Electronic Journal of Differential Equations, Vol. 2021 (2021), No. 85, pp. 1–12.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

SYMMETRY ANALYSIS FOR A SECOND-ORDER ORDINARY

DIFFERENTIAL EQUATION

SEBERT FENG

Abstract. In this article, we apply the Lie symmetry analysis to a second-

order nonlinear ordinary differential equation, which is a Liénard-type equation
with quadratic friction. We find the infinitesimal generators under certain

parametric conditions and apply them to construct canonical variables. Also
we present some formulas for the first integral for this equation.

1. Introduction

Nonlinear differential equations have a wide array of applications in many scien-
tific fields, and can model a lot of physical and biological phenomena. The study
of solutions to nonlinear differential equations has been an important topic in the
community of nonlinear sciences. However, it is not always possible to express
exact solutions of nonlinear differential equations explicitly in terms of elementary
functions. In some cases it is possible to find elementary functions that are con-
stant on solution curves, that is, elementary first integrals. These first integrals
allow us to occasionally find some useful properties that an explicit solution may
not reveal. Prelle-Singer [16] proposed a method for solving first-order ODEs so-
lutions in terms of elementary functions if such solutions exists. Duarte and his
co-authors [8] modified the technique developed by Prelle and Singer and applied
it to the second-order ODEs. Their approach was based on the conjecture that if
an elementary solution exists for the given second-order ODEs, then there exists
as least one elementary first integral I(x, y, y′) whose derivatives are all rational
functions of x, y, y′. Chandrasekar et al [4, 5] used an extended Prelle-Singer pro-
cedure applicable to identify integrable nonlinear oscillator systems and construct
integrating factors. Another two powerful techniques for studying explicit solutions
and first integrals of various differential equations are the Painlevé test [6, 7, 13],
and the Lie symmetry reduction method [2, 3, 11, 12, 14]. The latter method has
been applied in a variety of fields in the past decades.

We consider the force-free Duffing-van der Pol equation [15]

y′′ + (α+ βy2)y′ − γy + y3 = 0, (1.1)

2010 Mathematics Subject Classification. 34A05, 34L30.
Key words and phrases. Lie symmetry; Liénard equation; infinitesimal generator;

second-order ordinary differential equation; harmonic oscillator.
c©2021. This work is licensed under a CC BY 4.0 license.

Submitted January 18, 2021. Published October 15, 2021.

1



2 S. FENG EJDE-2021/85

where α, β, and γ are arbitrary parameters. It is integrable with the parametric
conditions α = 4/β and γ = −3/β2. Under the transformation

w = −ye(1/β)x, z = e−(2/β)x, (1.2)

equation (1.1) with restriction α = 4/β and γ = −3/β2 was shown to be trans-
formable into

w′′ − β2

2
w2w′ = 0,

which can then be integrated. Equation (1.1) arises in a model describing the
propagation of voltage pulses along a neuronal axon and has recently received much
attention from many authors. Similar results and more discussions can se found in
[10, 15].

In a parallel direction, while performing the invariance analysis of a similar kind
of problem, we find that not only (1.1) but also its generalized version

y′′ +
( 4

β
+ βy2

)
y′ +

3

β2
y + y3 + δy5 = 0, (1.3)

where δ is an arbitrary parameter, is invariant under the same set of Lie point sym-
metries. As a consequence one can use the same transformation (1.2) to trnasform
(1.3) into

w′′ − β2

2
w2w′ + δw5 = 0,

which is not so simple to integrate. However, we observe that this equation coin-
cides with the second equation in the so-called modified Emden equation (MEE)
hierarchy, investigated by Feix et al [9],

y′′ + yly′ + gy2l+1 = 0, l = 1, 2, . . . , n,

where g is an arbitrary parameter. In fact, they have shown that through a direct
transformation to a third-order equation, the above equation can be integrated to
obtain the general solution for the specific choice of the parameter g, namely, for
g = 1/(l + 2)2 [5, 9]. This provides us grounds for expecting that there should
be a number of integrable equations which also admits solutions which are both
oscillatory and non-oscillatory types in the class

y′′ + (k1y
q + k2)y′ + k3y

2q+1 + k4y
q+1 + λ1y = 0, q ∈ R. (1.4)

where k′is, i = 1, 2, 3, 4 and λ1 are arbitrary parameters. In this study, we restrict
our attention to this equation for its first integrals by means of the Lie symme-
try method. Equation (1.4) is a unified model for several ground-breaking physical
systems which includes simple harmonic oscillator, anharmonic oscillator, force-free
Helmholtz oscillator, force-free Duffing oscillator, MEE hierarchy, and the general-
ized DVP hierarchy, see [5, 15]. If k3 = 0, then equation (1.4) is the Duffing-van
der Pol-type oscillator. When q = 1, equation (1.4) becomes a more general MEE,

y′′ + (k1y + k2)y′ + k3y
3 + k4y

2 + λ1y = 0,

which provides us the force-free Helmholtz oscillator. When q = 2, equation (1.4)
reduces to the force-free Duffing-van der Pol oscillator.

Now, let us consider the usage of the Lie symmetry reduction to obtain the first
integrals of a second-order nonlinear ODEs. Symmetry is the key to solve differen-
tial equations. In this article, we study equation (1.4) to derive its first integrals
under certain parametric conditions by applying the Lie point symmetry reduction



EJDE-2021/85 SYMMETRY ANA74LYSIS FOR DIFFERENTIAL EQUATIONS 3

method. In fact, the exponent q determines the tangent vector, which induces the
infinitesimal generator. As a result of this, we can classify the integrable cases by
the different values of q. For certain value of parameters, we can find paramet-
ric conditions through the determining equations for the infinitesimal generator,
which enable us to construct the corresponding canonical coordinates. Through
the inverse transformation, we obtain the first integrals of equation (1.4).

2. Preliminaries

Let us briefly recall the Lie symmetry method [11, 14] for the ODEs of the form

y(n) = ω
(
x, y, y′, . . . , y(n−1)

)
, y(k) ≡ dky

dxk
. (2.1)

It is assumed that ω is (locally) a smooth function of all its arguments. We first
state the symmetry condition. A symmetry condition of (2.1) is a diffeomorphism
that maps the set of solutions of the ODE to itself. Any diffeomorphism,

Γ : (x, y) 7→ (x̂, ŷ),

maps smooth planar curves to smooth planar curves. On the plane, the diffeomor-
phism Γ generates a mapping on the derivatives y(k),

Γ :
(
x, y, y′, . . . , y(n)

)
7→
(
x̂, ŷ, ŷ′, . . . , ŷ(n)

)
,

where

ŷ(k) ≡ dkŷ

dx̂k
, k = 1, 2, . . . , n.

Using the chain rule, the function ŷ(k) can be written as

ŷ(k) =
dŷ(k−1)

dx̂
=
Dxŷ

k−1

Dxx̂
, k = 1, 2, . . . , n;

y(0) ≡ ŷ,
(2.2)

where D(x) is the total derivative with respect to x:

Dx = ∂x + y′∂y + y′′∂y′ + . . . .

We obtain the symmetry condition for ODE (2.1),

ŷ(n) = ω
(
x̂, ŷ, ŷ′, . . . , ŷ(n−1)

)
, (2.3)

where the function ŷ(k) is given by (2.2).
The action of a Lie symmetry maps every point on an orbit to a point on the same

orbit. Now consider the orbit through a non-invariant point (x, y). The tangent
vector to the orbit at the point (x̂, ŷ) is

(
ξ(x̂, ŷ), η(x̂, ŷ)

)
, where

dx̂

dε
= ξ(x̂, ŷ),

dŷ

dε
= η(x̂, ŷ).

In particular, the tangent vector at (x, y) is

(ξ(x, y), η(x, y)) =
(dx̂
dε

∣∣∣
ε=0

,
dŷ

dε

∣∣∣
ε=0

)
.

For almost all ODEs, the symmetry condition (2.3) is nonlinear. Lie symmetries
are obtained by linearizing (2.3) about ε = 0. It is usually easy to check whether
or not a given diffeomorphism is a symmetry of a particular ODE. Since the trivial



4 S. FENG EJDE-2021/85

symmetry condition corresponding to ε = 0 leaves every point unchanged, for ε
sufficiently close to 0, the prolonged Lie symmetries are of the form

x̂ = x+ εξ +O(ε2),

ŷ = y + εη +O(ε2),

ŷ(k) = y(k) + εη(k) +O(ε2), k ≥ 1.

(2.4)

Note that the superscript in η(k), (k = 1, 2, . . . , n) is merely an index; it does not
denote a derivative of η. Substituting (2.4) into the symmetry condition (2.3); the
O(ε) terms yield the linearized symmetry condition

η(n) = ξωx + ηωy + η(1)ωy′ + · · ·+ η(n−1)ωy(n−1) (2.5)

when (2.1) holds. The function η(k)(k = 1, 2, . . . , n) can be obtained from (2.2).
For k ≥ 1, we have

ŷ(k) =
Dxŷ

(k−1)

Dxx̂
=
y(k) + εDxη

(k−1) +O(ε2)

1 + εDxξ +O(ε2)

= y(k) + ε
(
Dxη

(k−1) − y(k)Dxξ
)

+O(ε2).

From (2.4), we obtain

η(k)
(
x, y, y′, . . . , y(k)

)
= Dxη

(k−1) − y(k)Dxξ. (2.6)

Now we consider the second-order ODE

y′′ = ω (x, y, y′) . (2.7)

The diffeomorphism of the form

(x̂, ŷ) =
(
x̂(x, y), ŷ(x, y)

)
is called a point symmetry. To find the Lie point symmetry of a second-order ODE,
we need to calculate η(1) and η(2) first. Since the functions ξ and η depend upon x
and y only, (2.6) gives

η(1) = ηx + (ηy − ξx)y′ − ξyy′2, (2.8)

η(2) = ηxx + (2ηxy − ξxx)y′ + (ηyy − 2ξxy)y′2 − ξyyy′3

+ (ηy − 2ξx − 3ξyy
′)y′′.

(2.9)

The linearized symmetry condition of (2.7) is obtained by substituting (2.8) and
(2.9) into (2.5) and then replacing y′′ by ω(x, y, y′). This gives

ηxx + (2ηxy − ξxx)y′ + (ηyy − 2ξxy)y′2 − ξyyy′3

= (−ηy + 2ξx + 3ξyy
′)ω + ξωx + ηωy + {ηx + (ηy − ξx)y′ − ξyy′2}ωy′ ,

(2.10)

which can be solved for many regular cases. Since ξ and η are independent of y′,
it follows that (2.10) can be decomposed into a system of PDEs, which are the
determining equations for the Lie point symmetries. Similarly, for higher-order
ODEs, we can also obtain the linearized symmetry condition, which usually looks
more complicated.



EJDE-2021/85 SYMMETRY ANA74LYSIS FOR DIFFERENTIAL EQUATIONS 5

3. Infinitesimal generator and canonical coordinates

Suppose that a first-order ODE has a one-parameter Lie group of symmetries,
whose tangent vector at (x, y) is (ξ, η). Then the partial differential operator

X = ξ(x, y)∂x + η(x, y)∂y

is called the infinitesimal generator of the Lie group. To deal with the action of
Lie symmetries on derivatives of order n or smaller, we introduce the prolonged
infinitesimal generator

X(n) = ξ∂x + η∂y + η(1)∂y′ + · · ·+ η(n)∂y(n) .

The coefficient of ∂y(n) is the O(ε) term in the expansion of ŷ(k), and so X(n)

is associated with the tangent vector in the space of variables (x, y, y′, . . . , y(n)).
We can use the prolonged infinitesimal generator to write the linearized symmetry
condition (2.5) in a compact form:

X(n)
(
y(n) − ω(x, y, y′, . . . , y(n−1))

)
= 0

when (2.1) holds.
Let L denote the set of all infinitesimal generators of one-parameter Lie groups

of point symmetries of an ODE of order n ≥ 2. The linearized symmetry condition
is linear in ξ and η, and so

X1, X2 ∈ L ⇒ c1X1 + c2X2 ∈ L ∀c1, c2 ∈ R.

Hence L is a vector space. The dimension of this vector space is the number of
arbitrary constants that appear in the general solution of the linearized symmetry
condition.

We know that if an ordinary differential equation admits an infinitesimal gener-
ator, then there exists a pair of variables

r = r(x, y) and s = s(x, y),

which are called canonical coordinates, with r and s (s 6= 0) being arbitrary par-
ticular solutions of the first-order linear partial equations

ξ(x, y)
∂r

∂x
+ η(x, y)

∂r

∂y
= 0,

ξ(x, y)
∂s

∂x
+ η(x, y)

∂s

∂y
= 1.

The change of coordinates should be invertible in some neighbourhood of (x, y), so
we impose the nondegeneracy condition

rxsy − rysx 6= 0.

Suppose that ξ(x, y) 6= 0. The invariant canonical coordinate r(x, y) is a first
integral of

dx

ξ(x, y)
=

dy

η(x, y)
.

The coordinate s(x, y) is obtained by the quadrature

s(x, y) =

∫
dx

ξ(x, y(r, x))
,



6 S. FENG EJDE-2021/85

where the integral is evaluated with r being treated as a constant. Similarly, if
ξ(x, y) = 0 and η(x, y) 6= 0, then

r = x and s =

∫
dy

η(x, y)
,

are canonical coordinates.

4. Main results

Let us consider (1.4) assuming that q is arbitrary. Chandrasekar et al [5] used
the extended Prelle-Singer procedure to identify the first integrals of equation (1.4).
Now, let us apply the method of Lie point symmetry [11, 14] to re-consider first
integrals of (1.4) under certain parametric conditions. Following the process to
determine the symmetries of a differential equation introduced in the preceding sec-
tion, we can obtain the linearized symmetry condition concerning equation (1.4).
Although (2.10) looks complicated, it is not difficult for us to solve ξ(x, y) and
η(x, y). Since the unknown functions do not depend on the derivative y′, after
setting the coefficients of the powers (y′)i (i = 0, 1, 2, 3) in (2.10) to zero, the lin-
earized symmetry condition (2.10) can be decomposed into the determining system
as follows

[y′]3 : ξyy = 0, (4.1)

[y′]2 : ηyy − 2ξxy = −2ξyk1y
q − 2ξyk2, (4.2)

[y′]1 : 2ηxy − ξxx = −ξx(k1y
q + k2)− 3k3ξyy

2q+1 − 3k4ξyy
q+1

− 3λ1ξyy − qk1ηyq−1,
(4.3)

[y′]0 : ηxx = −2k3ξxy
2q+1 − 2k4ξxy

q+1 − 2λ1ξxy + k3ηyy
2q+1

+ k4ηyy
q+1 + λ1ηyy − (2q + 1)k3ηy

2q

− (q + 1)k4ηy
q − λ1η − (k1y

q + k2)ηx.

(4.4)

The first equation (4.1) gives

ξ = a(x)y + b(x). (4.5)

Substituting (4.5) into (4.2), we have

η = − 2a(x)k1
(q + 1)(q + 2)

yq+2 + {a′(x)− a(x)k2}y2 + c(x)y + d(x), (4.6)

where q 6= −1 and q 6= −2. And a(x), b(x), c(x) and d(x) are functions of x to be
determined. Plugging (4.5)and(4.6) into (4.3) leads to

: −3k3a+
2k21q

(q + 1)(q + 2)
a = 0,

[yq+1] :
(q − 1)(q + 3)

q + 1
k1a
′ + 3k4a− k1k2qa = 0,

[yq] : k1b
′ + k1qc = 0,

[yq−1] : k1qd = 0,

[y] : a′′ − k2a′ + λ1a = 0,

[y0] : 2c′ − b′′ + k2b
′ = 0.

(4.7)



EJDE-2021/85 SYMMETRY ANA74LYSIS FOR DIFFERENTIAL EQUATIONS 7

Similarly, plugging (4.5) and (4.6) into (4.4), we find that

[y3q+2] : k1k3a = 0,

[y2q+2] : (2q − 1)k2k3a− k3(2q + 1)a′ +
2k21

(q + 1)(q + 2)
a′ − 2k1k4

(q + 1)(q + 2)
a = 0,

[y2q+1] : k3b
′ + qk3c = 0,

[y2q] : k3(2q + 1)d = 0,

[yq+2] :
q(q + 3)

(q + 1)(q + 2)
k1a
′′ − (q − 1)k2k4a+ (q + 1)k4a

′

+
2k1λ1
q + 2

a− q2 + 3q + 4

(q + 1)(q + 2)
k1k2a

′ = 0,

[yq+1] : 2k4b
′ + qk4c+ k1c

′ = 0,

[yq] : (q + 1)k4d+ k1d
′ = 0,

[y2] : a′′′ + k2λ1a+ λ1a
′ − k22a′ = 0,

[y] : c′′ + 2λ1b
′ + k2c

′ = 0,

[y0] : d′′ + λ1d+ k2d
′ = 0.

(4.8)
We assume that q 6= 0, 12 ,−

1
2 ,

1
3 ,−

1
3 , 1, 2. Since we can combine the coefficients

of y with the same power, under this assumption there is no equations with the
same power of yi. We need to carefully consider several cases.

Case 1: k1, k2, k3, k4, λ1 are arbitrary constants. The first equation in (4.8)
gives

k1k3a = 0,

which means a(x) = 0. The forth equation in (4.7) gives

k1qd = 0,

which means d(x) = 0. Then the determining system for b(x) and c(x) can be
reduced to

2c′ − b′′ + k2b
′ = 0, (4.9)

qc+ b′ = 0, (4.10)

c′′ + 2λ1b
′ + k2c

′ = 0, (4.11)

2k4b
′ + qk4c+ k1c

′ = 0. (4.12)

Substituting (4.10) into (4.9), we obtain

c(x) = c0e
k2q
q+2x, b(x) = − (q + 2)c0

k2
e

k2q
q+2x + c1,

where c0 and c1 are constants. Substituting b(x) and c(x) into (4.11) and (4.12),
we obtain the parametric conditions

λ1 =
q + 1

(q + 2)2
k22, k4 =

k1k2
q + 2

, (4.13)

respectively. Hence, the general solution of the linearized symmetry condition is

ξ = − (q + 2)c0
k2

e
k2q
q+2x + c1, η = c0e

k2q
q+2xy.



8 S. FENG EJDE-2021/85

By using ξ and η, every infinitesimal generator is of the form

χ = c0χ0 + c1χ1,

where

χ0 = − (q + 2)

k2
e

k2q
q+2x∂x + e

k2q
q+2xy∂y, χ1 = ∂x.

For the generator χ1, it is a homothety operator. Generally, it is hard to use this
operator to find the first integrals of complicated second-order nonlinear ODEs.
So we choose χ0, which is a translation operator, as a generator to get canonical
coordinates. Note that in the following cases, for simplicity, we assume that c0 = 1,
c1 = 0 to obtain the generator χ0.

The invariant canonical coordinate r(x, y) is a first integral of

dx

ξ(x, y)
=

dy

η(x, y)
.

Then we obtain

r(x, y) =
q + 2

k2
e

k2
q+2xy. (4.14)

The corresponding coordinate s(x, y) is obtained by the quadrature

s(x, y) =

∫
dx

ξ(x, y(r, x))
.

So we derive

s(x, y) =
1

q
e−

qk2
q+2x. (4.15)

Note that equations (4.14)and(4.15) can be rewritten in the parametric form

x = −q + 2

qk2
ln(qs), y =

k2
q + 2

(qs)
1
q r. (4.16)

Using the nonlinear transformations (4.16) yields

∂y

∂x
= − k22

(q + 2)2
q

q+1
q

(
rss

q+1
q +

1

q
rs

1
q

)
, (4.17)

∂2y

∂2x
=

k32
(q + 2)3

q
2q+1

q

(
rsss

2q+1
q +

q + 2

q
rss

q+1
q +

1

q2
rs

1
q

)
. (4.18)

Substituting (4.17) and (4.18) into (1.4), under parametric condition (4.13), we
obtain

rss =
k1k

q−1
2

(q + 2)q−1
rsr

q − k3k
2q−2
2

(q + 2)2q−2
r2q+1, (4.19)

which is integrated as

2k1√
k21 − 4(q + 1)k3

tanh−1
[k1(q + 2)kq2r

q+1 − 2k2(q + 1)(q + 2)qrs

(q + 2)kq2r
q+1
√
k21 − 4(q + 1)k3

]
+ ln

[
k3k

2q−2
2 r2q+2 − k1kq−12 (q + 2)q−1rs + (q + 1)(q + 2)2q−2r2s

]
= I,

(4.20)

where I is an arbitrary constant. Using the inverse transformation of (4.16), we
have

rs = −q + 2

k2
e

(q+1)k2
q+2 xy − (q + 2)2

k22
e

(q+1)k2
q+2 xy′, (4.21)

where k2y + (q + 2)y′ 6= 0.



EJDE-2021/85 SYMMETRY ANA74LYSIS FOR DIFFERENTIAL EQUATIONS 9

(i) If k3 6= k21
4(q+1) , substituting (4.21) into (4.20), we obtain the first integral of

equation (1.4) as follows

2k1√
k21 − 4(q + 1)k3

tanh−1
[k1yq+1 + 2k2(q+1)

q+2 y + 2(q + 1)y′√
k21 − 4(q + 1)k3yq+1

]
+

2q + 2

q + 2
k2x

+ ln
[
(q + 2)yq+1[k3(q + 2)yq+1 + k1k2y + k1(q + 2)y′] + (q + 1)[k2y + (q + 2)y′]2

]
= I,

where I is an arbitrary constant.

(ii) If k3 =
k21

4(q+1) , equation (4.19) can be integrated as

1

1− 2k2(q+2)q(q+1)rs
k1k

q
2(q+2)rq+1

+ ln
[
1− 2k2(q + 2)q(q + 1)rs

k1k
q
2(q + 2)rq+1

]
+ ln(rq+1) = I. (4.22)

Substituting (4.21) into (4.22), we obtain the first integral of equation (1.4) as
follows

q + 1

q + 2
k2x+ ln

[
k1y

q+1 + 2(q + 1)y′ +
2(q + 1)

q + 2
k2y
]

+
k1(q + 2)yq+1

k1(q + 2)yq+1 + 2(q + 1)k2y + 2(q + 1)(q + 2)y′
= I.

The above two formulas of first integrals for equation (1.4) are in good agreement
with the results presented in [5, 15].

Case 2: k3 = 0, k1, k2, k4, λ1 are arbitrary constants. In this case, (1.4)
becomes the Duffing van der Pol-type oscillator. The first integrals of this kind of
oscillator can also be found in [15].

By the first equation and the forth equation in (4.7), we obtain a(x) = 0 and
d(x) = 0. Then the determining system for b(x) and c(x) is the same as Case 1. So
we obtain

c(x) = c0e
k2q
q+2x, b(x) = − (q + 2)c0

k2
e

k2q
q+2x + c1,

where c0 and c1 are arbitrary constants under the parametric conditions

λ1 =
q + 1

(q + 2)2
k22, k4 =

k1k2
q + 2

. (4.23)

Using

ξ = −q + 2

k2
e

k2q
q+2x, η = e

k2q
q+2xy,

and combining (4.14) and (4.19), under the parametric condition (4.23), we deduce

rss =
k1k

q−1
2

(q + 2)q−1
rsr

q,

which is integrated as

rs =
k1k

q−1
2

(q + 1)(q + 2)q−1
rq+1 + I1, (4.24)



10 S. FENG EJDE-2021/85

where I1 is an arbitrary constant. Substituting equation (4.21) into (4.24), we
obtain the first integral of equation (1.4) as

e
(q+1)k2

q+2 x
(
y′ +

k2
q + 2

y +
k1
q + 1

yq+1
)

= I1,

where q 6= −1,−2.

Case 3: k1 = 0, k3 = 0 and k2, k4, λ1 are arbitrary constants. In this
case, (1.4) becomes the Duffing-type oscillator. The second equation in (4.7) gives
a(x) = 0. And the seventh equation in (4.8) gives d(x) = 0. The determining
system for b(x) and c(x) is reduced to

2c′ − b′′ + k2b
′ = 0, (4.25)

qc+ 2b′ = 0, (4.26)

c′′ + 2λ1b
′ + k2b

′ = 0. (4.27)

Substituting (4.26) into (4.25), we obtain

c(x) = c0e
qk2
q+4x, b(x) = − (q + 4)c0

2k2
e

qk2
q+4x + c1,

where c0 and c1 are arbitrary constants. In this case, we assume q 6= −4. If q = −4,
then we obtain c(x) = 0, and the equation (1.4) is partially integrable. Substituting
b(x) and c(x) into (4.27), we obtain one parametric condition

λ1 =
2(q + 2)

(q + 4)2
k22. (4.28)

For simplicity, we assume taht c0 = 1 and c1 = 0 which yield

ξ = −q + 4

2k2
e

qk2
q+4x, η = e

qk2
q+4xy.

Using ξ and η, we derive

r(x, y) =
q + 4

2k2
e

2k2
q+4xy, (4.29)

s(x, y) =
2

q
e−

qk2
q+4x. (4.30)

Formulas (4.29)and(4.30) can be rewritten to the parametric form

x = −q + 4

qk2
ln
(qs

2

)
, y =

2k2
q + 4

(qs
2

)2/q
r. (4.31)

Using the nonlinear transformation (4.31) yields

∂y

∂x
= − 2qk22

(q + 4)2

(q
2

)2/q (
rss

q+2
q +

2

q
rs2/q

)
, (4.32)

∂2y

∂2x
=

2q2k32
(q + 4)3

(q
2

)2/q [
rsss

2q+2
q +

(
1 +

4

q

)
rss

q+2
q +

4

q2
rs2/q

]
. (4.33)

Substituting (4.32) and (4.33) into (1.4), under the parametric condition (4.28), we
obtain

rss = −k4
( 2k2
q + 4

)q−2
rq+1,



EJDE-2021/85 SYMMETRY ANA74LYSIS FOR DIFFERENTIAL EQUATIONS 11

which is integrated as

r2s = − 2k4
q + 2

( 2k2
q + 4

)q−2
rq+2 + I2, (4.34)

where I2 is an arbitrary constant. Using the inverse transformation of (4.31), we
can deduce

rs = −q + 4

2k2
e

k2(q+2)
q+4 x

(
y +

q + 4

2k2
y′
)
, (4.35)

where 2k2y + (q + 4)y′ 6= 0. Substituting equation (4.35) into (4.34), we can re-
produce the first integral of equation (1.4) as follows

e
2(q+2)k2

q+4 x
[y′2

2
+

2k2
(q + 4)

yy′ +
2k22

(q + 4)2
y2 +

k4
q + 2

yq+2
]

= I2.

Case 4: k1 = 0, k4 = 0 and k2, k3, λ1 are arbitrary constants. The first
equation in (4.7) and the forth equation in (4.8) give a(x) = 0 and d(x) = 0. Then
we can solve for b(x) and c(x) from the system

2c′ − b′′ + k2b
′ = 0, (4.36)

qc+ b′ = 0, (4.37)

c′′ + 2λ1b
′ + k2c

′ = 0. (4.38)

Substituting (4.37) into (4.36), we obtain

c(x) = c0e
qk2
q+2x, b(x) = − (q + 2)c0

k2
e

qk2
q+2x + c1,

where c0 and c1 are arbitrary constants. Substituting b(x) and c(x) into (4.38), we
obtain one parametric condition

λ1 =
q + 1

(q + 2)2
k22. (4.39)

For simplicity, we assume that c0 = 1 and c1 = 0. Then

ξ = −q + 2

k2
e

qk2
q+2x, η = e

qk2
(q+2)

xy.

By (4.14) and (4.19), under the parametric condition (4.39) we derive

rss = − k3k
2q−2
2

(q + 2)2q−2
r2q+1,

which is integrated as

r2s = − k3k
2q−2
2

(q + 1)(q + 2)2q−2
r2q+2 + I3, (4.40)

where I3 is an arbitrary constant. Substituting equation (4.21)into(4.40), we obtain
the first integral of equation (1.4) as

e
(2q+2)k2

q+2 x
[ k22

2(q + 2)2
y2 +

k2
q + 2

yy′ +
y′2

2
+

k3
2q + 2

y2q+2
]

= I3.

Note that all first integrals described herein agree well with those presented in
the literature [5, 8, 15] etc. For other cases of k1, k2, k3, k4 and λ1, the original
equation is partially integrable or the first integral can be derived directly without
applying the Lie point symmetry. But the values of q may effect the first integrals



12 S. FENG EJDE-2021/85

of equation (1.4). For example, if q = −1 or −2, the tangent vector η(x, y) is
undefined. However, we can still classify the first integrals of (1.4) in this case by
applying the Lie symmetry method as well as the theory of vector fields. In the
subsequent work, we will continue to consider this problem for various values of q
specifically and its applications.

Acknowledgments. The author would like to thank X.Y. Li and E. Suazo for
their fruitful discussions and comments.

References

[1] J. A. Almendral, M. F. Sanjuan; Integrability and symmetries for the helmholtz oscillator

with friction, J. Phys. A (Math. Gen.), 36 (2003), 695–710.
[2] D. J. Arrigo; Symmetry Analysis of Differential Equations, Wiley, New Jersey, 2014.

[3] G. W. Bluman, S. C. Anco; Symmetry and Integration Methods for Differential Equations,

Springer, New York, 2002.
[4] V. K. Chandrasckar, M. Senthilvelan, M. Lakshmanan; On the complete integrability and

linearization of certain second-order nonlinear ordinary differential equations, Proc. R. Soc.

Lond. Ser. A, 461 (2005), 2451–2476.
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tonomous nonlinear ordinary differential equations, Commun. Nonlinear Sci. Numer. Simul.,

16 (2011), 1127–1134.

[8] L. G. S. Duarte, S. E. S. Duarte, A. C. P. da Mota, J. E. F. Skeca; Solving second-order
ordinary differential equations by extending the prelle-singer method, J. Phys. A (Math.

Gen.), 34 (2001), 3015–3024.

[9] M. R. Feix, C. Geronimi, L. Cairo, P. G. L. Leach, R. L. Lemmer, S. Bouquet; On the
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rescaling, J. Phys. A (Math. Gen.), 30 (1997), 7347-7461.

[10] P. Holmes, D. Rand; Phase portraits and bifurcations of the non-linear oscillator: ẍ+ (α+
γx2)ẋ+ βx+ δx3 = 0, Int. J. Non-Linear Mech., 15 (1980), 449–458.

[11] P. E. Hydon; Symmetry Methods for Differential Equations, Cambridge University Press,
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[12] N. H. Ibragimov; Handbook of Lie Group Analysis of Differential Equations, CRC Press,

Boca Raton, Vol. III, 1995.
[13] E. L. Ince; Ordinary differential equations, Dover, New York, 1956.
[14] P. J. Olver; Applications of Lie Groups to Differential Equations, Springer Verlag, 1991.

[15] A. D. Polyanin, V. F. Zaitsev; Handbook of Exact Solutions For Ordinary Differential Equa-
tions, 2nd edition, Chapman & Hall/CRC, London, 2003.

[16] M. Prelle, M. Singer; Elementary first integrals of differential equations, Trans. Am. Math.

Soc. 279 (1983) 215–229.

Sebert Feng
The Science Academy of South Texas, Mercedes, Texas 78570, USA

Email address: feng14xi@gmail.com


	1. Introduction
	2. Preliminaries
	3. Infinitesimal generator and canonical coordinates
	4. Main results
	Acknowledgments

	References