -x^{k}\right )\\ f_2^2 f_0 &=-\left (-x^{k} k -x^{k}\right )^{2} \arccos \left (x \right )^{n} \lambda \end{align*}

Substituting the above terms back in equation (2) gives

\[ \left (-x^{k} k -x^{k}\right ) u^{\prime \prime }\left (x \right )-\left (-\frac {x^{k} k^{2}}{x}-\frac {x^{k} k}{x}+\lambda x \,x^{k} \arccos \left (x \right )^{n} \left (-x^{k} k -x^{k}\right )\right ) u^{\prime }\left (x \right )-\left (-x^{k} k -x^{k}\right )^{2} \arccos \left (x \right )^{n} \lambda u \left (x \right ) = 0 \]

Entering second order change of variable on \(y\) method 2 solverIn normal form the ode

\begin{align*} \left (-x^{k} k -x^{k}\right ) \left (\frac {d^{2}u}{d x^{2}}\right )-\left (-\frac {x^{k} k^{2}}{x}-\frac {x^{k} k}{x}+\lambda x \,x^{k} \arccos \left (x \right )^{n} \left (-x^{k} k -x^{k}\right )\right ) \left (\frac {d u}{d x}\right )-\left (-x^{k} k -x^{k}\right )^{2} \arccos \left (x \right )^{n} \lambda u = 0\tag {1} \end{align*}

Becomes

\begin{align*} \frac {d^{2}u}{d x^{2}}+p \left (x \right ) \left (\frac {d u}{d x}\right )+q \left (x \right ) u&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\arccos \left (x \right )^{n} x^{k +1} \lambda -\frac {k}{x}\\ q \left (x \right )&=x^{k} \lambda \left (k +1\right ) \arccos \left (x \right )^{n} \end{align*}

Applying change of variables on the depndent variable \(u = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(u\).

\begin{align*} \frac {d^{2}}{d x^{2}}v \left (x \right )+\left (\frac {2 n}{x}+p \right ) \left (\frac {d}{d x}v \left (x \right )\right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end{align*}

Let the coefficient of \(v \left (x \right )\) above be zero. Hence

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end{align*}

Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives

\begin{align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n \left (-\arccos \left (x \right )^{n} x^{k +1} \lambda -\frac {k}{x}\right )}{x}+x^{k} \lambda \left (k +1\right ) \arccos \left (x \right )^{n}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=k +1 \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} \frac {d^{2}}{d x^{2}}v \left (x \right )+\left (\frac {2 k +2}{x}-\arccos \left (x \right )^{n} x^{k +1} \lambda -\frac {k}{x}\right ) \left (\frac {d}{d x}v \left (x \right )\right )&=0 \\ \frac {d^{2}}{d x^{2}}v \left (x \right )+\frac {\left (k +2-x^{k +2} \lambda \arccos \left (x \right )^{n}\right ) \left (\frac {d}{d x}v \left (x \right )\right )}{x}&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = \frac {d}{d x}v \left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} \frac {d}{d x}u \left (x \right )+\frac {\left (k +2-x^{k +2} \lambda \arccos \left (x \right )^{n}\right ) u \left (x \right )}{x} = 0 \tag {8} \\ \end{align*}

The above is now solved for \(u \left (x \right )\). Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} \frac {d}{d x}u \left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}\\ &= {\mathrm e}^{\int -\frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,{\mathrm e}^{\int -\frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\int -\frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\int -\frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}\) gives the final solution

\[ u \left (x \right ) = {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 \]

Now that \(u \left (x \right )\) is known, then

\begin{align*} \frac {d}{d x}v \left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_2\\ &= \int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 d x +c_2 \end{align*}

Hence

\begin{align*} u&= v \left (x \right ) x^{n}\\ &= \left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 d x +c_2 \right ) x^{k +1}\\ &= x^{k +1} \left (c_1 \int {\mathrm e}^{\int \frac {\arccos \left (x \right )^{n} x^{k +1} \lambda x -k -2}{x}d x}d x +c_2 \right )\\ \end{align*}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 \,x^{k +1}+\frac {\left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 d x +c_2 \right ) x^{k +1} \left (k +1\right )}{x} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u \left (-x^{k} k -x^{k}\right )} \\ y &= -\frac {\left ({\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 \,x^{k +1}+\frac {\left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 d x +c_2 \right ) x^{k +1} \left (k +1\right )}{x}\right ) x^{-k -1}}{\left (-x^{k} k -x^{k}\right ) \left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} c_1 d x +c_2 \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left ({\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} x^{k +1}+\frac {\left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}d x +c_3 \right ) x^{k +1} \left (k +1\right )}{x}\right ) x^{-k -1}}{\left (-x^{k} k -x^{k}\right ) \left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}d x +c_3 \right )} \]

Summary of solutions found

\begin{align*} y &= -\frac {\left ({\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x} x^{k +1}+\frac {\left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}d x +c_3 \right ) x^{k +1} \left (k +1\right )}{x}\right ) x^{-k -1}}{\left (-x^{k} k -x^{k}\right ) \left (\int {\mathrm e}^{\int \frac {-k -2+x^{k +2} \lambda \arccos \left (x \right )^{n}}{x}d x}d x +c_3 \right )} \\ \end{align*}

2.15.3.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

14.230 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime }&=-\left (k +1\right ) x^{k} y^{2}+\lambda \arccos \left (x \right )^{n} \left (x^{k +1} y-1\right ) \\ \end{align*}

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

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

Shows that

\begin{align*} f_0(x) & =-\arccos \left (x \right )^{n} \lambda \\ f_1(x) & =\lambda x \,x^{k} \arccos \left (x \right )^{n}\\ f_2(x) &=-x^{k} k -x^{k} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = x^{-k -1} \]

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

\[ y = x^{-k -1}+\frac {{\mathrm e}^{\int \left (2 x^{-k -1} \left (-x^{k} k -x^{k}\right )+\lambda x \,x^{k} \arccos \left (x \right )^{n}\right )d x}}{c_1 -\int {\mathrm e}^{\int \left (2 x^{-k -1} \left (-x^{k} k -x^{k}\right )+\lambda x \,x^{k} \arccos \left (x \right )^{n}\right )d x} \left (-x^{k} k -x^{k}\right )d x} \]

Summary of solutions found

\begin{align*} y &= x^{-k -1}+\frac {{\mathrm e}^{\int \left (2 x^{-k -1} \left (-x^{k} k -x^{k}\right )+\lambda x \,x^{k} \arccos \left (x \right )^{n}\right )d x}}{c_1 -\int {\mathrm e}^{\int \left (2 x^{-k -1} \left (-x^{k} k -x^{k}\right )+\lambda x \,x^{k} \arccos \left (x \right )^{n}\right )d x} \left (-x^{k} k -x^{k}\right )d x} \\ \end{align*}

2.15.3.3 ✓ Maple. Time used: 0.013 (sec). Leaf size: 150

ode:=diff(y(x),x) = -(1+k)*x^k*y(x)^2+lambda*arccos(x)^n*(x^(1+k)*y(x)-1); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {x^{-k -1} \left (x^{k +1} {\mathrm e}^{\int \frac {\lambda \arccos \left (x \right )^{n} x^{k +1} x -2 k -2}{x}d x}+\left (k +1\right ) \int x^{k} {\mathrm e}^{\lambda \int x^{k +1} \arccos \left (x \right )^{n}d x -2 \left (k +1\right ) \int \frac {1}{x}d x}d x +c_1 \right )}{\int x^{k} {\mathrm e}^{\lambda \int x^{k +1} \arccos \left (x \right )^{n}d x -2 \left (k +1\right ) \int \frac {1}{x}d x}d x k +\int x^{k} {\mathrm e}^{\lambda \int x^{k +1} \arccos \left (x \right )^{n}d x -2 \left (k +1\right ) \int \frac {1}{x}d x}d x +c_1} \]

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: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE, diff(diff(y(x),x),x) = (lambda*arccos(x)^n* 
x^(k+1)*x+k)/x*diff(y(x),x)-lambda*arccos(x)^n*x^k*(k+1)*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)] 
      checking if the LODE is missing y 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int\ 
(r(x), dx)) * 2F1([a1, a2], [b1], f) 
      -> Trying changes of variables to rationalize or make the ODE simpler 
      <- unable to find a useful change of variables 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         trying 2nd order exact linear 
         trying symmetries linear in x and y(x) 
         trying to convert to a linear ODE with constant coefficients 
         trying 2nd order, integrating factor of the form mu(x,y) 
         trying a symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under \ 
a power @ Moebius 
         -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(\ 
int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
         -> Trying changes of variables to rationalize or make the ODE simpler 
         <- unable to find a useful change of variables 
            trying a symmetry of the form [xi=0, eta=F(x)] 
         trying to convert to an ODE of Bessel type 
   -> Trying a change of variables to reduce to Bernoulli 
   -> Calling odsolve with the ODE, diff(y(x),x)-((-x^k*k-x^k)*y(x)^2+y(x)+ 
lambda*arccos(x)^n*x^(k+1)*y(x)*x-x^2*lambda*arccos(x)^n)/x, y(x), explicit 
      *** Sublevel 2 *** 
      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: 
         trying Riccati_symmetries 
      trying inverse_Riccati 
      trying 1st order ODE linearizable_by_differentiation 
   -> trying a symmetry pattern of the form [F(x)*G(y), 0] 
   -> trying a symmetry pattern of the form [0, F(x)*G(y)] 
   -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] 
trying inverse_Riccati 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
   -> Computing symmetries using: way = 4 
   -> Computing symmetries using: way = 2 
   -> Computing symmetries using: way = 6 
[0, exp(-Int((x^(k+2)*lambda*arccos(x)^n-2*k-2)/x,x))*(y-1/(x^k)/x)^2] 
   <- successful computation of symmetries. 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\left (k +1\right ) x^{k} y \left (x \right )^{2}+\lambda \arccos \left (x \right )^{13429} \left (x^{k +1} y \left (x \right )-1\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\left (k +1\right ) x^{k} y \left (x \right )^{2}+\lambda \arccos \left (x \right )^{13429} \left (x^{k +1} y \left (x \right )-1\right ) \end {array} \]

2.15.3.4 ✗ Mathematica

ode=D[y[x],x]==-(k+1)*x^k*y[x]^2+\[Lambda]*ArcCos[x]^n*(x^(k+1)*y[x]-1); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.15.3.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
k = symbols("k") 
lambda_ = symbols("lambda_") 
n = symbols("n") 
y = Function("y") 
ode = Eq(-lambda_*(x**(k + 1)*y(x) - 1)*acos(x)**n + x**k*(k + 1)*y(x)**2 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out