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exact differential equation∗
pahio†
2013-03-22 1:06:39
Let R be a region in R2 and let the functions X : R → R, Y : R → R have
continuous partial derivatives in R. The first order differential equation
X(x, y) + Y (x, y)
dy
= 0
dx
or
X(x, y)dx + Y (x, y)dy = 0
(1)
is called an exact differential equation, if the condition
∂X
∂Y
=
∂y
∂x
(2)
is true in R.
By (2), the left hand side of (1) is the total differential of a function, there
is a function f : R → R such that the equation (1) reads
d f (x, y) = 0,
whence its general integral is
f (x, y) = C.
The solution function f can be calculated as the line integral
Z
P
f (x, y) :=
[X(x, y) dx + Y (x, y) dy]
(3)
P0
along any curve γ connecting an arbitrarily chosen point P0 = (x0 , y0 ) and the
point P = (x, y) in the region R (the integrating factor is now ≡ 1).
∗ hExactDifferentialEquationi created: h2013-03-2i by: hpahioi version: h40648i Privacy
setting: h1i hResulti h34A05i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
Example. Solve the differential equation
2x
y 2 − 3x2
dx +
dy = 0.
3
y
y4
This equation is exact, since
∂ 2x
6x
∂ y 2 − 3x2
=
−
=
.
∂y y 3
y4
∂x
y4
If we use as the integrating way the broken line from (0, 1) to (x, 1) and from
this to (x, y), the integral (3) is simply
Z x
Z y 2
2x
y − 3x2
x2 1
1 x2
x2 1
2
2
dx+
dy
=
−
+1
=
x
−
+
+1−x
=
− +1.
3
y4
y3 y
y y3
y3 y
0 1
1
Thus we have the general integral
x2
1
−
= C
y3
y
of the given differential equation.
2
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