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Patrick Rea
Artikel: Exact differential equation GudangMovies21 Rebahinxxi
In mathematics, an exact differential equation or total differential equation is a certain kind of ordinary differential equation which is widely used in physics and engineering.
Definition
Given a simply connected and open subset D of
R
2
{\displaystyle \mathbb {R} ^{2}}
and two functions I and J which are continuous on D, an implicit first-order ordinary differential equation of the form
I
(
x
,
y
)
d
x
+
J
(
x
,
y
)
d
y
=
0
,
{\displaystyle I(x,y)\,dx+J(x,y)\,dy=0,}
is called an exact differential equation if there exists a continuously differentiable function F, called the potential function, so that
∂
F
∂
x
=
I
{\displaystyle {\frac {\partial F}{\partial x}}=I}
and
∂
F
∂
y
=
J
.
{\displaystyle {\frac {\partial F}{\partial y}}=J.}
An exact equation may also be presented in the following form:
I
(
x
,
y
)
+
J
(
x
,
y
)
y
′
(
x
)
=
0
{\displaystyle I(x,y)+J(x,y)\,y'(x)=0}
where the same constraints on I and J apply for the differential equation to be exact.
The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function
F
(
x
0
,
x
1
,
.
.
.
,
x
n
−
1
,
x
n
)
{\displaystyle F(x_{0},x_{1},...,x_{n-1},x_{n})}
, the exact or total derivative with respect to
x
0
{\displaystyle x_{0}}
is given by
d
F
d
x
0
=
∂
F
∂
x
0
+
∑
i
=
1
n
∂
F
∂
x
i
d
x
i
d
x
0
.
{\displaystyle {\frac {dF}{dx_{0}}}={\frac {\partial F}{\partial x_{0}}}+\sum _{i=1}^{n}{\frac {\partial F}{\partial x_{i}}}{\frac {dx_{i}}{dx_{0}}}.}
= Example
=The function
F
:
R
2
→
R
{\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} }
given by
F
(
x
,
y
)
=
1
2
(
x
2
+
y
2
)
+
c
{\displaystyle F(x,y)={\frac {1}{2}}(x^{2}+y^{2})+c}
is a potential function for the differential equation
x
d
x
+
y
d
y
=
0.
{\displaystyle x\,dx+y\,dy=0.\,}
First-order exact differential equations
= Identifying first-order exact differential equations
=Let the functions
M
{\textstyle M}
,
N
{\textstyle N}
,
M
y
{\textstyle M_{y}}
, and
N
x
{\textstyle N_{x}}
, where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region
R
:
α
<
x
<
β
,
γ
<
y
<
δ
{\textstyle R:\alpha
. Then the differential equation
M
(
x
,
y
)
+
N
(
x
,
y
)
d
y
d
x
=
0
{\displaystyle M(x,y)+N(x,y){\frac {dy}{dx}}=0}
is exact if and only if
M
y
(
x
,
y
)
=
N
x
(
x
,
y
)
{\displaystyle M_{y}(x,y)=N_{x}(x,y)}
That is, there exists a function
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
, called a potential function, such that
ψ
x
(
x
,
y
)
=
M
(
x
,
y
)
and
ψ
y
(
x
,
y
)
=
N
(
x
,
y
)
{\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}
So, in general:
M
y
(
x
,
y
)
=
N
x
(
x
,
y
)
⟺
{
∃
ψ
(
x
,
y
)
ψ
x
(
x
,
y
)
=
M
(
x
,
y
)
ψ
y
(
x
,
y
)
=
N
(
x
,
y
)
{\displaystyle M_{y}(x,y)=N_{x}(x,y)\iff {\begin{cases}\exists \psi (x,y)\\\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}
Proof
The proof has two parts.
First, suppose there is a function
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
such that
ψ
x
(
x
,
y
)
=
M
(
x
,
y
)
and
ψ
y
(
x
,
y
)
=
N
(
x
,
y
)
{\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}
It then follows that
M
y
(
x
,
y
)
=
ψ
x
y
(
x
,
y
)
and
N
x
(
x
,
y
)
=
ψ
y
x
(
x
,
y
)
{\displaystyle M_{y}(x,y)=\psi _{xy}(x,y){\text{ and }}N_{x}(x,y)=\psi _{yx}(x,y)}
Since
M
y
{\displaystyle M_{y}}
and
N
x
{\displaystyle N_{x}}
are continuous, then
ψ
x
y
{\displaystyle \psi _{xy}}
and
ψ
y
x
{\displaystyle \psi _{yx}}
are also continuous which guarantees their equality.
The second part of the proof involves the construction of
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
and can also be used as a procedure for solving first-order exact differential equations. Suppose that
M
y
(
x
,
y
)
=
N
x
(
x
,
y
)
{\displaystyle M_{y}(x,y)=N_{x}(x,y)}
and let there be a function
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
for which
ψ
x
(
x
,
y
)
=
M
(
x
,
y
)
and
ψ
y
(
x
,
y
)
=
N
(
x
,
y
)
{\displaystyle \psi _{x}(x,y)=M(x,y){\text{ and }}\psi _{y}(x,y)=N(x,y)}
Begin by integrating the first equation with respect to
x
{\displaystyle x}
. In practice, it doesn't matter if you integrate the first or the second equation, so long as the integration is done with respect to the appropriate variable.
∂
ψ
∂
x
(
x
,
y
)
=
M
(
x
,
y
)
{\displaystyle {\frac {\partial \psi }{\partial x}}(x,y)=M(x,y)}
ψ
(
x
,
y
)
=
∫
M
(
x
,
y
)
d
x
+
h
(
y
)
{\displaystyle \psi (x,y)=\int M(x,y)\,dx+h(y)}
ψ
(
x
,
y
)
=
Q
(
x
,
y
)
+
h
(
y
)
{\displaystyle \psi (x,y)=Q(x,y)+h(y)}
where
Q
(
x
,
y
)
{\displaystyle Q(x,y)}
is any differentiable function such that
Q
x
=
M
{\displaystyle Q_{x}=M}
. The function
h
(
y
)
{\displaystyle h(y)}
plays the role of a constant of integration, but instead of just a constant, it is function of
y
{\displaystyle y}
, since
M
{\displaystyle M}
is a function of both
x
{\displaystyle x}
and
y
{\displaystyle y}
and we are only integrating with respect to
x
{\displaystyle x}
.
Now to show that it is always possible to find an
h
(
y
)
{\displaystyle h(y)}
such that
ψ
y
=
N
{\displaystyle \psi _{y}=N}
.
ψ
(
x
,
y
)
=
Q
(
x
,
y
)
+
h
(
y
)
{\displaystyle \psi (x,y)=Q(x,y)+h(y)}
Differentiate both sides with respect to
y
{\displaystyle y}
.
∂
ψ
∂
y
(
x
,
y
)
=
∂
Q
∂
y
(
x
,
y
)
+
h
′
(
y
)
{\displaystyle {\frac {\partial \psi }{\partial y}}(x,y)={\frac {\partial Q}{\partial y}}(x,y)+h'(y)}
Set the result equal to
N
{\displaystyle N}
and solve for
h
′
(
y
)
{\displaystyle h'(y)}
.
h
′
(
y
)
=
N
(
x
,
y
)
−
∂
Q
∂
y
(
x
,
y
)
{\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)}
In order to determine
h
′
(
y
)
{\displaystyle h'(y)}
from this equation, the right-hand side must depend only on
y
{\displaystyle y}
. This can be proven by showing that its derivative with respect to
x
{\displaystyle x}
is always zero, so differentiate the right-hand side with respect to
x
{\displaystyle x}
.
∂
N
∂
x
(
x
,
y
)
−
∂
∂
x
∂
Q
∂
y
(
x
,
y
)
⟺
∂
N
∂
x
(
x
,
y
)
−
∂
∂
y
∂
Q
∂
x
(
x
,
y
)
{\displaystyle {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial x}}{\frac {\partial Q}{\partial y}}(x,y)\iff {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial }{\partial y}}{\frac {\partial Q}{\partial x}}(x,y)}
Since
Q
x
=
M
{\displaystyle Q_{x}=M}
,
∂
N
∂
x
(
x
,
y
)
−
∂
M
∂
y
(
x
,
y
)
{\displaystyle {\frac {\partial N}{\partial x}}(x,y)-{\frac {\partial M}{\partial y}}(x,y)}
Now, this is zero based on our initial supposition that
M
y
(
x
,
y
)
=
N
x
(
x
,
y
)
{\displaystyle M_{y}(x,y)=N_{x}(x,y)}
Therefore,
h
′
(
y
)
=
N
(
x
,
y
)
−
∂
Q
∂
y
(
x
,
y
)
{\displaystyle h'(y)=N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)}
h
(
y
)
=
∫
(
N
(
x
,
y
)
−
∂
Q
∂
y
(
x
,
y
)
)
d
y
{\displaystyle h(y)=\int {\left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right)dy}}
ψ
(
x
,
y
)
=
Q
(
x
,
y
)
+
∫
(
N
(
x
,
y
)
−
∂
Q
∂
y
(
x
,
y
)
)
d
y
+
C
{\displaystyle \psi (x,y)=Q(x,y)+\int \left(N(x,y)-{\frac {\partial Q}{\partial y}}(x,y)\right)\,dy+C}
And this completes the proof.
= Solutions to first-order exact differential equations
=First-order exact differential equations of the form
M
(
x
,
y
)
+
N
(
x
,
y
)
d
y
d
x
=
0
{\displaystyle M(x,y)+N(x,y){\frac {dy}{dx}}=0}
can be written in terms of the potential function
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
∂
ψ
∂
x
+
∂
ψ
∂
y
d
y
d
x
=
0
{\displaystyle {\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0}
where
{
ψ
x
(
x
,
y
)
=
M
(
x
,
y
)
ψ
y
(
x
,
y
)
=
N
(
x
,
y
)
{\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}
This is equivalent to taking the total derivative of
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
.
∂
ψ
∂
x
+
∂
ψ
∂
y
d
y
d
x
=
0
⟺
d
d
x
ψ
(
x
,
y
(
x
)
)
=
0
{\displaystyle {\frac {\partial \psi }{\partial x}}+{\frac {\partial \psi }{\partial y}}{\frac {dy}{dx}}=0\iff {\frac {d}{dx}}\psi (x,y(x))=0}
The solutions to an exact differential equation are then given by
ψ
(
x
,
y
(
x
)
)
=
c
{\displaystyle \psi (x,y(x))=c}
and the problem reduces to finding
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
.
This can be done by integrating the two expressions
M
(
x
,
y
)
d
x
{\displaystyle M(x,y)\,dx}
and
N
(
x
,
y
)
d
y
{\displaystyle N(x,y)\,dy}
and then writing down each term in the resulting expressions only once and summing them up in order to get
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
.
The reasoning behind this is the following. Since
{
ψ
x
(
x
,
y
)
=
M
(
x
,
y
)
ψ
y
(
x
,
y
)
=
N
(
x
,
y
)
{\displaystyle {\begin{cases}\psi _{x}(x,y)=M(x,y)\\\psi _{y}(x,y)=N(x,y)\end{cases}}}
it follows, by integrating both sides, that
{
ψ
(
x
,
y
)
=
∫
M
(
x
,
y
)
d
x
+
h
(
y
)
=
Q
(
x
,
y
)
+
h
(
y
)
ψ
(
x
,
y
)
=
∫
N
(
x
,
y
)
d
y
+
g
(
x
)
=
P
(
x
,
y
)
+
g
(
x
)
{\displaystyle {\begin{cases}\psi (x,y)=\int M(x,y)\,dx+h(y)=Q(x,y)+h(y)\\\psi (x,y)=\int N(x,y)\,dy+g(x)=P(x,y)+g(x)\end{cases}}}
Therefore,
Q
(
x
,
y
)
+
h
(
y
)
=
P
(
x
,
y
)
+
g
(
x
)
{\displaystyle Q(x,y)+h(y)=P(x,y)+g(x)}
where
Q
(
x
,
y
)
{\displaystyle Q(x,y)}
and
P
(
x
,
y
)
{\displaystyle P(x,y)}
are differentiable functions such that
Q
x
=
M
{\displaystyle Q_{x}=M}
and
P
y
=
N
{\displaystyle P_{y}=N}
.
In order for this to be true and for both sides to result in the exact same expression, namely
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
, then
h
(
y
)
{\displaystyle h(y)}
must be contained within the expression for
P
(
x
,
y
)
{\displaystyle P(x,y)}
because it cannot be contained within
g
(
x
)
{\displaystyle g(x)}
, since it is entirely a function of
y
{\displaystyle y}
and not
x
{\displaystyle x}
and is therefore not allowed to have anything to do with
x
{\displaystyle x}
. By analogy,
g
(
x
)
{\displaystyle g(x)}
must be contained within the expression
Q
(
x
,
y
)
{\displaystyle Q(x,y)}
.
Ergo,
Q
(
x
,
y
)
=
g
(
x
)
+
f
(
x
,
y
)
and
P
(
x
,
y
)
=
h
(
y
)
+
d
(
x
,
y
)
{\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+d(x,y)}
for some expressions
f
(
x
,
y
)
{\displaystyle f(x,y)}
and
d
(
x
,
y
)
{\displaystyle d(x,y)}
.
Plugging in into the above equation, we find that
g
(
x
)
+
f
(
x
,
y
)
+
h
(
y
)
=
h
(
y
)
+
d
(
x
,
y
)
+
g
(
x
)
⇒
f
(
x
,
y
)
=
d
(
x
,
y
)
{\displaystyle g(x)+f(x,y)+h(y)=h(y)+d(x,y)+g(x)\Rightarrow f(x,y)=d(x,y)}
and so
f
(
x
,
y
)
{\displaystyle f(x,y)}
and
d
(
x
,
y
)
{\displaystyle d(x,y)}
turn out to be the same function. Therefore,
Q
(
x
,
y
)
=
g
(
x
)
+
f
(
x
,
y
)
and
P
(
x
,
y
)
=
h
(
y
)
+
f
(
x
,
y
)
{\displaystyle Q(x,y)=g(x)+f(x,y){\text{ and }}P(x,y)=h(y)+f(x,y)}
Since we already showed that
{
ψ
(
x
,
y
)
=
Q
(
x
,
y
)
+
h
(
y
)
ψ
(
x
,
y
)
=
P
(
x
,
y
)
+
g
(
x
)
{\displaystyle {\begin{cases}\psi (x,y)=Q(x,y)+h(y)\\\psi (x,y)=P(x,y)+g(x)\end{cases}}}
it follows that
ψ
(
x
,
y
)
=
g
(
x
)
+
f
(
x
,
y
)
+
h
(
y
)
{\displaystyle \psi (x,y)=g(x)+f(x,y)+h(y)}
So, we can construct
ψ
(
x
,
y
)
{\displaystyle \psi (x,y)}
by doing
∫
M
(
x
,
y
)
d
x
{\displaystyle \int M(x,y)\,dx}
and
∫
N
(
x
,
y
)
d
y
{\displaystyle \int N(x,y)\,dy}
and then taking the common terms we find within the two resulting expressions (that would be
f
(
x
,
y
)
{\displaystyle f(x,y)}
) and then adding the terms which are uniquely found in either one of them –
g
(
x
)
{\displaystyle g(x)}
and
h
(
y
)
{\displaystyle h(y)}
.
Second-order exact differential equations
The concept of exact differential equations can be extended to second-order equations. Consider starting with the first-order exact equation:
I
(
x
,
y
)
+
J
(
x
,
y
)
d
y
d
x
=
0
{\displaystyle I(x,y)+J(x,y){dy \over dx}=0}
Since both functions
I
(
x
,
y
)
{\displaystyle I(x,y)}
,
J
(
x
,
y
)
{\displaystyle J(x,y)}
are functions of two variables, implicitly differentiating the multivariate function yields
d
I
d
x
+
(
d
J
d
x
)
d
y
d
x
+
d
2
y
d
x
2
(
J
(
x
,
y
)
)
=
0
{\displaystyle {dI \over dx}+\left({dJ \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}
Expanding the total derivatives gives that
d
I
d
x
=
∂
I
∂
x
+
∂
I
∂
y
d
y
d
x
{\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}
and that
d
J
d
x
=
∂
J
∂
x
+
∂
J
∂
y
d
y
d
x
{\displaystyle {dJ \over dx}={\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}}
Combining the
d
y
d
x
{\textstyle {dy \over dx}}
terms gives
∂
I
∂
x
+
d
y
d
x
(
∂
I
∂
y
+
∂
J
∂
x
+
∂
J
∂
y
d
y
d
x
)
+
d
2
y
d
x
2
(
J
(
x
,
y
)
)
=
0
{\displaystyle {\partial I \over \partial x}+{dy \over dx}\left({\partial I \over \partial y}+{\partial J \over \partial x}+{\partial J \over \partial y}{dy \over dx}\right)+{d^{2}y \over dx^{2}}(J(x,y))=0}
If the equation is exact, then
∂
J
∂
x
=
∂
I
∂
y
{\textstyle {\partial J \over \partial x}={\partial I \over \partial y}}
. Additionally, the total derivative of
J
(
x
,
y
)
{\displaystyle J(x,y)}
is equal to its implicit ordinary derivative
d
J
d
x
{\textstyle {dJ \over dx}}
. This leads to the rewritten equation
∂
I
∂
x
+
d
y
d
x
(
∂
J
∂
x
+
d
J
d
x
)
+
d
2
y
d
x
2
(
J
(
x
,
y
)
)
=
0
{\displaystyle {\partial I \over \partial x}+{dy \over dx}\left({\partial J \over \partial x}+{dJ \over dx}\right)+{d^{2}y \over dx^{2}}(J(x,y))=0}
Now, let there be some second-order differential equation
f
(
x
,
y
)
+
g
(
x
,
y
,
d
y
d
x
)
d
y
d
x
+
d
2
y
d
x
2
(
J
(
x
,
y
)
)
=
0
{\displaystyle f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}
If
∂
J
∂
x
=
∂
I
∂
y
{\displaystyle {\partial J \over \partial x}={\partial I \over \partial y}}
for exact differential equations, then
∫
(
∂
I
∂
y
)
d
y
=
∫
(
∂
J
∂
x
)
d
y
{\displaystyle \int \left({\partial I \over \partial y}\right)\,dy=\int \left({\partial J \over \partial x}\right)\,dy}
and
∫
(
∂
I
∂
y
)
d
y
=
∫
(
∂
J
∂
x
)
d
y
=
I
(
x
,
y
)
−
h
(
x
)
{\displaystyle \int \left({\partial I \over \partial y}\right)\,dy=\int \left({\partial J \over \partial x}\right)\,dy=I(x,y)-h(x)}
where
h
(
x
)
{\displaystyle h(x)}
is some arbitrary function only of
x
{\displaystyle x}
that was differentiated away to zero upon taking the partial derivative of
I
(
x
,
y
)
{\displaystyle I(x,y)}
with respect to
y
{\displaystyle y}
. Although the sign on
h
(
x
)
{\displaystyle h(x)}
could be positive, it is more intuitive to think of the integral's result as
I
(
x
,
y
)
{\displaystyle I(x,y)}
that is missing some original extra function
h
(
x
)
{\displaystyle h(x)}
that was partially differentiated to zero.
Next, if
d
I
d
x
=
∂
I
∂
x
+
∂
I
∂
y
d
y
d
x
{\displaystyle {dI \over dx}={\partial I \over \partial x}+{\partial I \over \partial y}{dy \over dx}}
then the term
∂
I
∂
x
{\displaystyle {\partial I \over \partial x}}
should be a function only of
x
{\displaystyle x}
and
y
{\displaystyle y}
, since partial differentiation with respect to
x
{\displaystyle x}
will hold
y
{\displaystyle y}
constant and not produce any derivatives of
y
{\displaystyle y}
. In the second-order equation
f
(
x
,
y
)
+
g
(
x
,
y
,
d
y
d
x
)
d
y
d
x
+
d
2
y
d
x
2
(
J
(
x
,
y
)
)
=
0
{\displaystyle f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}
only the term
f
(
x
,
y
)
{\displaystyle f(x,y)}
is a term purely of
x
{\displaystyle x}
and
y
{\displaystyle y}
. Let
∂
I
∂
x
=
f
(
x
,
y
)
{\displaystyle {\partial I \over \partial x}=f(x,y)}
. If
∂
I
∂
x
=
f
(
x
,
y
)
{\displaystyle {\partial I \over \partial x}=f(x,y)}
, then
f
(
x
,
y
)
=
d
I
d
x
−
∂
I
∂
y
d
y
d
x
{\displaystyle f(x,y)={dI \over dx}-{\partial I \over \partial y}{dy \over dx}}
Since the total derivative of
I
(
x
,
y
)
{\displaystyle I(x,y)}
with respect to
x
{\displaystyle x}
is equivalent to the implicit ordinary derivative
d
I
d
x
{\displaystyle {dI \over dx}}
, then
f
(
x
,
y
)
+
∂
I
∂
y
d
y
d
x
=
d
I
d
x
=
d
d
x
(
I
(
x
,
y
)
−
h
(
x
)
)
+
d
h
(
x
)
d
x
{\displaystyle f(x,y)+{\partial I \over \partial y}{dy \over dx}={dI \over dx}={d \over dx}(I(x,y)-h(x))+{dh(x) \over dx}}
So,
d
h
(
x
)
d
x
=
f
(
x
,
y
)
+
∂
I
∂
y
d
y
d
x
−
d
d
x
(
I
(
x
,
y
)
−
h
(
x
)
)
{\displaystyle {dh(x) \over dx}=f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))}
and
h
(
x
)
=
∫
(
f
(
x
,
y
)
+
∂
I
∂
y
d
y
d
x
−
d
d
x
(
I
(
x
,
y
)
−
h
(
x
)
)
)
d
x
{\displaystyle h(x)=\int \left(f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))\right)\,dx}
Thus, the second-order differential equation
f
(
x
,
y
)
+
g
(
x
,
y
,
d
y
d
x
)
d
y
d
x
+
d
2
y
d
x
2
(
J
(
x
,
y
)
)
=
0
{\displaystyle f(x,y)+g\left(x,y,{dy \over dx}\right){dy \over dx}+{d^{2}y \over dx^{2}}(J(x,y))=0}
is exact only if
g
(
x
,
y
,
d
y
d
x
)
=
d
J
d
x
+
∂
J
∂
x
=
d
J
d
x
+
∂
J
∂
x
{\displaystyle g\left(x,y,{dy \over dx}\right)={dJ \over dx}+{\partial J \over \partial x}={dJ \over dx}+{\partial J \over \partial x}}
and only if the below expression
∫
(
f
(
x
,
y
)
+
∂
I
∂
y
d
y
d
x
−
d
d
x
(
I
(
x
,
y
)
−
h
(
x
)
)
)
d
x
=
∫
(
f
(
x
,
y
)
−
∂
(
I
(
x
,
y
)
−
h
(
x
)
)
∂
x
)
d
x
{\displaystyle \int \left(f(x,y)+{\partial I \over \partial y}{dy \over dx}-{d \over dx}(I(x,y)-h(x))\right)\,dx=\int \left(f(x,y)-{\partial \left(I(x,y)-h(x)\right) \over \partial x}\right)\,dx}
is a function solely of
x
{\displaystyle x}
. Once
h
(
x
)
{\displaystyle h(x)}
is calculated with its arbitrary constant, it is added to
I
(
x
,
y
)
−
h
(
x
)
{\displaystyle I(x,y)-h(x)}
to make
I
(
x
,
y
)
{\displaystyle I(x,y)}
. If the equation is exact, then we can reduce to the first-order exact form which is solvable by the usual method for first-order exact equations.
I
(
x
,
y
)
+
J
(
x
,
y
)
d
y
d
x
=
0
{\displaystyle I(x,y)+J(x,y){dy \over dx}=0}
Now, however, in the final implicit solution there will be a
C
1
x
{\displaystyle C_{1}x}
term from integration of
h
(
x
)
{\displaystyle h(x)}
with respect to
x
{\displaystyle x}
twice as well as a
C
2
{\displaystyle C_{2}}
, two arbitrary constants as expected from a second-order equation.
= Example
=Given the differential equation
(
1
−
x
2
)
y
″
−
4
x
y
′
−
2
y
=
0
{\displaystyle (1-x^{2})y''-4xy'-2y=0}
one can always easily check for exactness by examining the
y
″
{\displaystyle y''}
term. In this case, both the partial and total derivative of
1
−
x
2
{\displaystyle 1-x^{2}}
with respect to
x
{\displaystyle x}
are
−
2
x
{\displaystyle -2x}
, so their sum is
−
4
x
{\displaystyle -4x}
, which is exactly the term in front of
y
′
{\displaystyle y'}
. With one of the conditions for exactness met, one can calculate that
∫
(
−
2
x
)
d
y
=
I
(
x
,
y
)
−
h
(
x
)
=
−
2
x
y
{\displaystyle \int (-2x)\,dy=I(x,y)-h(x)=-2xy}
Letting
f
(
x
,
y
)
=
−
2
y
{\displaystyle f(x,y)=-2y}
, then
∫
(
−
2
y
−
2
x
y
′
−
d
d
x
(
−
2
x
y
)
)
d
x
=
∫
(
−
2
y
−
2
x
y
′
+
2
x
y
′
+
2
y
)
d
x
=
∫
(
0
)
d
x
=
h
(
x
)
{\displaystyle \int \left(-2y-2xy'-{d \over dx}(-2xy)\right)\,dx=\int (-2y-2xy'+2xy'+2y)\,dx=\int (0)\,dx=h(x)}
So,
h
(
x
)
{\displaystyle h(x)}
is indeed a function only of
x
{\displaystyle x}
and the second-order differential equation is exact. Therefore,
h
(
x
)
=
C
1
{\displaystyle h(x)=C_{1}}
and
I
(
x
,
y
)
=
−
2
x
y
+
C
1
{\displaystyle I(x,y)=-2xy+C_{1}}
. Reduction to a first-order exact equation yields
−
2
x
y
+
C
1
+
(
1
−
x
2
)
y
′
=
0
{\displaystyle -2xy+C_{1}+(1-x^{2})y'=0}
Integrating
I
(
x
,
y
)
{\displaystyle I(x,y)}
with respect to
x
{\displaystyle x}
yields
−
x
2
y
+
C
1
x
+
i
(
y
)
=
0
{\displaystyle -x^{2}y+C_{1}x+i(y)=0}
where
i
(
y
)
{\displaystyle i(y)}
is some arbitrary function of
y
{\displaystyle y}
. Differentiating with respect to
y
{\displaystyle y}
gives an equation correlating the derivative and the
y
′
{\displaystyle y'}
term.
−
x
2
+
i
′
(
y
)
=
1
−
x
2
{\displaystyle -x^{2}+i'(y)=1-x^{2}}
So,
i
(
y
)
=
y
+
C
2
{\displaystyle i(y)=y+C_{2}}
and the full implicit solution becomes
C
1
x
+
C
2
+
y
−
x
2
y
=
0
{\displaystyle C_{1}x+C_{2}+y-x^{2}y=0}
Solving explicitly for
y
{\displaystyle y}
yields
y
=
C
1
x
+
C
2
1
−
x
2
{\displaystyle y={\frac {C_{1}x+C_{2}}{1-x^{2}}}}
Higher-order exact differential equations
The concepts of exact differential equations can be extended to any order. Starting with the exact second-order equation
d
2
y
d
x
2
(
J
(
x
,
y
)
)
+
d
y
d
x
(
d
J
d
x
+
∂
J
∂
x
)
+
f
(
x
,
y
)
=
0
{\displaystyle {d^{2}y \over dx^{2}}(J(x,y))+{dy \over dx}\left({dJ \over dx}+{\partial J \over \partial x}\right)+f(x,y)=0}
it was previously shown that equation is defined such that
f
(
x
,
y
t
)
=
d
h
t
(
x
)
d
x
+
d
d
x
(
I
(
x
,
y
)
−
h
(
x
)
)
−
∂
J
∂
x
d
y
d
x
{\displaystyle f(x,yt)={dht(x) \over dx}+{d \over dx}(I(x,y)-h(x))-{\partial J \over \partial x}{dy \over dx}}
Implicit differentiation of the exact second-order equation
n
{\displaystyle n}
times will yield an
(
n
+
2
)
{\displaystyle (n+2)}
th-order differential equation with new conditions for exactness that can be readily deduced from the form of the equation produced. For example, differentiating the above second-order differential equation once to yield a third-order exact equation gives the following form
d
3
y
d
x
3
(
J
(
x
,
y
)
)
+
d
2
y
d
x
2
d
J
d
x
+
d
2
y
d
x
2
(
d
J
d
x
+
∂
J
∂
x
)
+
d
y
d
x
(
d
2
J
d
x
2
+
d
d
x
(
∂
J
∂
x
)
)
+
d
f
(
x
,
y
)
d
x
=
0
{\displaystyle {d^{3}y \over dx^{3}}(J(x,y))+{d^{2}y \over dx^{2}}{dJ \over dx}+{d^{2}y \over dx^{2}}\left({dJ \over dx}+{\partial J \over \partial x}\right)+{dy \over dx}\left({d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)\right)+{df(x,y) \over dx}=0}
where
d
f
(
x
,
y
)
d
x
=
d
2
h
(
x
)
d
x
2
+
d
2
d
x
2
(
I
(
x
,
y
)
−
h
(
x
)
)
−
d
2
y
d
x
2
∂
J
∂
x
−
d
y
d
x
d
d
x
(
∂
J
∂
x
)
=
F
(
x
,
y
,
d
y
d
x
)
{\displaystyle {df(x,y) \over dx}={d^{2}h(x) \over dx^{2}}+{d^{2} \over dx^{2}}(I(x,y)-h(x))-{d^{2}y \over dx^{2}}{\partial J \over \partial x}-{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)=F\left(x,y,{dy \over dx}\right)}
and where
F
(
x
,
y
,
d
y
d
x
)
{\displaystyle F\left(x,y,{dy \over dx}\right)}
is a function only of
x
,
y
{\displaystyle x,y}
and
d
y
d
x
{\displaystyle {dy \over dx}}
. Combining all
d
y
d
x
{\displaystyle {dy \over dx}}
and
d
2
y
d
x
2
{\displaystyle {d^{2}y \over dx^{2}}}
terms not coming from
F
(
x
,
y
,
d
y
d
x
)
{\displaystyle F\left(x,y,{dy \over dx}\right)}
gives
d
3
y
d
x
3
(
J
(
x
,
y
)
)
+
d
2
y
d
x
2
(
2
d
J
d
x
+
∂
J
∂
x
)
+
d
y
d
x
(
d
2
J
d
x
2
+
d
d
x
(
∂
J
∂
x
)
)
+
F
(
x
,
y
,
d
y
d
x
)
=
0
{\displaystyle {d^{3}y \over dx^{3}}(J(x,y))+{d^{2}y \over dx^{2}}\left(2{dJ \over dx}+{\partial J \over \partial x}\right)+{dy \over dx}\left({d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)\right)+F\left(x,y,{dy \over dx}\right)=0}
Thus, the three conditions for exactness for a third-order differential equation are: the
d
2
y
d
x
2
{\displaystyle {d^{2}y \over dx^{2}}}
term must be
2
d
J
d
x
+
∂
J
∂
x
{\displaystyle 2{dJ \over dx}+{\partial J \over \partial x}}
, the
d
y
d
x
{\displaystyle {dy \over dx}}
term must be
d
2
J
d
x
2
+
d
d
x
(
∂
J
∂
x
)
{\displaystyle {d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)}
and
F
(
x
,
y
,
d
y
d
x
)
−
d
2
d
x
2
(
I
(
x
,
y
)
−
h
(
x
)
)
+
d
2
y
d
x
2
∂
J
∂
x
+
d
y
d
x
d
d
x
(
∂
J
∂
x
)
{\displaystyle F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2}}(I(x,y)-h(x))+{d^{2}y \over dx^{2}}{\partial J \over \partial x}+{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)}
must be a function solely of
x
{\displaystyle x}
.
= Example
=Consider the nonlinear third-order differential equation
y
y
‴
+
3
y
′
y
″
+
12
x
2
=
0
{\displaystyle yy'''+3y'y''+12x^{2}=0}
If
J
(
x
,
y
)
=
y
{\displaystyle J(x,y)=y}
, then
y
″
(
2
d
J
d
x
+
∂
J
∂
x
)
{\displaystyle y''\left(2{dJ \over dx}+{\partial J \over \partial x}\right)}
is
2
y
′
y
″
{\displaystyle 2y'y''}
and
y
′
(
d
2
J
d
x
2
+
d
d
x
(
∂
J
∂
x
)
)
=
y
′
y
″
{\displaystyle y'\left({d^{2}J \over dx^{2}}+{d \over dx}\left({\partial J \over \partial x}\right)\right)=y'y''}
which together sum to
3
y
′
y
″
{\displaystyle 3y'y''}
. Fortunately, this appears in our equation. For the last condition of exactness,
F
(
x
,
y
,
d
y
d
x
)
−
d
2
d
x
2
(
I
(
x
,
y
)
−
h
(
x
)
)
+
d
2
y
d
x
2
∂
J
∂
x
+
d
y
d
x
d
d
x
(
∂
J
∂
x
)
=
12
x
2
−
0
+
0
+
0
=
12
x
2
{\displaystyle F\left(x,y,{dy \over dx}\right)-{d^{2} \over dx^{2}}\left(I(x,y)-h(x)\right)+{d^{2}y \over dx^{2}}{\partial J \over \partial x}+{dy \over dx}{d \over dx}\left({\partial J \over \partial x}\right)=12x^{2}-0+0+0=12x^{2}}
which is indeed a function only of
x
{\displaystyle x}
. So, the differential equation is exact. Integrating twice yields that
h
(
x
)
=
x
4
+
C
1
x
+
C
2
=
I
(
x
,
y
)
{\displaystyle h(x)=x^{4}+C_{1}x+C_{2}=I(x,y)}
. Rewriting the equation as a first-order exact differential equation yields
x
4
+
C
1
x
+
C
2
+
y
y
′
=
0
{\displaystyle x^{4}+C_{1}x+C_{2}+yy'=0}
Integrating
I
(
x
,
y
)
{\displaystyle I(x,y)}
with respect to
x
{\displaystyle x}
gives that
x
5
5
+
C
1
x
2
+
C
2
x
+
i
(
y
)
=
0
{\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+i(y)=0}
. Differentiating with respect to
y
{\displaystyle y}
and equating that to the term in front of
y
′
{\displaystyle y'}
in the first-order equation gives that
i
′
(
y
)
=
y
{\displaystyle i'(y)=y}
and that
i
(
y
)
=
y
2
2
+
C
3
{\displaystyle i(y)={y^{2} \over 2}+C_{3}}
. The full implicit solution becomes
x
5
5
+
C
1
x
2
+
C
2
x
+
C
3
+
y
2
2
=
0
{\displaystyle {x^{5} \over 5}+C_{1}x^{2}+C_{2}x+C_{3}+{y^{2} \over 2}=0}
The explicit solution, then, is
y
=
±
C
1
x
2
+
C
2
x
+
C
3
−
2
x
5
5
{\displaystyle y=\pm {\sqrt {C_{1}x^{2}+C_{2}x+C_{3}-{\frac {2x^{5}}{5}}}}}
See also
Exact differential
Inexact differential equation
References
Further reading
Boyce, William E.; DiPrima, Richard C. (1986). Elementary Differential Equations (4th ed.). New York: John Wiley & Sons, Inc. ISBN 0-471-07894-8
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Artikel Terkait "exact differential equation"
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