2.2.28 Problem 31

Internal problem ID [13234]
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 : 31
Date solved : Wednesday, December 31, 2025 at 12:18:29 PM
CAS classification : [_Riccati]

2.2.28.1 Solved using first_order_ode_riccati

12.371 (sec)

Entering first order ode riccati solver

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

In canonical form the ODE is

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

Substituting the above terms back in equation (2) gives

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

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

Where

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

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\left (a b \,x^{m +n}+x^{n} a c -\frac {n}{x}\right ) \xi \left (x \right )\right )' + \left (b \,x^{m +n -1} m a \xi \left (x \right )\right ) &= 0\\ \frac {-b a \left (\left (\frac {d}{d x}\xi \left (x \right )\right ) x +n \xi \left (x \right )\right ) x^{m +n +1}-a c \left (\left (\frac {d}{d x}\xi \left (x \right )\right ) x +n \xi \left (x \right )\right ) x^{n +1}+\left (\frac {d^{2}}{d x^{2}}\xi \left (x \right )\right ) x^{2}+n \left (\left (\frac {d}{d x}\xi \left (x \right )\right ) x -\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 {-b a \left (\xi ^{\prime } x +n \xi \right ) x^{m +n +1}-a c \left (\xi ^{\prime } x +n \xi \right ) x^{n +1}+\xi ^{\prime \prime } x^{2}+n \left (\xi ^{\prime } x -\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 )&=-a b \,x^{m +n}-x^{n} a c +\frac {n}{x}\\ q \left (x \right )&=-n \left (a b \,x^{m +n -1}+a c \,x^{n -1}+\frac {1}{x^{2}}\right ) \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 (-a b \,x^{m +n}-x^{n} a c +\frac {n}{x}\right )}{x}-n \left (a b \,x^{m +n -1}+a c \,x^{n -1}+\frac {1}{x^{2}}\right )&=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}-a b \,x^{m +n}-x^{n} a c \right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\left (-\frac {n}{x}-a b \,x^{m +n}-x^{n} a c \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}-a b \,x^{m +n}-x^{n} a c \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^{m +n +1}+a \,x^{n +1} c +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^{m +n +1}+a \,x^{n +1} c +n}{x}d x}\\ &= x^{-n} {\mathrm e}^{-\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} \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 \,x^{-n} {\mathrm e}^{-\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,x^{-n} {\mathrm e}^{-\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}&= \int {0 \,dx} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(x^{-n} {\mathrm e}^{-\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}\) gives the final solution

\[ u \left (x \right ) = x^{n} {\mathrm e}^{\frac {a \,x^{n +1} \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right )}{\left (n +1\right ) \left (m +n +1\right )}} c_1 \]

Simplifying the above gives

\begin{align*} u \left (x \right ) &= x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} 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 x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2 \end{align*}

Hence

\begin{align*} \xi &= v \left (x \right ) x^{n}\\ &= \left (\int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2 \right ) x^{-n}\\ &= \left (c_1 \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}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 (a b \,x^{m +n}+x^{n} a c -\frac {n}{x}-\frac {\left (x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} c_1 \,x^{-n}-\frac {\left (\int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2 \right ) x^{-n} n}{x}\right ) x^{n}}{\int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}} 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 {\left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {\left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2}d x}\\ &= {\mathrm e}^{\int \frac {\left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2}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 {\left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2}d x}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\int \frac {\left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2}d x}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\int \frac {\left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2}d x}\) gives the final solution

\[ u = {\mathrm e}^{\int -\frac {\left (\int c_1 a \,x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x \left (b \,x^{m}+c \right )-{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{\int x^{n} {\mathrm e}^{\frac {a \,x^{n +1} \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right )}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2}d x} c_3 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} u &= {\mathrm e}^{\int -\frac {\left (\int c_1 a \,x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x \left (b \,x^{m}+c \right )-{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{\int x^{n} {\mathrm e}^{\frac {a \,x^{n +1} \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right )}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2}d x} c_3 \\ \end{align*}

The constants can be merged to give

\[ u = {\mathrm e}^{\int -\frac {\left (\int c_1 a \,x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x \left (b \,x^{m}+c \right )-{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{\int x^{n} {\mathrm e}^{\frac {a \,x^{n +1} \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right )}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2}d x} \]

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = -\frac {{\mathrm e}^{\int -\frac {\left (\int c_1 a \,x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x \left (b \,x^{m}+c \right )-{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{\int x^{n} {\mathrm e}^{\frac {a \,x^{n +1} \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right )}{\left (n +1\right ) \left (m +n +1\right )}} c_1 d x +c_2}d x} \left (a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )\right ) x^{n}}{c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2} \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 {a c_1 \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}} c_1 +a c_2 \left (b \,x^{m}+c \right )}{\left (c_1 \int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_2 \right ) a} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = \frac {a \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}+a c_4 \left (b \,x^{m}+c \right )}{\left (\int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_4 \right ) a} \]

Summary of solutions found

\begin{align*} y &= \frac {a \left (b \,x^{m}+c \right ) \int x^{n} {\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}d x -{\mathrm e}^{\frac {a \left (b \left (n +1\right ) x^{m +n +1}+x^{n +1} c \left (m +n +1\right )\right )}{\left (m +n +1\right ) \left (n +1\right )}}+a c_4 \left (b \,x^{m}+c \right )}{\left (\int x^{n} {\mathrm e}^{\frac {a x \left (b \left (n +1\right ) x^{m}+c \left (m +n +1\right )\right ) x^{n}}{\left (n +1\right ) \left (m +n +1\right )}}d x +c_4 \right ) a} \\ \end{align*}

2.2.28.2 ✗ Maple

ode:=diff(y(x),x) = a*x^n*y(x)^2-a*x^n*(b*x^m+c)*y(x)+b*m*x^(m-1); 
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+m)*a*b*x+x^n* 
a*c*x-n)/x*diff(y(x),x)-a*x^n*b*m*x^(m-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 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Kummer 
            -> hyper3: Equivalence to 1F1 under a power @ Moebius 
         -> hypergeometric 
            -> heuristic approach 
            -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
         -> Mathieu 
            -> Equivalence to the rational form of Mathieu ODE under a power @\ 
 Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      -> 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 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            -> Whittaker 
               -> hyper3: Equivalence to 1F1 under a power @ Moebius 
            -> hypergeometric 
               -> heuristic approach 
               -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebi\ 
us 
            -> Mathieu 
               -> Equivalence to the rational form of Mathieu ODE under a powe\ 
r @ Moebius 
         -> Heun: Equivalence to the GHE or one of its 4 confluent cases under \ 
a power @ Moebius 
         -> 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)-(a*x^n*y(x)^2+y(x)+(-x^(n+m)*a 
*b-x^n*a*c)*y(x)*x+x^2*b*m*x^(m-1))/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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \,x^{13234} y \left (x \right )^{2}-a \,x^{13234} \left (b \,x^{m}+c \right ) y \left (x \right )+b m \,x^{m -1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=a \,x^{13234} y \left (x \right )^{2}-a \,x^{13234} \left (b \,x^{m}+c \right ) y \left (x \right )+b m \,x^{m -1} \end {array} \]

2.2.28.3 ✓ Mathematica. Time used: 47.552 (sec). Leaf size: 353

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

\begin{align*} y(x)&\to \frac {b m \left (b x^m+c\right )^2 \left (1+c_1 \int _1^x\frac {\exp \left (a K[1]^{n+1} \left (\frac {b K[1]^m}{m+n+1}+\frac {c}{n+1}\right )\right ) K[1]^{m-1}}{\left (b K[1]^m+c\right )^2}dK[1]\right )}{b c_1 m \left (b x^m+c\right ) \int _1^x\frac {\exp \left (a K[1]^{n+1} \left (\frac {b K[1]^m}{m+n+1}+\frac {c}{n+1}\right )\right ) K[1]^{m-1}}{\left (b K[1]^m+c\right )^2}dK[1]+c_1 e^{a x^{n+1} \left (\frac {b x^m}{m+n+1}+\frac {c}{n+1}\right )}+b^2 m x^m+b c m}\\ y(x)&\to \frac {b m \left (b x^m+c\right )^2 \int _1^x\frac {\exp \left (a K[1]^{n+1} \left (\frac {b K[1]^m}{m+n+1}+\frac {c}{n+1}\right )\right ) K[1]^{m-1}}{\left (b K[1]^m+c\right )^2}dK[1]}{b m \left (b x^m+c\right ) \int _1^x\frac {\exp \left (a K[1]^{n+1} \left (\frac {b K[1]^m}{m+n+1}+\frac {c}{n+1}\right )\right ) K[1]^{m-1}}{\left (b K[1]^m+c\right )^2}dK[1]+e^{a x^{n+1} \left (\frac {b x^m}{m+n+1}+\frac {c}{n+1}\right )}} \end{align*}

2.2.28.4 ✗ Sympy

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

NotImplementedError : The given ODE a*b*x**(m + n)*y(x) + a*c*x**n*y(x) - a*x**n*y(x)**2 - b*m*x**(m - 1) + Derivative(y(x), x) cannot be solved by the factorable group method