We disclose a simple and straightforward method of solving ordinary or linear partial differential equations of any order and apply it to solve the generalized Euler-Tricomi equation. The method is easier than classical methods and also didactic. Date: Jan, 10, 202

1. 1
New Information on the Generalized Euler-Tricomi
Equation
1Lohans de Oliveira Miranda; 2Lossian Barbosa Bacelar Miranda
1Universidad Europea del Atlântico, Espanha, lohansmiranda@gmail.com
2IFPI, Brazil, lossianm@gmail.com
Date: 10/01/2022
Abstract. We disclose a simple and straightforward method of solving
ordinary or linear partial differential equations of any order and apply it
to solve the generalized Euler-Tricomi equation. The method is easier
than classical methods and also didactic.
1 Preliminaries
Next, we transcribe what was already established in [3].
“Consider:
1) 𝑥
⃗ = (𝑥1, 𝑥2, … , 𝑥𝑛) ∈ 𝐴, 𝐴 open set of ℝ𝑛
;
2) 𝑘 ∈ 𝐼𝑛 = {1, 2, 3, … , 𝑛};
3) 𝑢: 𝐴 → ℝ, differentiable function of order 𝑘, with continuous
derivatives (1)
Let us consider the “𝑘-dimensional Hessian matrix” given by
𝐻 = (
𝜕𝑘
𝑢(𝑥
⃗)
𝜕𝑥𝑖1
𝜕𝑥𝑖2
… 𝜕𝑥𝑖𝑘
) (2)
From 𝐻, let us consider the following system, being 𝑏𝑖1𝑖2…𝑖𝑘
(𝑥
⃗) and
𝑓𝑖1𝑖2…𝑖𝑘
(𝑥
⃗) differentiable functions of order 𝑘, with continuous derivatives,
and
𝑓𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝑏𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
being well defined in 𝐴:
(𝑏𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝜕𝑘
𝑢(𝑥
⃗)
𝜕𝑥𝑖1
𝜕𝑥𝑖2
… 𝜕𝑥𝑖𝑘
) = (𝑓𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)) (3)
Let us denote:
𝑔𝑖1𝑖2…𝑖𝑘
(𝑥
⃗) =
𝑓𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝑏𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
(4)
So (3) will be written as
(
𝜕𝑘
𝑢(𝑥
⃗)
𝜕𝑥𝑖1
𝜕𝑥𝑖2
… 𝜕𝑥𝑖𝑘
) = (𝑔𝑖1…𝑖𝑘
(𝑥
⃗)) (5)
Observation 1. Repeated applications of the Fundamental Theorem of
Calculus for each of the 𝑛𝑘
partial differential equations

2. 2
𝜕𝑘
𝑢(𝑥
⃗)
𝜕𝑥𝑖1
𝜕𝑥𝑖2
… 𝜕𝑥𝑖𝑘
= 𝑔𝑖1…𝑖𝑘
(𝑥
⃗) (6)
give us the 𝑛𝑘
solutions
𝑢𝑖1𝑖2…𝑖𝑘
(𝑥
⃗) = ∭ … ∫ 𝑔𝑖1…𝑖𝑘
(𝑥
⃗)𝜕𝑥𝑖1
𝜕𝑥𝑖2
… 𝜕𝑥𝑖𝑘−1
+
∑ 𝑐𝑠,𝑖1…𝑖𝑘
𝑘−1
𝑠=1
∏ 𝑥𝑖𝜃
𝑘−1
𝜃=𝑠+1
+ 𝑐𝑘−1,𝑖1𝑖2…𝑖𝑘
(7)
Here, the c (under indexed) are real or complex numbers. Obviously,
𝜕𝑘
𝑢𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝜕𝑥𝑗1
𝜕𝑥𝑗2
… 𝜕𝑥𝑗
= 𝑔𝑖1𝑖2…𝑖𝑘
(𝑥
⃗) (8)
if (𝑖1, 𝑖2, … , 𝑖𝑘) = (𝑗1, 𝑗2, … , 𝑗𝑘).
Now, consider the function
𝑢
̃(𝑥
⃗) = ∑ 𝑢𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝑖1,𝑖2,…,𝑖𝑘∈𝐼𝑛
(9)"
Now we can announce the main results.
2. Main results
Again, we transcribe what was done in [3]:
“Proposition 1. In the hypotheses established above, if for (𝑖1, 𝑖2, … , 𝑖𝑘) ≠
(𝑗1, 𝑗2, … , 𝑗𝑘) we have
𝜕𝑘
𝑢𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝜕𝑥𝑗1
𝜕𝑥𝑗2
… 𝜕𝑥𝑗
= 0, (10)
then 𝑢
̃(𝑥
⃗) = ∑ 𝑢𝑖1𝑖2…𝑖𝑘
(𝑥
⃗)
𝑖1,𝑖2,…,𝑖𝑘∈𝐼𝑛
defined in (9) will be the solution of 𝑛𝑘
partial differential equations defined in (3), or alternatively in (6). In
particular, 𝑢
̃(𝑥
⃗) will be a solution of the 2𝑛𝑘
− 1 differential equations
defined by the sums of the elements of all non-empty subsets of the set
𝐵 = {
𝜕𝑘
𝑢(𝑥
⃗)
𝜕𝑥𝑖1
𝜕𝑥𝑖2
… 𝜕𝑥𝑖𝑘
; 𝑖1,𝑖2, … , 𝑖𝑘 ∈ 𝐼𝑛 } (11)
Demonstration. It is an immediate consequence of the construction of
𝑢
̃(𝑥
⃗) and of the hypothesis (𝑖1, 𝑖2, … , 𝑖𝑘) ≠ (𝑗1, 𝑗2, … , 𝑗𝑘).
Observation 2. The thesis of Proposition 1 can still be obtained even if the
assumptions established in (10) are not satisfied. To do so, it is enough
to find the unknown functions involved that satisfy the required integral
equations”.
Theorem 1. Let:

3. 3
∑ 𝑓𝑘
𝑛
𝑘=1
(𝑥)
𝑑𝑘
𝑦
𝑑𝑥𝑘
(𝑥) = ∑ ℎ𝑘
𝑛
𝑘=1
(𝑥) (12)
𝑑𝑘
𝑦
𝑑𝑥𝑘
(𝑥) =
ℎ𝑘(𝑥)
𝑓𝑘(𝑥)
≝ 𝑔𝑘(𝑥); 1 ≤ 𝑘 ≤ 𝑛 (13)
Consider
𝑦𝑘(𝑥) = ∫ ∫ … ∫ 𝑔𝑘(𝑥) 𝑑𝑥𝑑𝑥 … 𝑑𝑥 + (∑ 𝑎𝑘−1𝑥𝑘−1
𝑛
𝑘=1
) ; 1 ≤ 𝑘 ≤ 𝑛 (14)
𝑦
̃(𝑥) ≝ ∑ 𝑦𝑘
𝑛
𝑘=1
(𝑥); 1 ≤ 𝑘 ≤ 𝑛 (15 )
𝑑𝑖
𝑦𝑘
𝑑𝑥𝑖
(𝑥) = {
𝑦𝑘(𝑥), 𝑖 = 𝑘
0, 𝑖 ≠ 𝑘
(16)
Then, 𝑦
̃(𝑥) is solution of ∑ 𝑓𝑘
𝑛
𝑘=1 (𝑥)
𝑑𝑘𝑦
𝑑𝑥𝑘
(𝑥) = ∑ ℎ𝑘
𝑛
𝑘=1 (𝑥) and of the 2𝑛
− 1
ordinary differential equations that we can form with their 𝑛 terms.
Proof: It is an immediate consequence of the construction of 𝑦
̃(𝑥) and the
hypothesis (16).
Observation 1. As in Observation 2 of [3], if the hypotheses (16) are not
satisfied, we can still, in the present case, try to find a solution for the
ordinary differential equation from the search for some auxiliary
functions, solutions of integral equations in a single variable.
Observation 2. Imitating what was done in [3], we can note that the
method established there also applies to any linear partial differential
equations. In such a way that the method is applicable to any linear
differential equations, whether ordinary or partial.
3. Aplication: Generalized Euler-Tricomi equation (
𝝏𝟐𝒖
𝝏𝒙𝟐
+ 𝒈(𝒙)
𝝏𝟐𝒖
𝝏𝒚𝟐
= 𝟎)
In order 2, the method established above solves the main partial
differential equations of Physics and Technology. Here, we'll make a brief
application to the generalized Euler-Tricomi equation,
𝜕2
𝑢
𝜕𝑥2
+ 𝑔(𝑥)
𝜕2
𝑢
𝜕𝑦2
= 0 (17)

4. 4
(
𝜕2
𝑢
𝜕𝑥2
0
0 𝑔(𝑥)
𝜕2
𝑢
𝜕𝑦2
)
= (
𝑓(𝑥, 𝑦) 0
0 −𝑓(𝑥, 𝑦)
) (18)
Here we have:
𝑢11(𝑥, 𝑦) = ∬ 𝑓 𝑑𝑥𝑑𝑥 + 𝑘1(𝑦)𝑥 + 𝑘2(𝑦) (19)
𝑢22(𝑥, 𝑦) = − ∬
𝑓
𝑔(𝑥)
𝑑𝑦𝑑𝑦 + 𝑣(𝑥)𝑦 + 𝑣2(𝑥) (20)
𝑢
̃ = ∬ 𝑓 𝑑𝑥𝑑𝑥 −
1
𝑔(𝑥)
∬ 𝑓(𝑥, 𝑦) 𝑑𝑦𝑑𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 + 𝑑 (21)
𝜕2
𝑢
̃
𝜕𝑥2
= 𝑓 − 𝑔(𝑥)−3
∬ (𝑔2
𝜕2
𝑓
𝜕𝑥2
− 𝑓(𝑥, 𝑦)𝑔(𝑥)𝑔′′
(𝑥) − 2𝑔𝑔′
𝜕𝑓
𝜕𝑥
+ 2𝑓𝑔′𝑔′) 𝑑𝑦𝑑𝑦 (22)
𝜕2
𝑢
̃
𝜕𝑦2
= −𝑔−1
𝑓 + ∬
𝜕2
𝑓
𝜕𝑦2
𝑑𝑥𝑑𝑥 (23)
Logo,
Then,
𝜕2
𝑢
̃
𝜕𝑥2
+ 𝑔
𝜕2
𝑢
̃
𝜕𝑦2
=
−𝑔−3
∬ (𝑔2
𝜕2
𝑓
𝜕𝑥2
− 𝑓𝑔𝑔′′
− 2𝑔𝑔′
𝜕𝑓
𝜕𝑥
+ 2𝑓𝑔′
𝑔′
) 𝑑𝑦𝑑𝑦 + 𝑔 ∬
𝜕2
𝑓
𝜕𝑦2
𝑑𝑥𝑑𝑥 (24)
From this last equation, let us note that if
𝑔 ∬
𝜕2
𝑓
𝜕𝑦2
𝑑𝑥𝑑𝑥 = 𝑔−3
∬ (𝑔2
𝜕2
𝑓
𝜕𝑥2
− 𝑓𝑔𝑔′′
− 2𝑔𝑔′
𝜕𝑓
𝜕𝑥
+ 2𝑓𝑔′
𝑔′
) 𝑑𝑦𝑑𝑦 (25)
then we have
𝜕2𝑢
𝜕𝑥2
+ 𝑥
𝜕2𝑢
𝜕𝑦2
= 0.
Finding all pairs (𝑓(𝑥, 𝑦), 𝑔(𝑥)) satisfying this integral equation would give
the maximum the method could give.
Comment 3.1. (Looking for 𝑓 in the form 𝑓(𝑥, 𝑦) = 𝐹(𝑥)𝐺(𝑦)). Given a
function 𝑔(𝑥) in any domain, it does not seem an easy task to find the
corresponding function 𝑓(𝑥, 𝑦) that satisfies equation (25). Here, we will
look for them in the form of separable variables or also exponential.
Making 𝑓(𝑥, 𝑦) = 𝐹(𝑥)𝐺(𝑦) in (25) we have

5. 5
𝐺′′(𝑦)
∬ 𝐺(𝑦) 𝑑𝑦𝑑𝑦
=
𝑔(𝑔𝐹′′
− 2𝑔′
𝐹′
) + (2𝑔′
𝑔′
− 𝑔𝑔′′)𝐹
𝑔4 ∬ 𝐹(𝑥) 𝑑𝑥𝑑𝑥
= 𝑘 (26)
where 𝑘 is any constant.
With the notations
𝛼(𝑥) = ∬ 𝐹(𝑥) 𝑑𝑥𝑑𝑥 (27)
𝛽(𝑦) = ∬ 𝐺(𝑦) 𝑑𝑦𝑑𝑦 (28)
we will have
𝑔(𝑔𝛼′′′′
− 2𝑔′
𝛼′′′) + (2𝑔′
𝑔′
− 𝑔𝑔′′)𝛼′′
− 𝑘𝑔4
𝛼 = 0 (29)
𝛽′′′′
− 𝑘𝛽 = 0 (30)
(29) reduces to
𝑔(𝑔𝛼′′′′
− 2𝑔′
𝛼′′′) − 𝑘𝑔4
𝛼 = 0 (31)
case
2𝑔′
𝑔′
− 𝑔𝑔′′
= 0 (32)
(32) is equivalent to
𝑔 =
𝑐2
𝑐1 + 𝑥
(33)
Replacing (33) in (31) we will have
𝑐2(𝑐1 + 𝑥)2
𝛼′′′′
+ 2(𝑐1 + 𝑥)𝛼′′′
− 𝑘𝑐2
3
𝛼 = 0 (34)
Let: 𝑐2 ≝ 𝑏; 𝑐1 + 𝑥 ≝ 𝑧; 𝑘𝑐2
3
≝ 𝑎; 𝑦(𝑧) = 𝛼(𝑥(𝑧)). Then, (34) becomes
𝑏𝑧2
𝑦′′′′(𝑧) + 2𝑧𝑦′′′(𝑧) − 𝑎𝑦(𝑧) = 0 (35)
The solution of (35) por Wolfram Alpha is:
𝑦(𝑧) = (
𝑖
4
)
3−2/𝑏
𝑏
1
𝑏
−3/2
𝑐̃1𝑎
3
2
−1/𝑏
𝑧3−2/𝑏
𝐹3
0 (;
3
2
−
1
𝑏
, 2 −
1
𝑏
,
5
2
−
1
𝑏
;
𝑎𝑧2
16𝑏
)+
𝑎𝑐̃3𝑧2
𝐹3
0 (;
3
2
, 2,
1
2
+
1
𝑏
;
𝑎𝑧2
16𝑏
)
𝑏
+
𝑖√𝑎𝑐̃2𝑧 𝐹3
0 (;
1
2
,
3
2
,
1
𝑏
;
𝑎𝑧2
16𝑏
)
4√𝑏
+
𝑐̃4𝐺0,4
2,0
(
𝑎𝑧2
16𝑏
|0,1,
1
2
,
3
2
−
1
𝑏
) (36)
where:
𝐹
𝑞
𝑝 (𝑎1, 𝑎2, … , 𝑎𝑝; 𝑏1, 𝑏2, … , 𝑏𝑞; 𝑧) is the “generalized hypergeometric
function”,

6. 6
𝐺𝑝,𝑞
𝑚,𝑛
(𝑧 |𝑏1,𝑏2,…,𝑏𝑞
𝑎1,𝑎2,…,𝑎𝑝
) is the “Meijer G-function” and 𝑐̃𝑖, 𝑖 = 1,2,3,4, are
constant.
From 𝑧 = 𝑥 + 𝑐1, 𝛼(𝑥) = 𝑦(𝑥 + 𝑐1) and 𝛼(𝑥) = ∬ 𝐹(𝑥) 𝑑𝑥𝑑𝑥 it is concluded
that 𝐹(𝑥) = 𝑦′′(𝑥 + 𝑐1). Then, 𝑓(𝑥, 𝑦) = 𝐹(𝑥)𝐺(𝑦) = 𝑦′′(𝑥 + 𝑐1)𝛽′′(𝑦). The
solution of the generalized Euler-Tricomi equation will then be:
𝑢
̃ = ∬ 𝑦′′(𝑥 + 𝑐1)𝛽′′(𝑦)𝑑𝑥𝑑𝑥 −
1
𝑔(𝑥)
∬ 𝑦′′(𝑥 + 𝑐1)𝛽′′(𝑦) 𝑑𝑦𝑑𝑦 + 𝑎𝑥𝑦 + 𝑏𝑥 + 𝑐𝑦 +
𝑑 = 𝛽′′(𝑦)𝑦(𝑥 + 𝑐1) −
1
𝑔(𝑥)
𝑦′′(𝑥 + 𝑐1)𝛽(𝑦) (37)
where
𝛽(𝑦) = 𝑑2𝑒− √𝑘
4
𝑦
+ 𝑑4𝑒 √𝑘
4
𝑦
+ 𝑑3𝑠𝑒𝑛(√𝑘
4
𝑦) + 𝑑1𝑐𝑜𝑠(√𝑘
4
𝑦) (38)
is the solution of (30).
Comment 3.2. Equation (29) is atypical, as it is a “differential equation”
in two unknown functions α(x) and g(x), which must be searched coupled
in the same equation and with equal domains. Usual techniques do not
allow you to find your solutions easily. Next, we'll try one more specific
situation.
Comment 3.3. (Hypotheses 𝑔𝛼′′′′
− 2𝑔′
𝛼′′′
). From
𝑔𝛼′′′′
− 2𝑔′
𝛼′′′
(39)
results
𝛼′′′′
𝛼′′′
= 2
𝑔′
𝑔
(40)
Let
𝛼′′′
= 𝜃 (41)
Then, we have, from (40)
𝑑𝜃
𝑑𝑥
𝜃
= 2
𝑑𝑔
𝑑𝑥
𝑔
(42)
and of this,
∫
1
𝜃
𝑑𝜃 = ∫
1
𝑔
𝑑𝑔 (43)
𝜃 = 𝑔2
(44)
From this, from (29), (30) and (41) follow
(2𝑔′
𝑔′
− 𝑔𝑔′′)𝛼′′ − 𝑘𝑔4
𝛼 = 0 (45)
𝛼′′′(𝑥) = 𝑔2(𝑥) (46)

7. 7
𝛼 = ∭ 𝑔2
(𝑥) 𝑑𝑥𝑑𝑥𝑑𝑥 + 𝑐1𝑥2
+ 𝑐2𝑥 + 𝑐3 (47)
Note that the solution to the generalized Euler-Tricomi equation, in this
case, is conditioned on finding the difficult solution of the nonlinear
equation (45).
The various cases presented above, plus the historical facts lead us to
formalize the following concept.
4. “Paraordinary” differential equations
The manuals define ordinary differential equations as follows: “An
equation with a function in one independent variable as unknown, containing not
only the unknown function itself, but also its derivatives of various orders” [1].
If in the equation there is more than one unknown function with the same
independent variable in the same domain (along with their respective
derivatives), we will use the name “paraordinary” differential equation.
The solutions of these “paraequations” are explicit or implicit functions
involving these unknown functions, which satisfy the “paraequation”,
that is, when substituted in the “paraequation” they transform it into a
true identity.
Example (Derivative of linear momentum).
Consider the linear momentum equation:
𝑑
𝑑𝑡
(𝑚(𝑡)𝑣(𝑡)) = 𝐹(𝑡); 𝑣(𝑡) =
𝑑𝑥
𝑑𝑡
(𝑡) = 𝑣(𝑡) (48)
with 𝑥(𝑡) position vector along the real line, 𝑚(𝑡) the variable mass and
𝑣(𝑡) the speed of the particle at the instant 𝑡.
By derivation of (48), we have
𝑚(𝑡)
𝑑2
𝑥
𝑑𝑡2
(𝑡) +
𝑑𝑚
𝑑𝑡
(𝑡)
𝑑𝑥
𝑑𝑡
(𝑡) = 2𝐹(𝑡) − 𝐹(𝑡) (49)
Consider 𝑚(𝑡) > 0, ∀𝑡 ∈ ℝ. We have: 2 ∬
𝐹(𝑡)
𝑚(𝑡)
𝑑𝑡𝑑𝑡 is solution of 𝑚(𝑡)
𝑑2𝑥
𝑑𝑡2
(𝑡) =
2𝐹(𝑡) and − ∫
𝐹(𝑡)
𝑑𝑚
𝑑𝑡
(𝑡)
𝑑𝑡 is solution of
𝑑𝑚
𝑑𝑡
(𝑡)
𝑑𝑥
𝑑𝑡
(𝑡) = −𝐹(𝑡).
𝑢
̃(𝑡) = − ∫
𝐹(𝑡)
𝑑𝑚
𝑑𝑡
(𝑡)
𝑑𝑡 + 2 ∬
𝐹(𝑡)
𝑚(𝑡)
𝑑𝑡𝑑𝑡 is solution of (49) if
𝑑𝑚
𝑑𝑡
(𝑡) ∫
2𝐹(𝑡)
𝑚(𝑡)
𝑑𝑡 −
𝑑𝐹
𝑑𝑡
(𝑡)𝑚(𝑡)
𝑑𝑚
𝑑𝑡
(𝑡) − 𝐹(𝑡)𝑚(𝑡)
𝑑2
𝑚
𝑑𝑡2 (𝑡)
𝑑𝑚
𝑑𝑡
(𝑡)
𝑑𝑚
𝑑𝑡
(𝑡)
= 0 (50)
From solutions for (50) results solutions for (49).

8. 8
Conclusion
The presented method partially solves the generalized Euler-
Tricomi equation in a wide range, with the possibility of nine parameters.
The method complements the classic results of [2]. The exposition above
makes it clear that less restrictive conditions on the function 𝑓(𝑥, 𝑦), that
the separation of variables 𝑓(𝑥, 𝑦) = 𝐹(𝑥)𝐺(𝑦), can generate other families
of solutions.
References
[1]. Differential equation, ordinary. Encyclopedia of Mathematics. URL:
http://encyclopediaofmath.org/index.php?title=Differential_equation,_o
rdinary&oldid=50981
[2]. GERALD B. FOLLAND. Introduction to Partial Differential Equations,
2nd ed. Princeton University Press, New Jersey, 1995.
[3]. MIRANDA, Lohans de O. and MIRANDA, Lossian B. B. One Solution
for Many Linear Partial Differential Equations With Terms of Equal
Orders, Journal of Nepal Mathematical Society (JNMS), Vol. 4, Issue 2
(2021).
[4]. MIRANDA, Lohans de O. and MIRANDA, Lossian B. B. Novas
Informações Sobre a Equação de Euler-Tricomi. pt.slideshare.net. Visto
em 04.01.2022.
[5]. MIRANDA, Lohans de O. and MIRANDA, Lossian B. B. Novas
Informações Sobre a Equação Generalizada de Euler-Tricomi.
https://pt.slideshare.net/lossian/novas-informaes-sobre-a-equao-
generalizada-de-eulertricomi. Seen on 10.01.2022.
[6]. Tricomi equation. Encyclopedia of Mathematics. URL:
http://encyclopediaofmath.org/index.php?title=Tricomi_equation&oldid
=33467 . In Dec, 24, 2021.