Introduction

The aim of the first part of the article is to describe a unique reduction effect of the general Riccati-type differential equation [1,2,3,4]. The originality of this effect consists in the uniqueness of solution of respective differential equation. This is perhaps the most surprising property of the presented reduction. Let us get to the merits of the discussed problem. In the paper, we show that if the function \(y_1\) is any particular solution of the differential equation

$$L[y]=q(x)y^2+r(x),$$

where \(L[\cdot ]\) is some linear differential operatorFootnote 1 and \(q, r \in C(I)\), \(I\subset \mathbb {R}\) is an open interval, then for every \(N\in \mathbb {N}\), \(N\ge 3\), there exists a uniquely determined polynomial \(P\in \mathbb {R}[y]\), \(\deg P=N\), \(P(0)=P'(0)=P''(0)=0\), such that the linear substitution \(y=\frac{N}{2}y_1+z\) reduces the differential equation

$$L[y]=q(x)y^2+\frac{N}{2}r(x)+P(y)$$

to the following general Bernoulli-type equation of the form

$$L[z]-\frac{N}{N-1}q(x)y_1z=a_N\frac{q(x)}{y_1^{N-2}}z^N,$$

where \(a_N=\frac{(-1)^N}{N-1}\left( \frac{2}{N}\right) ^{N-1}\). We begin with a simple illustration of this fact.

Lemma 1

Assume that the function \(y_1\ne 0\) is a solution of the Bernoulli differential equation

$$y'+p(x)y=q(x)y^2.$$

Then, the substitution \(y=\frac{3}{2} y_1+z\) reduces the equation

$$y'+p(x)y=q(x)y^2-\frac{2}{9}\frac{q(x)}{y_1}y^3$$

to the Bernoulli equation.

Proof

We have

$$\begin{aligned} y'+p(x)y&=\frac{3}{2} y_1'+z'+\frac{3}{2} p(x)y_1+p(x)z,\\ q(x)y^2-\frac{2}{9}\frac{q(x)}{y_1}y^3&=q(x)\left( \frac{3}{2} y_1+z\right) ^2-\frac{2}{9} \frac{q(x)}{y_1(x)}\left( \frac{3}{2} y_1+z\right) ^3\\&=\frac{9}{4} q(x)y_1^2+3q(x)y_1z+q(x)z^2-\frac{3}{4} q(x)y_1^2-\frac{3}{2} q(x)y_1z-q(x)z^2-\frac{2}{9}\frac{q(x)}{y_1} z^3\\&=\frac{3}{2}q(x)y_1^2+\frac{3}{2} q(x)y_1z-\frac{2}{9} \frac{q(x)}{y_1}z^3, \end{aligned}$$

which implies

$$\begin{aligned} \frac{3}{2} y_1'+z'+\frac{3}{2} p(x)y_1+p(x)z=\frac{3}{2} q(x)y_1^2+\frac{3}{2} q(x)y_1z-\frac{2}{9} \frac{q(x)}{y_1}z^3. \end{aligned}$$
(1)

By the assumption, we have

$$y_1'+p(x)y_1=q(x)y_1^2,$$

therefore, from Eq. 1, we get the Bernoulli equation

$$z'+\left( p(x)-\frac{3}{2} q(x)y_1\right) z=-\frac{2}{9} \frac{q(x)}{y_1}z^3.$$

Example 1

Let us consider the following equation

$$\begin{aligned} y'+y=y^2-\frac{2}{9} y^3. \end{aligned}$$
(2)

Since the function \(y\equiv 1\) is a solution of the differential equation \(y'+y=y^2\), the substitution \(y=\frac{3}{2} +z\) reduces Eq. 2 to the form

$$z'-\frac{1}{2} z=-\frac{2}{9} z^3,$$

which possesses the solutions of the form

$$z=\pm \frac{3}{2} \sqrt{\frac{e^x}{e^x+C}},$$

where \(C\in [0,\infty ).\)

Generalization

The following generalization of Lemma 1 holds.

Theorem 1

Let \(y_1\) be a particular solution of the differential equation

$$y'+p(x)y=q(x)y^2,$$

where \(p,q\in C(I)\) and \(I\subset \mathbb {R}\) is a non-empty and non-degenerate interval. If \(y_1\ne 0\), then, for every \(N\in \mathbb {N}\), \(N\ge 3\), the substitution \(y=\frac{N}{2}y_1+z\) reduces the following differential equation

$$\begin{aligned} y'+p(x)y=\sum _{k=2}^Na_k\frac{q(x)}{y_1^{k-2}}y^k, \end{aligned}$$
(3)

where

$$\begin{aligned} a_{N-k}=\frac{(-1)^{N-k}}{N-1}\left( {\begin{array}{c}N\\ k\end{array}}\right) \left( \frac{2}{N}\right) ^{N-k-1}\qquad k=0,1,\ldots , N-2, \end{aligned}$$
(4)

to the Bernoulli equation of the form

$$z'+\left( p(x)-\frac{N}{N-1} q(x) y_1\right) z=a_N\frac{q(x)}{y_1^{N-2}}z^N.$$

Remark 1

We note that Eq. 4 implies the identity

$$\begin{aligned} a_{N-k}:=(-1)^k\left( {\begin{array}{c}N\\ k\end{array}}\right) \left( \frac{N}{2}\right) ^k a_N \end{aligned}$$
(5)

for each \(k=1,\ldots , N-2\), \(a_2:=1\).

Sketch of the proof

The proof comes down to verification of the following identities for the coefficients \(a_i\)

$$\begin{aligned} \sum _{i=0}^ka_{N-i}\left( \frac{N}{2}\right) ^{k-i}\left( {\begin{array}{c}N-i\\ k-i\end{array}}\right) =0 \end{aligned}$$
(6)

for every \(k=1,\ldots , N-2\) and

$$\begin{aligned} \left( \frac{N}{2} \right) ^2+\sum _{i=0}^{N-3}a_{N-i}\left( \frac{N}{2}\right) ^{N-i}=\frac{N}{2}. \end{aligned}$$
(7)

Identities Eq. 6 result (after the substitution \(y=\frac{N}{2} y_1+z\) in Eq. 3 and the expansion of the powers) from the form of the coefficients at \(z^{N-k}\) for \(k=1,\ldots , N-2\), respectively, identity Eq. 7 corresponds to the coefficient at \(z^0\).

Notice that identity Eq. 6 is equivalent to the convolution identity

$$\sum _{i=0}^k (-1)^i\left( {\begin{array}{c}N\\ i\end{array}}\right) \left( {\begin{array}{c}N-i\\ k-i\end{array}}\right) =0,$$

which after applying the equality

$$\left( {\begin{array}{c}N\\ i\end{array}}\right) \left( {\begin{array}{c}N-i\\ k-i\end{array}}\right) =\frac{N!}{i!(k-i)!}=\frac{N!}{k!}\left( {\begin{array}{c}k\\ i\end{array}}\right) , \qquad 0\le i\le k,$$

implies

$$\sum _{i=0}^k(-1)^i\left( {\begin{array}{c}k\\ i\end{array}}\right) =0,$$

which gives \((1-1)^k=0.\) Therefore, identity Eq. 6 is true.

Wherein, Eq. 7 is equivalent to the identity

$$\left( \frac{N}{2}\right) ^2+\sum _{i=0}^{N-3}(-1)^i\left( {\begin{array}{c}N\\ i\end{array}}\right) \left( \frac{N}{2}\right) ^Na_N=\frac{N}{2}$$

that is

$$\begin{aligned} \left( \frac{N}{2}\right) ^2+\left( \frac{N}{2}\right) ^Na_N\left( (1-1)^N-\sum _{i=N-2}^N(-1)^i\left( {\begin{array}{c}N\\ i\end{array}}\right) \right) =\frac{N}{2},\\ \left( \frac{N}{2}\right) ^2+\frac{(-1)^N}{N-1}\cdot \frac{N}{2}\left( -(-1)^N \left( {\begin{array}{c}N\\ 2\end{array}}\right) +(-1)^N\left( {\begin{array}{c}N\\ 1\end{array}}\right) -(-1)^N\right) =\frac{N}{2},\\ \left( \frac{N}{2}\right) ^2-\frac{1}{N-1}\cdot \frac{N}{2} \cdot \frac{N(N-1)}{2}+\frac{1}{N-1} \cdot \frac{N}{2}(N-1)=\frac{N}{2},\\ 0=0, \end{aligned}$$

which finishes the proof.

The above theorem is also attractive because of the following fact.

Theorem 2

Under the assumptions of Theorem 1, if for some \(\beta \in \mathbb {R}\) the substitution \(y=\beta y_1+z\) reduces the following differential equation

$$y'+p(x)y=\sum _{k=2}^N\alpha _k(x)y^k,$$

where \(\alpha _k\in C(I)\), \(k=2,\ldots ,N\) and \(\alpha _2(x)=q(x)\) to the Bernoulli equation of the form

$$z'+\tilde{p}(x)z=\tilde{q}(x)z^N,$$

then the coefficients \(\beta\) and \(\alpha _k(x)\) for \(k=3,\ldots ,N\) are the same as in Theorem 1 that is \(\beta =\frac{N}{2}\) and

$$\alpha _k(x)=a_k\frac{q(x)}{y_1^{k-2}}, \qquad k=3,\ldots , N.$$

Proof

The statement results directly from the calculations presented in the early stage of the proof of Theorem 1 leading to the generating of formulae Eqs. 6 and 7.

Remark 2

It is worth pointing out that both Theorem 1 and Theorem 2 can be generalized to the case when the linear differential operator

$$L_1[y]=y'+p(x)y$$

is replaced by the general linear differential operator

$$L_n[y]=\sum _{r=0}^np_r(x)y^{(r)},$$

where \(p_r\in C(I)\), for \(r=0,1,\ldots ,n\), \(n\in \mathbb {N}\), and, what is interesting, the right side of the respective equation remains the same as in Eq. 3.

There is one more possibility of generalization of Theorem 1. Namely, Theorem 1 can be generalized to the case of Riccati equations and even to the case of general Riccati equations.

Theorem 3

Let \(y_1\) be a particular solution of the following n-th order general Riccati equations

$$\sum _{s=0}^n p_s(x)y^{(s)}=q(x)y^2+r(x),$$

where \(p_s,q,r\in C(I)\), \(s=0,1,\ldots ,n\), and \(I\in \mathbb {R}\) is a certain non-empty and non-degenerate interval. If \(y_1\ne 0\) for \(x\in I\), then for every \(N\in \mathbb {N}\), \(N\ge 3\), the substitution \(y=\frac{N}{2}y_1+z\) reduces the differential equation of the form

$$\sum _{s=0}^n p_s(x)y^{(s)}=\frac{N}{2}r(x)+\sum _{k=2}^N a_k\frac{q(x)}{y_1^{k-2}}y^k,$$

where coefficients \(a_k\) are defined as in Theorem 1, to the following general Bernoulli-type equation

$$\sum _{s=1}^n p_s(x)z^{(s)}+\left( p_0(x)-\frac{N}{N-1}q(x)y_1\right) z=a_N\frac{q(x)}{y_1^{N-2}}z^N.$$

Proof

The proof is analogous to the proof of Theorem 1.

Example 2

Let \(y_1\) be a solution of the following classical Riccati equation

$$\begin{aligned} y'=a(y^2+x^{-2}), \qquad a\in \mathbb {R}\setminus \{0\}, \quad 1-4a^2\ge 0 \quad \text {i.e. } a\in \left[ -\frac{1}{2},\frac{1}{2}\right] , \end{aligned}$$
(8)

which possesses “two” special solutions \(y_1=\frac{1\pm \sqrt{1-4a^2}}{-2ax}\) - see Remark 3. Then, the substitution \(y=2y_1+z\) reduces the following differential equation

$$y'=\frac{2a}{x^2}+ay^2-\frac{a}{3y_1}y^3+\frac{a}{24y_1^2}y^4$$

to the Bernoulli equation

$$z'-\frac{4}{3}ay_1z=\frac{a}{24y_1^2}z^4.$$

Remark 3

(see [5]) The classical Riccati equations \(y'=a(y^2+x^n)\), where \(n\in \left\{ \frac{4k}{1-2k}:k\in \mathbb {Z}\right\} \cup \{-2\}\), which obey the Eq. 8, were solved by Daniel Bernoulli and Jacopo Riccati in 1724 by the substitution

$$\begin{aligned} y=\frac{1}{x^2v}-\frac{1}{ax}, \end{aligned}$$
(9)

which implies

$$y'=-\frac{2}{x^3v}-\frac{v'}{x^2v^2}+\frac{1}{ax^2}$$

and

$$y'=a(y^2+x^n) \quad \Leftrightarrow \quad v'=-a\left( \frac{1}{x^2}+x^{n+2}v^2\right) .$$

Hence, if \(n=-2\) and we assume (the special assumption) that \(v(x)=-y(x)\), then by Eq. 9

$$-v=\frac{1}{x^2v}-\frac{1}{ax},$$

i.e.

$$v^2-\frac{v}{ax}+\frac{1}{x^2}=0 \quad \Rightarrow \quad v=\frac{1\pm \sqrt{1-4a^2}}{2ax} \quad \Rightarrow \quad y=\frac{1\pm \sqrt{1-4a^2}}{-2ax},$$

which are two special solutions of the differential equation

$$y'=a(y^2+x^{-2}).$$

Some special Riccati equations

The authors were also interested in nonlinear substitutions in Riccati equations. In addition to the known results, for example, the fundamental fact that the Riccati equation can be rewritten as linear second order homogeneous differential equation, we were also interested in specific cases absent from the known literature, see [6,7,8,9,10]. We will present four original results.

Theorem 4

If almost canonical form of Riccati equation is separable differential equation

$$\begin{aligned} f'+\alpha (x)=\beta (x)f^2, \end{aligned}$$

where \(\alpha , \beta \in C(I)\), \(\alpha (x)\ne 0\) for every \(x\in I\), \(I \subset \mathbb {R}\) is an interval, and

$$\begin{aligned} \frac{\beta (x)}{\alpha (x)}\equiv -y_0^2,\quad y_0\in \mathbb {R}, \end{aligned}$$

and a function f(x) is a solution of this differential equation, \(f(x)\ne 0\) for every \(x \in I\), then function:

$$\begin{aligned} y(x)=\frac{1}{f(x)}-y_0 \end{aligned}$$

is a solution of Riccati differential equation:

$$\begin{aligned} y'+a(x)(y+y_0)=b(x)y^2, \end{aligned}$$

where \(a,b \in C(I)\), \(b(x) \ne 0\) for every \(x \in I\), whenever

$$\begin{aligned} \frac{a(x)}{b(x)}\equiv -2y_0. \end{aligned}$$

Proof

Setting \(y=\frac{1}{f}-y_0\) we get:

$$\begin{aligned} y'+a(x)(y+y_0)=b(x)y^2\iff -\frac{f'}{f^2}+\frac{a(x)}{f}=b(x)\left( \frac{1}{f^2}-\frac{2y_0}{f}+y_0^2\right) , \end{aligned}$$

and since \(\frac{a(x)}{b(x)}=-2y_0\), we get

$$\begin{aligned}&-f'=b(x)+y_0^2f^2b(x)=b(x)(1+y_0^2f^2)\iff \frac{df}{1+y_0^2f^2}=-b(x)dx\iff \\ {}&\iff \frac{1}{y_0}\arctan (y_0f)=-\int b(x)dx. \end{aligned}$$

\(\square\)

The inverse version of Theorem 4 also holds.

Theorem 5

If \(y_0\in \mathbb {R}\) and function y(x) is particular solution of Riccati differential equation

$$\begin{aligned} y'+a(x)(y+y_0)=b(x)y^2 \end{aligned}$$

where \(a,b \in C(I)\), \(I \subset \mathbb {R}\) is an interval, \(b(x)\ne 0\) for every \(x\in I\), and

$$\begin{aligned} \frac{a(x)}{b(x)}\equiv -2y_0, \end{aligned}$$

then function

$$\begin{aligned} f(x)=\frac{1}{y-y_0} \end{aligned}$$

is a solution of the following Riccati equation:

$$\begin{aligned} f'+\alpha (x)=\beta (x)f^2, \end{aligned}$$

where \(\alpha , \beta \in C(I),\,\alpha (x)\ne 0\) for every \(x\in I\), whenever

$$\begin{aligned} \frac{\beta (x)}{\alpha (x)}\equiv -y_0^2. \end{aligned}$$

This equation is then a separable differential equation.

Proof

It follows easily from proof of Theorem 4.

Theorem 6

Let \(a,b\in C(I), \, I \subset \mathbb {R}\) be an interval, \(y_0\in \mathbb {R}, \, \alpha (x):=\exp \left( \int (2y_0b(x)-a(x))dx\right)\) and let \(y\ne 0\) be a solution of the following almost canonical Riccati differential equation:

$$\begin{aligned} y'=-b(x)\alpha (x)+\frac{y_0}{\alpha (x)}\left( a(x)-y_0b(x)\right) y^2. \end{aligned}$$

Then, the function:

$$\begin{aligned} f(x):=\frac{\alpha (x)}{y(x)}+y_0 \end{aligned}$$

is a solution of Bernoulli equation:

$$\begin{aligned} f'+a(x)f=b(x)f^2. \end{aligned}$$

Proof

We have:

$$\begin{aligned} f'=\frac{\alpha 'y-\alpha y'}{y^2} \end{aligned}$$

which implies:

$$\begin{aligned}&\hspace{4cm}f'+a(x)f=b(x)f^2\\ \nonumber&\iff \frac{\alpha 'y-\alpha y'}{y^2}+a(x)\left( \frac{\alpha (x)}{y(x)}+y_0\right) =b(x)\left( \frac{\alpha (x)}{y(x)}+y_0\right) ^2\\ {}&\nonumber \iff \alpha 'y-\alpha y'+a(x)\alpha (x)y+y_0a(x)y^2=\\ {}&\nonumber \hspace{1cm}=b(x)(\alpha (x))^2+2y_0\alpha (x)b(x)y(x)+b(x)y_0^2(y(x))^2. \end{aligned}$$
(10)

We also find:

$$\begin{aligned}&\alpha '+a(x)\alpha (x)=2y_0\alpha (x)b(x)\iff \alpha '=\alpha (2y_0b(x)-a(x))\\ {}&\iff \frac{d\alpha }{\alpha }=(2y_0b(x)-a(x))dx\iff \alpha =\exp \left( \int (2y_0b(x)-a(x))dx\right) , \end{aligned}$$

which, by the assumption on the form \(\alpha\), gives:

$$\begin{aligned} (10)\iff -\alpha y'=b(x)(\alpha (x))^2+y_0(b(x)y_0-a(x))y^2. \end{aligned}$$

At the end of this section, we will present one more result, which we will illustrate with an appropriate example.

Theorem 7

Let \(a,b\in C(I),\,c,d\in C^1(I)\), where \(I\subset \mathbb {R}\) is a nonempty and nontrivial interval. If \(d(x)\not \equiv 0\) for every \(x\in I\), then the substitution \(y(x)=f(x)-\frac{c(x)}{3d(x)}\) reduces both differential equations:

$$\begin{aligned} y'+a(x)y+b(x)=c(x)y^2+d(x)y^3, \end{aligned}$$

and

$$\begin{aligned} y''+a(x)y'+b(x)y=c(x)y^2+d(x)y^3 \end{aligned}$$

to the Riccati equation and the Riccati equation of second order, respectively.

Proof

Substituting \(y(x)=f(x)-\frac{c(x)}{3d(x)}\) we find:

$$\begin{aligned}&y'+a(x)y+b(x)=c(x)y^2+d(x)y^3\iff \\ {}&\iff f'(x)-\left( \frac{c(x)}{3d(x)}\right) '+a(x)f(x)-a(x)\frac{c(x)}{3d(x)}+b(x)=\\ {}&\hspace{1cm}=c(x)\left( f^2-\frac{2c}{3d}f+\frac{c^2}{9d^2}\right) +d(x)\left( f^3-\frac{c}{d}f^2+\frac{c^2}{3d^2}f-\frac{c^3}{27d^3}\right) \iff \\ {}&\iff f'+\left( a+\frac{c^2}{3d}\right) f+\left( b-\frac{ac}{3d}-\left( \frac{c}{3d}\right) '-\frac{c^2}{9d^2}+\frac{c^3}{27d^2}\right) =df^3, \end{aligned}$$

and

$$\begin{aligned}&\hspace{3cm}y''+a(x)y'+b(x)y=c(x)y^2+d(x)y^3\iff \\ {}&\iff f''+af'+\left( b+\frac{c^2}{3d}\right) f+\left( -\left( \frac{c}{3d}\right) ''-\left( \frac{c}{3d}\right) 'a-\frac{c}{3d}b-\frac{2c^3}{27d^2}\right) =df^3. \end{aligned}$$

Example 3

The substitution \(y(x)=f(x)-\tan x\) reduces the following general–type Riccati equation:

$$\begin{aligned} y'-(3\tan x+\cot x)y=3y^2+\cot xy^3 \end{aligned}$$
(11)

to the Bernoulli differential equation:

$$\begin{aligned} f'-\cot xf=\cot xf^3 \end{aligned}$$

which possesses the solutions:

$$\begin{aligned} f(x)=\frac{\sin x}{\sqrt{\frac{1}{2}\cos 2x+C}}, \end{aligned}$$

where \(C\in \mathbb {R}\) and either \(x\in \mathbb {R}\) whenever \(C>\frac{1}{2}\) or \(\frac{1}{2}\cos 2x+C>0\) if \(C\in \left( -\frac{1}{2},\frac{1}{2}\right]\).