2.2.66 Problem 69

Internal problem ID [13272]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power Functions
Problem number : 69
Date solved : Sunday, January 18, 2026 at 07:04:07 PM
CAS classification : [[_homogeneous, `class D`], _rational, _Riccati]

2.2.66.1 Solved using first_order_ode_homog_type_D2

0.349 (sec)

Entering first order ode homog type D2 solver

\begin{align*} \left (a \,x^{2}+b x +e \right ) \left (y^{\prime } x -y\right )-y^{2}+x^{2}&=0 \\ \end{align*}

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} \left (a \,x^{2}+b x +e \right ) \left (\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x -u \left (x \right ) x \right )-u \left (x \right )^{2} x^{2}+x^{2} = 0 \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {\left (u \left (x \right )-1\right ) \left (u \left (x \right )+1\right )}{a \,x^{2}+b x +e} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (u \left (x \right )-1\right ) \left (u \left (x \right )+1\right )}{a \,x^{2}+b x +e}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{a \,x^{2}+b x +e}\\ g(u) &= \left (u -1\right ) \left (u +1\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{\left (u -1\right ) \left (u +1\right )}\,du} &= \int { \frac {1}{a \,x^{2}+b x +e} \,dx} \\ \end{align*}

\[ -\frac {\ln \left (u \left (x \right )+1\right )}{2}+\frac {\ln \left (u \left (x \right )-1\right )}{2}=\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (u -1\right ) \left (u +1\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=-1\\ u \left (x \right )&=1 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {\ln \left (u \left (x \right )+1\right )}{2}+\frac {\ln \left (u \left (x \right )-1\right )}{2} &= \frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}+c_1 \\ u \left (x \right ) &= -1 \\ u \left (x \right ) &= 1 \\ \end{align*}

Converting \(-\frac {\ln \left (u \left (x \right )+1\right )}{2}+\frac {\ln \left (u \left (x \right )-1\right )}{2} = \frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}+c_1\) back to \(y\) gives

\begin{align*} -\frac {\ln \left (\frac {y}{x}+1\right )}{2}+\frac {\ln \left (\frac {y}{x}-1\right )}{2} = \frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}+c_1 \end{align*}

Converting \(u \left (x \right ) = -1\) back to \(y\) gives

\begin{align*} y = -x \end{align*}

Converting \(u \left (x \right ) = 1\) back to \(y\) gives

\begin{align*} y = x \end{align*}

Simplifying the above gives

\begin{align*} -\frac {\ln \left (\frac {y+x}{x}\right )}{2}+\frac {\ln \left (\frac {y-x}{x}\right )}{2} &= \frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}+c_1 \\ y &= -x \\ y &= x \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= x \\ y &= \frac {x \left (1+{\mathrm e}^{-\frac {2 \left (c_1 \sqrt {4 a e -b^{2}}+2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )\right )}{\sqrt {4 a e -b^{2}}}}\right )}{-1+{\mathrm e}^{-\frac {2 \left (c_1 \sqrt {4 a e -b^{2}}+2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )\right )}{\sqrt {4 a e -b^{2}}}}} \\ y &= -x \\ \end{align*}

Summary of solutions found

\begin{align*} y &= x \\ y &= \frac {x \left (1+{\mathrm e}^{-\frac {2 \left (c_1 \sqrt {4 a e -b^{2}}+2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )\right )}{\sqrt {4 a e -b^{2}}}}\right )}{-1+{\mathrm e}^{-\frac {2 \left (c_1 \sqrt {4 a e -b^{2}}+2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )\right )}{\sqrt {4 a e -b^{2}}}}} \\ y &= -x \\ \end{align*}

2.2.66.2 Solved using first_order_ode_riccati

0.669 (sec)

Entering first order ode riccati solver

\begin{align*} \left (a \,x^{2}+b x +e \right ) \left (y^{\prime } x -y\right )-y^{2}+x^{2}&=0 \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {a \,x^{2} y+b x y+y^{2}+y e -x^{2}}{\left (a \,x^{2}+b x +e \right ) x} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \frac {x a y}{a \,x^{2}+b x +e}+\frac {b y}{a \,x^{2}+b x +e}+\frac {y^{2}}{\left (a \,x^{2}+b x +e \right ) x}+\frac {y e}{\left (a \,x^{2}+b x +e \right ) x}-\frac {x}{a \,x^{2}+b x +e} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=-\frac {x}{a \,x^{2}+b x +e}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {1}{\left (a \,x^{2}+b x +e \right ) x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{\left (a \,x^{2}+b x +e \right ) x}} \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 {2 a x +b}{\left (a \,x^{2}+b x +e \right )^{2} x}-\frac {1}{\left (a \,x^{2}+b x +e \right ) x^{2}}\\ f_1 f_2 &=\frac {1}{\left (a \,x^{2}+b x +e \right ) x^{2}}\\ f_2^2 f_0 &=-\frac {1}{\left (a \,x^{2}+b x +e \right )^{3} x} \end{align*}

Substituting the above terms back in equation (2) gives

\[ \frac {u^{\prime \prime }\left (x \right )}{\left (a \,x^{2}+b x +e \right ) x}+\frac {\left (2 a x +b \right ) u^{\prime }\left (x \right )}{\left (a \,x^{2}+b x +e \right )^{2} x}-\frac {u \left (x \right )}{\left (a \,x^{2}+b x +e \right )^{3} x} = 0 \]

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

\begin{align*} \frac {\frac {d^{2}u}{d x^{2}}}{\left (a \,x^{2}+b x +e \right ) x}+\frac {\left (2 a x +b \right ) \left (\frac {d u}{d x}\right )}{\left (a \,x^{2}+b x +e \right )^{2} x}-\frac {u}{\left (a \,x^{2}+b x +e \right )^{3} x} = 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 )&=\frac {2 a x +b}{a \,x^{2}+b x +e}\\ q \left (x \right )&=-\frac {1}{\left (a \,x^{2}+b x +e \right )^{2}} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }u \left (\tau \right )\right )+q_{1} u \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\frac {d^{2}}{d x^{2}}\tau \left (x \right )+p \left (x \right ) \left (\frac {d}{d x}\tau \left (x \right )\right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

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

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int \frac {2 a x +b}{a \,x^{2}+b x +e}d x}d x\\ &= \int e^{-\ln \left (a \,x^{2}+b x +e \right )} \,dx\\ &= \int \frac {1}{a \,x^{2}+b x +e}d x\\ &= \frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\\ &= \frac {-\frac {1}{\left (a \,x^{2}+b x +e \right )^{2}}}{\frac {1}{\left (a \,x^{2}+b x +e \right )^{2}}}\\ &= -1\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+q_{1} u \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )-u \left (\tau \right )&=0 \end{align*}

The above ode is now solved for \(u \left (\tau \right )\).Entering second order linear constant coefficient ode solver

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A u''(\tau ) + B u'(\tau ) + C u(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=-1\). Let the solution be \(u \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }-{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ \lambda ^{2}-1 = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form. Using the quadratic formula the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=-1\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-1\right )}\\ &= \pm 1 \end{align*}

Hence

\begin{align*} \lambda _1 &= + 1 \\ \lambda _2 &= - 1 \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= 1 \\ \lambda _2 &= -1 \\ \end{align*}

Since the roots are distinct, the solution is

\begin{align*} u \left (\tau \right ) &= c_1 e^{\lambda _1 \tau } + c_2 e^{\lambda _2 \tau } \\ u \left (\tau \right ) &= c_1 e^{\left (1\right )\tau } +c_2 e^{\left (-1\right )\tau } \\ \end{align*}

Or

\[ u \left (\tau \right ) =c_1 \,{\mathrm e}^{\tau }+c_2 \,{\mathrm e}^{-\tau } \]

The above solution is now transformed back to \(u\) using (6) which results in

\[ u = c_1 \,{\mathrm e}^{\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_2 \,{\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}} \]

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {4 c_1 a \,{\mathrm e}^{\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}}{\left (4 a e -b^{2}\right ) \left (\frac {\left (2 a x +b \right )^{2}}{4 a e -b^{2}}+1\right )}-\frac {4 c_2 a \,{\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}}{\left (4 a e -b^{2}\right ) \left (\frac {\left (2 a x +b \right )^{2}}{4 a e -b^{2}}+1\right )} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{\frac {u}{\left (a \,x^{2}+b x +e \right ) x}} \\ y &= -\frac {\left (\frac {4 c_1 a \,{\mathrm e}^{\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}}{\left (4 a e -b^{2}\right ) \left (\frac {\left (2 a x +b \right )^{2}}{4 a e -b^{2}}+1\right )}-\frac {4 c_2 a \,{\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}}{\left (4 a e -b^{2}\right ) \left (\frac {\left (2 a x +b \right )^{2}}{4 a e -b^{2}}+1\right )}\right ) \left (a \,x^{2}+b x +e \right ) x}{c_1 \,{\mathrm e}^{\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_2 \,{\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (\frac {4 a \,{\mathrm e}^{\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}}{\left (4 a e -b^{2}\right ) \left (\frac {\left (2 a x +b \right )^{2}}{4 a e -b^{2}}+1\right )}-\frac {4 c_3 a \,{\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}}{\left (4 a e -b^{2}\right ) \left (\frac {\left (2 a x +b \right )^{2}}{4 a e -b^{2}}+1\right )}\right ) \left (a \,x^{2}+b x +e \right ) x}{{\mathrm e}^{\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_3 \,{\mathrm e}^{-\frac {2 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}} \]

Simplifying the above gives

\begin{align*} y &= \frac {\left (-{\mathrm e}^{\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_3 \right ) x}{{\mathrm e}^{\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_3} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\left (-{\mathrm e}^{\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_3 \right ) x}{{\mathrm e}^{\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+c_3} \\ \end{align*}

2.2.66.3 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.070 (sec)

Entering first order ode riccati guess solver

\begin{align*} \left (a \,x^{2}+b x +e \right ) \left (y^{\prime } x -y\right )-y^{2}+x^{2}&=0 \\ \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) & =-\frac {x}{a \,x^{2}+b x +e}\\ f_1(x) & =\frac {1}{x}\\ f_2(x) &=\frac {1}{\left (a \,x^{2}+b x +e \right ) x} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = -x \]

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

\[ y = \frac {x \left ({\mathrm e}^{-\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}-2 c_1 \right )}{{\mathrm e}^{-\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+2 c_1} \]

Summary of solutions found

\begin{align*} y &= \frac {x \left ({\mathrm e}^{-\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}-2 c_1 \right )}{{\mathrm e}^{-\frac {4 \arctan \left (\frac {2 a x +b}{\sqrt {4 a e -b^{2}}}\right )}{\sqrt {4 a e -b^{2}}}}+2 c_1} \\ \end{align*}

2.2.66.4 ✓ Maple. Time used: 0.005 (sec). Leaf size: 58

ode:=(a*x^2+b*x+e)*(x*diff(y(x),x)-y(x))-y(x)^2+x^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = -\tanh \left (\frac {c_1 \sqrt {4 e a -b^{2}}+2 \arctan \left (\frac {2 a x +b}{\sqrt {4 e a -b^{2}}}\right )}{\sqrt {4 e a -b^{2}}}\right ) x \]

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 homogeneous D 
<- homogeneous successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \,x^{2}+b x +e \right ) \left (x \left (\frac {d}{d x}y \left (x \right )\right )-y \left (x \right )\right )-y \left (x \right )^{2}+x^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) a \,x^{2}+y \left (x \right ) b x +y \left (x \right )^{2}+y \left (x \right ) e -x^{2}}{\left (a \,x^{2}+b x +e \right ) x} \end {array} \]

2.2.66.5 ✓ Mathematica. Time used: 0.162 (sec). Leaf size: 53

ode=(a*x^2+b*x+e)*(x*D[y[x],x]-y[x])-y[x]^2+x^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

2.2.66.6 ✗ Sympy

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

Timed Out
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0