2.8.3 Problem 12

Internal problem ID [13355]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-2
Problem number : 12
Date solved : Sunday, January 18, 2026 at 07:29:10 PM
CAS classification : [_Riccati]

2.8.3.1 Solved using first_order_ode_riccati

1.982 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime }&=a \,x^{n} y^{2}-a b \,x^{n +1} \ln \left (x \right ) y+b \ln \left (x \right )+b \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= a \,x^{n} y^{2}-a b \,x^{n +1} \ln \left (x \right ) y+b \ln \left (x \right )+b \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

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

With Riccati ODE standard form

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

Shows that \(f_0(x)=b \ln \left (x \right )+b\), \(f_1(x)=-a b \,x^{n} x \ln \left (x \right )\) and \(f_2(x)=x^{n} a\). Let

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

Substituting the above terms back in equation (2) gives

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

Entering second order ode lagrange adjoint equation method solverIn normal form the ode

\begin{align*} x^{n} a \left (\frac {d^{2}u}{d x^{2}}\right )-\left (\frac {a n \,x^{n}}{x}-a^{2} b \,x^{2 n} x \ln \left (x \right )\right ) \left (\frac {d u}{d x}\right )+x^{2 n} a^{2} \left (b \ln \left (x \right )+b \right ) 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&=r \left (x \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {a b \,x^{n +2} \ln \left (x \right )-n}{x}\\ q \left (x \right )&=a b \,x^{n} \left (\ln \left (x \right )+1\right )\\ r \left (x \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {\left (a b \,x^{n +2} \ln \left (x \right )-n \right ) \xi \left (x \right )}{x}\right )' + \left (a b \,x^{n} \left (\ln \left (x \right )+1\right ) \xi \left (x \right )\right ) &= 0\\ \frac {\left (\frac {d^{2}}{d x^{2}}\xi \left (x \right )\right ) x^{2}-\ln \left (x \right ) x^{n +2} \xi \left (x \right ) a b n -a b \,x^{n +3} \ln \left (x \right ) \left (\frac {d}{d x}\xi \left (x \right )\right )+n \left (x \left (\frac {d}{d x}\xi \left (x \right )\right )-\xi \left (x \right )\right )}{x^{2}}&= 0 \end{align*}

Which is solved for \(\xi (x)\). Entering second order change of variable on \(y\) method 2 solverIn normal form the ode

\begin{align*} \frac {\xi ^{\prime \prime } x^{2}-\ln \left (x \right ) x^{n +2} \xi a b n -a b \,x^{n +3} \ln \left (x \right ) \xi ^{\prime }+n \left (x \xi ^{\prime }-\xi \right )}{x^{2}} = 0\tag {1} \end{align*}

Becomes

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

Where

\begin{align*} p \left (x \right )&=-x^{n +1} \ln \left (x \right ) a b +\frac {n}{x}\\ q \left (x \right )&=-x^{n} \ln \left (x \right ) a b n -\frac {n}{x^{2}} \end{align*}

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

\begin{align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \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 (-x^{n +1} \ln \left (x \right ) a b +\frac {n}{x}\right )}{x}-x^{n} \ln \left (x \right ) a b n -\frac {n}{x^{2}}&=0 \tag {5} \end{align*}

Solving (5) for \(n\) gives

\begin{align*} n&=-n \tag {6} \end{align*}

Substituting this value in (3) gives

\begin{align*} v^{\prime \prime }\left (x \right )+\left (-\frac {n}{x}-x^{n +1} \ln \left (x \right ) a b \right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (-\frac {n}{x}-x^{n +1} \ln \left (x \right ) a b \right ) v^{\prime }\left (x \right )&=0 \tag {7} \\ \end{align*}

Using the substitution

\begin{align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end{align*}

Then (7) becomes

\begin{align*} u^{\prime }\left (x \right )+\left (-\frac {n}{x}-x^{n +1} \ln \left (x \right ) a b \right ) u \left (x \right ) = 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*} u^{\prime }\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 {a b \,x^{n +2} \ln \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 {a b \,x^{n +2} \ln \left (x \right )+n}{x}d x}\\ &= {\mathrm e}^{\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {-a b \,x^{n +2}-n^{2}-2 n}{n +2}} \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}^{\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {-a b \,x^{n +2}-n^{2}-2 n}{n +2}}\right ) &= 0 \end{align*}

Integrating gives

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

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {-a b \,x^{n +2}-n^{2}-2 n}{n +2}}\) gives the final solution

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

Simplifying the above gives

\begin{align*} u \left (x \right ) &= {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 \\ \end{align*}

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

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

Hence

\begin{align*} \xi &= v \left (x \right ) x^{n}\\ &= \left (\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2 \right ) x^{-n}\\ &= \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right ) x^{-n}\\ \end{align*}

The original ode now reduces to first order ode

\begin{align*} \xi \left (x \right ) u^{\prime }-u \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) u&=\int \xi \left (x \right ) r \left (x \right )d x\\ u^{\prime }+u \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

Or

\begin{align*} u^{\prime }+u \left (\frac {a b \,x^{n +2} \ln \left (x \right )-n}{x}-\frac {\left ({\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 \,x^{-n}-\frac {\left (\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2 \right ) x^{-n} n}{x}\right ) x^{n}}{\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2}\right )&=0 \end{align*}

Which is now a first order ode. This is now solved for \(u\). Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} u^{\prime } + q(x)u &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )}d x}\\ &= {\mathrm e}^{\int -\frac {-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )}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 {-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )}d x}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\int -\frac {-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )}d x}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\int -\frac {-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )}d x}\) gives the final solution

\[ u = {\mathrm e}^{\int \frac {\int -c_1 a b \,x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x \ln \left (x \right ) x^{n +2}+c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}-a b \,x^{n +2} \ln \left (x \right ) c_2}{\left (\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2 \right ) x}d x} c_3 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} u &= {\mathrm e}^{\int \frac {\int -c_1 a b \,x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x \ln \left (x \right ) x^{n +2}+c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}-a b \,x^{n +2} \ln \left (x \right ) c_2}{\left (\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2 \right ) x}d x} c_3 \\ \end{align*}

The constants can be merged to give

\[ u = {\mathrm e}^{\int \frac {\int -c_1 a b \,x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x \ln \left (x \right ) x^{n +2}+c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}-a b \,x^{n +2} \ln \left (x \right ) c_2}{\left (\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2 \right ) x}d x} \]

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {{\mathrm e}^{\int \frac {\int -c_1 a b \,x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x \ln \left (x \right ) x^{n +2}+c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}-a b \,x^{n +2} \ln \left (x \right ) c_2}{\left (\int {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}} x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} c_1 d x +c_2 \right ) x}d x} \left (-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}\right )}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right )} \end{equation}

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

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u \,x^{n} a} \\ y &= -\frac {\left (-\ln \left (x \right ) x^{n +2} c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x a b -a b \,x^{n +2} \ln \left (x \right ) c_2 +c_1 \,x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}\right ) x^{-n}}{x \left (c_1 \int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_2 \right ) a} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\left (-\int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x x^{n +2} \ln \left (x \right ) a b -b \ln \left (x \right ) a \,x^{n +2} c_4 +x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}\right ) x^{-n}}{\left (\int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_4 \right ) x a} \]

Summary of solutions found

\begin{align*} y &= -\frac {\left (-\int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x x^{n +2} \ln \left (x \right ) a b -b \ln \left (x \right ) a \,x^{n +2} c_4 +x^{\frac {a b \,x^{n +2}+n^{2}+3 n +2}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}\right ) x^{-n}}{\left (\int x^{\frac {a b \,x^{n +2}+n^{2}+2 n}{n +2}} {\mathrm e}^{-\frac {x^{n +2} a b}{\left (n +2\right )^{2}}}d x +c_4 \right ) x a} \\ \end{align*}

2.8.3.2 ✗ Maple

ode:=diff(y(x),x) = a*x^n*y(x)^2-a*b*x^(n+1)*ln(x)*y(x)+b*ln(x)+b; 
dsolve(ode,y(x), singsol=all);
 

\[ \text {No solution found} \]

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) = -(x^(n+1)*ln(x)*a*b* 
x-n)/x*diff(y(x),x)-a*x^n*b*(ln(x)+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 
         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 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 
      <- 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 
            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 und\ 
er a power @ Moebius 
            -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = e\ 
xp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
               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 
         <- 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)-(a*x^n*y(x)^2+y(x)-a*b*x^(n+1) 
*ln(x)*y(x)*x+x^2*(b*ln(x)+b))/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
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \,x^{13355} y \left (x \right )^{2}-a b \,x^{13356} \ln \left (x \right ) y \left (x \right )+b \ln \left (x \right )+b \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \,x^{13355} y \left (x \right )^{2}-a b \,x^{13356} \ln \left (x \right ) y \left (x \right )+b \ln \left (x \right )+b \end {array} \]

2.8.3.3 ✗ Mathematica

ode=D[y[x],x]==a*x^n*y[x]^2-a*b*x^(n+1)*Log[x]*y[x]+b*Log[x]+b; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Not solved

2.8.3.4 ✗ Sympy

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

NotImplementedError : The given ODE a*b*x**(n + 1)*y(x)*log(x) - a*x**n*y(x)**2 - b*log(x) - b + Der
 

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

classify_ode(ode,func=y(x)) 
 
('lie_group',)