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.
Given a simply connected and open subset D of
R2
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,[1] [2] so that
\partialF | |
\partialx |
=I
\partialF | |
\partialy |
=J.
An exact equation may also be presented in the following form:
I(x,y)+J(x,y)y'(x)=0
The nomenclature of "exact differential equation" refers to the exact differential of a function. For a function
F(x0,x1,...,xn-1,xn)
x0
dF | = | |
dx0 |
\partialF | |
\partialx0 |
n | |
+\sum | |
i=1 |
\partialF | |
\partialxi |
dxi | |
dx0 |
.
The function
F:R2\toR
F(x,y)=
1 | |
2 |
(x2+y2)+c
is a potential function for the differential equation
xdx+ydy=0.
Let the functions , , , and , where the subscripts denote the partial derivative with respect to the relative variable, be continuous in the region . Then the differential equation
M(x,y)+N(x,y)
dy | |
dx |
=0
is exact if and only if
My(x,y)=Nx(x,y)
That is, there exists a function
\psi(x,y)
\psix(x,y)=M(x,y)and\psiy(x,y)=N(x,y)
So, in general:
My(x,y)=Nx(x,y)\iff \begin{cases} \exists\psi(x,y)\\ \psix(x,y)=M(x,y)\\ \psiy(x,y)=N(x,y) \end{cases}
The proof has two parts.
First, suppose there is a function
\psi(x,y)
\psix(x,y)=M(x,y)and\psiy(x,y)=N(x,y)
It then follows that
My(x,y)=\psixy(x,y)andNx(x,y)=\psiyx(x,y)
Since
My
Nx
\psixy
\psiyx
The second part of the proof involves the construction of
\psi(x,y)
My(x,y)=Nx(x,y)
\psi(x,y)
\psix(x,y)=M(x,y)and\psiy(x,y)=N(x,y)
Begin by integrating the first equation with respect to
x
where
Q(x,y)
Qx=M
h(y)
y
M
x
y
x
Now to show that it is always possible to find an
h(y)
\psiy=N
Differentiate both sides with respect to
y
Set the result equal to
N
h'(y)
In order to determine
h'(y)
y
x
x
Since
Qx=M
My(x,y)=Nx(x,y)
Therefore,
And this completes the proof.
First order exact differential equations of the form
can be written in terms of the potential function
\psi(x,y)
where
This is equivalent to taking the exact differential of
\psi(x,y)
The solutions to an exact differential equation are then given by
and the problem reduces to finding
\psi(x,y)
This can be done by integrating the two expressions
M(x,y)dx
N(x,y)dy
\psi(x,y)
The reasoning behind this is the following. Since
it follows, by integrating both sides, that
Therefore,
where
Q(x,y)
P(x,y)
Qx=M
Py=N
In order for this to be true and for both sides to result in the exact same expression, namely
\psi(x,y)
h(y)
P(x,y)
g(x)
y
x
x
g(x)
Q(x,y)
Ergo,
for some expressions
f(x,y)
d(x,y)
f(x,y)
d(x,y)
Since we already showed that
it follows that
So, we can construct
\psi(x,y)
\int{M(x,y)dx}
\int{N(x,y)dy}
f(x,y)
g(x)
h(y)
The concept of exact differential equations can be extended to second order equations.[3] Consider starting with the first-order exact equation:
I\left(x,y\right)+J\left(x,y\right){dy\overdx}=0
Since both functions
J\left(x,y\right)
{dI\overdx}+\left({dJ\overdx}\right){dy\overdx}+{d2y\overdx2}\left(J\left(x,y\right)\right)=0
Expanding the total derivatives gives that
{dI\overdx}={\partialI\over\partialx}+{\partialI\over\partialy}{dy\overdx}
and that
{dJ\overdx}={\partialJ\over\partialx}+{\partialJ\over\partialy}{dy\overdx}
Combining the terms gives
{\partialI\over\partialx}+{dy\overdx}\left({\partialI\over\partialy}+{\partialJ\over\partialx}+{\partialJ\over\partialy}{dy\overdx}\right)+{d2y\overdx2}\left(J\left(x,y\right)\right)=0
If the equation is exact, then Additionally, the total derivative of
J\left(x,y\right)
{\partialI\over\partialx}+{dy\overdx}\left({\partialJ\over\partialx}+{dJ\overdx}\right)+{d2y\overdx2}\left(J\left(x,y\right)\right)=0
Now, let there be some second-order differential equation
f\left(x,y\right)+g\left(x,y,{dy\overdx}\right){dy\overdx}+{d2y\overdx2}\left(J\left(x,y\right)\right)=0
If
{\partialJ\over\partialx}={\partialI\over\partialy}
\int\left({\partialI\over\partialy}\right)dy=\int\left({\partialJ\over\partialx}\right)dy
and
\int\left({\partialI\over\partialy}\right)dy=\int\left({\partialJ\over\partialx}\right)dy=I\left(x,y\right)-h\left(x\right)
where
h\left(x\right)
x
I\left(x,y\right)
y
h\left(x\right)
I\left(x,y\right)
h\left(x\right)
Next, if
{dI\overdx}={\partialI\over\partialx}+{\partialI\over\partialy}{dy\overdx}
then the term
{\partialI\over\partialx}
x
y
x
y
y
f\left(x,y\right)+g\left(x,y,{dy\overdx}\right){dy\overdx}+{d2y\overdx2}\left(J\left(x,y\right)\right)=0
only the term
f\left(x,y\right)
x
y
{\partialI\over\partialx}=f\left(x,y\right)
{\partialI\over\partialx}=f\left(x,y\right)
f\left(x,y\right)={dI\overdx}-{\partialI\over\partialy}{dy\overdx}
Since the total derivative of
I\left(x,y\right)
x
{dI\overdx}
f\left(x,y\right)+{\partialI\over\partialy}{dy\overdx}={dI\overdx}={d\overdx}\left(I\left(x,y\right)-h\left(x\right)\right)+{dh\left(x\right)\overdx}
So,
{dh\left(x\right)\overdx}=f\left(x,y\right)+{\partialI\over\partialy}{dy\overdx}-{d\overdx}\left(I\left(x,y\right)-h\left(x\right)\right)
and
h\left(x\right)=\int\left(f\left(x,y\right)+{\partialI\over\partialy}{dy\overdx}-{d\overdx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx
Thus, the second order differential equation
f\left(x,y\right)+g\left(x,y,{dy\overdx}\right){dy\overdx}+{d2y\overdx2}\left(J\left(x,y\right)\right)=0
is exact only if
g\left(x,y,{dy\overdx}\right)={dJ\overdx}+{\partialJ\over\partialx}={dJ\overdx}+{\partialJ\over\partialx}
\int\left(f\left(x,y\right)+{\partialI\over\partialy}{dy\overdx}-{d\overdx}\left(I\left(x,y\right)-h\left(x\right)\right)\right)dx=\int\left(f\left(x,y\right)-{\partial\left(I\left(x,y\right)-h\left(x\right)\right)\over\partialx}\right)dx
is a function solely of
x
h\left(x\right)
I\left(x,y\right)-h\left(x\right)
I\left(x,y\right)
I\left(x,y\right)+J\left(x,y\right){dy\overdx}=0
Now, however, in the final implicit solution there will be a
C1x
h\left(x\right)
x
C2
Given the differential equation
\left(1-x2\right)y''-4xy'-2y=0
one can always easily check for exactness by examining the
y''
1-x2
x
-2x
-4x
y'
\int\left(-2x\right)dy=I\left(x,y\right)-h\left(x\right)=-2xy
Letting
f\left(x,y\right)=-2y
\int\left(-2y-2xy'-{d\overdx}\left(-2xy\right)\right)dx=\int\left(-2y-2xy'+2xy'+2y\right)dx=\int\left(0\right)dx=h\left(x\right)
So,
h\left(x\right)
x
h\left(x\right)=C1
I\left(x,y\right)=-2xy+C1
2\right)y'=0 | |
-2xy+C | |
1+\left(1-x |
Integrating
I\left(x,y\right)
x
2y+C | |
-x | |
1x+i\left(y\right)=0 |
where
i\left(y\right)
y
y
y'
-x2+i'\left(y\right)=1-x2
So,
i\left(y\right)=y+C2
C1x+C
2y=0 | |
2+y-x |
Solving explicitly for
y
y=
C1x+C2 | |
1-x2 |
The concepts of exact differential equations can be extended to any order. Starting with the exact second order equation
{d2y\overdx2}\left(J\left(x,y\right)\right)+{dy\overdx}\left({dJ\overdx}+{\partialJ\over\partialx}\right)+f\left(x,y\right)=0
it was previously shown that equation is defined such that
f\left(x,y\right)={dh\left(x\right)\overdx}+{d\overdx}\left(I\left(x,y\right)-h\left(x\right)\right)-{\partialJ\over\partialx}{dy\overdx}
Implicit differentiation of the exact second-order equation
n
\left(n+2\right)
{d3y\overdx3}\left(J\left(x,y\right)\right)+{d2y\overdx2}{dJ\overdx}+{d2y\overdx2}\left({dJ\overdx}+{\partialJ\over\partialx}\right)+{dy\overdx}\left({d2J\overdx2}+{d\overdx}\left({\partialJ\over\partialx}\right)\right)+{df\left(x,y\right)\overdx}=0
where
{df\left(x,y\right)\overdx}={d2h\left(x\right)\overdx2}+{d2\overdx2}\left(I\left(x,y\right)-h\left(x\right)\right)-{d2y\overdx2}{\partialJ\over\partialx}-{dy\overdx}{d\overdx}\left({\partialJ\over\partialx}\right)=F\left(x,y,{dy\overdx}\right)
F\left(x,y,{dy\overdx}\right)
x,y
{dy\overdx}
{dy\overdx}
{d2y\overdx2}
F\left(x,y,{dy\overdx}\right)
{d3y\overdx3}\left(J\left(x,y\right)\right)+{d2y\overdx2}\left(2{dJ\overdx}+{\partialJ\over\partialx}\right)+{dy\overdx}\left({d2J\overdx2}+{d\overdx}\left({\partialJ\over\partialx}\right)\right)+F\left(x,y,{dy\overdx}\right)=0
Thus, the three conditions for exactness for a third-order differential equation are: the
{d2y\overdx2}
2{dJ\overdx}+{\partialJ\over\partialx}
{dy\overdx}
{d2J\overdx2}+{d\overdx}\left({\partialJ\over\partialx}\right)
F\left(x,y,{dy\overdx}\right)-{d2\overdx2}\left(I\left(x,y\right)-h\left(x\right)\right)+{d2y\overdx2}{\partialJ\over\partialx}+{dy\overdx}{d\overdx}\left({\partialJ\over\partialx}\right)
must be a function solely of
x
Consider the nonlinear third-order differential equation
yy'''+3y'y''+12x2=0
If
J\left(x,y\right)=y
y''\left(2{dJ\overdx}+{\partialJ\over\partialx}\right)
2y'y''
y'\left({d2J\overdx2}+{d\overdx}\left({\partialJ\over\partialx}\right)\right)=y'y''
3y'y''
F\left(x,y,{dy\overdx}\right)-{d2\overdx2}\left(I\left(x,y\right)-h\left(x\right)\right)+{d2y\overdx2}{\partialJ\over\partialx}+{dy\overdx}{d\overdx}\left({\partialJ\over\partialx}\right)=12x2-0+0+0=12x2
which is indeed a function only of
x
4+C | |
h\left(x\right)=x | |
1x+C |
2=I\left(x,y\right)
4+C | |
x | |
1x+C |
2+yy'=0
Integrating
I\left(x,y\right)
x
{x5\over
2+C | |
5}+C | |
2x+i\left(y\right)=0 |
y
y'
i'\left(y\right)=y
i\left(y\right)={y2\over2}+C3
{x5\over
2+C | |
5}+C | |
2x+C |
2\over | |
3+{y |
2}=0
The explicit solution, then, is
2+C | |
y=\pm\sqrt{C | |
2x+C |
|