Download Lecture 10: Exact Equations

LECTURE 10
Exact Equations
We shall now present another technique for solving first order, non-linear, ordinary differential equations.
This technique is a generalization of the one we used for separable equations. To motivate it, let us consider
first the equation
(10.1)
ψ (x, y) = C
which fixes “implicitly” y as a function of x, because, at least in principle we should be able to solve (10.1)
for y. On the other hand, we can also associate with (10.1) a differential equation by regarding y as a
function of x and differentiating (10.1) with respect to x:
⇒
d
(ψ(x, y) = C )
dx
(10.2)
∂ψ ∂ψ dy
+
=0
∂x ∂y dx
Here’s an explicit example. Consider the equation of a circle:
x2 + y 2 = R 2 .
(10.3)
If we differentiate this equation as in (10.2) we get
2x + 2y
or
(10.4)
y +
dy
=0
dx
x
=0
y
which is a first order non-linear ODE. Now the (algebraic) solutions of (10.3) are given by
y = ± R 2 − x2 .
(10.5)
Note that, as self-consistency demands, the functions
y ( x) = ± R 2 − x 2
are also solutions of the differential equation (10.4); e.g.,
d
dx
R2 − x2
+ √ 2x
R
−
x2
= √ −2 x
R
−
x2
+ √ 2x
R
− x2
=0 .
Thus, to every algebraic relation of the form (10.1) we have an associated ODE (10.2), and the solutions
y(x) of the ODE (10.2) will correspond to solutions of the original algebraic relations.
What we will attempt next is to try to reverse this procedure. That is to say, we will try to solve differential
equations of the form (10.2) by first identifying the corresponding algebraic relation (10.1), and then solving
(algebraically) for y.
The problem, however, is this: given a general non-linear, first order differential equation
(10.6)
M (x, y) + N (x, y)y = 0
how do we recognize that it’s of the form (10.2), and more to the point, if it is of the form (10.2), how do
we identify the function
that defines the corresponding algebraic relation?
ψ
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10. EXACT EQUATIONS
33
It turns out that there is a rather simple criteria for seeing if a differential equation of the form (10.3) can
be re-expressed in the form (10.2).
MN
Suppose that the functions , , ∂M
∂y , and ∂N
∂x are all continuous in a region R = {(x, y) |
α < x < β , γ < y < δ }. Then
M (x, y) + N (x, y)y = 0
(10.7)
can be re-expressed in the form
∂ψ ∂ψ +
y =0
∂x ∂y
at each point in R if and only if
∂M ∂N
(10.8)
=
∂y
∂x
at each point in R.
Theorem 10.1.
In other words, there exists a function
ψ
(10.9)
such that
M
=
N
=
∂ψ
∂x
∂ψ
∂y
if and only if (10.8) is satisfied. If (10.8) is satisfied then we say the differential equation is exact.
necessity of condition (10.8) is easy to see; for if M and N satisfy (10.9) then we have
The
∂M
∂y
∂ ∂ψ
= ∂y
= ∂ ∂ψ = ∂N
.
∂x ∂x ∂y
∂x
The sufficiency of condition (10.9) is a little more difficult to prove; but it amounts to a short calculation
which more or less follows the pattern of the examples given below (see the textbook for more details).
Example 10.2.
(2x + 3) + (2y − 2)y = 0
We have
M (x, y) = 2x + 3
N (x, y) = 2y − 2
and
∂M
= 0 = ∂N
∂y
∂x
so this equation is exact. We now try to find a function ψ such that its partial derivative with respect to x
is M and its partial derivative with respect to y is N . Now the most general function ψ one can have such
that
∂ψ
= 2x + 3
∂x
would be something of the form
ψ = x2 + 3 x + h 1 ( y ) .
(10.10)
(You should verify for yourself why this must be so.) Similarly, in order for
we must have
(10.11)
∂ψ
∂y
= 2y − 2
ψ = y2 − 2y + h2 (x) .
10. EXACT EQUATIONS
34
We now ask the question: is there any way we can make the forms (10.10) and (10.11) compatible with one
another. Certainly, we simply set
h 1 (y ) = y 2
h 2 ( x) = x2
So we set
ψ = x 2 + 3x + y 2 − 2y .
A general solution of the original differential equation is then constructed by solving
ψ = x2 + 3x + y2 − 2y = C
for y. One obtains
4 − 4(x2 + 3x − C )
2
±
.
y=
2
Example
10.3.
dy
dx
ax − by = bx − cy
This equation does not seem to be exact. However, if we re-write it in the form
−ax + by + (bx − cy)y = 0
it is clearly exact since
∂M
∂
∂
= ∂y
(−ax + by) = b = ∂x
(bx − cy) = ∂N
.
∂y
∂x
We therefore try to construct a function ψ such that
∂ψ
(10.12)
= −ax + by
∂x
and
∂ψ
(10.13)
= bx − cy .
∂y
Now in order for (10.12) to hold, ψ must be of the form
1
ψ = − ax2 + bxy + h1 (y)
(10.14)
2
and for (10.13) to hold we must have
1
ψ = bxy − cy2 + h2 (x) .
(10.15)
2
In order to satisfy both (10.14) and (10.15) we can take
1
1
ψ = − ax2 + bxy − cy2 .
(10.16)
2
2
Setting
1
1
ψ = − ax2 + bxy − cy2 = k
2
2
and solving for y yields
bx ± b2 x2 − c (ax2 + 2k)
.
y=
c
2
(The constant k corresponds to initial conditions: k = cy2o
.)