exact differential equation

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

dy- X (x, y) + Y(x, y)dx = 0

X(x, y)dx + Y (x, y)dy = 0

(1)

is called an exact differential equation, if the condition

∂X ∂Y ----= ---- ∂y ∂x

is true in R.

Then there is a function f : R → ℝ such that the equation (1) has the form

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

f(x, y) := ∫ _P₀^P[X(x, y) dx + Y (x, y) dy]

(2)

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).

Example. Solve the differential equation

2 2 2x-dx + y--−-3x--dy = 0. y3 y4

This equation is exact, since

2 2 -∂-2x- = − 6x-= -∂-y--−-3x-. ∂y y3 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 (2) is simply

∫ ∫ x 2x y y2 − 3x2 x2 1 2 1 x2 2 x2 1 13-dx + ---y4---dy = y3-− y-+ 1 = x − y-+ y3-+ 1 − x = y3-− y-+ 1. 0 1

Thus we have the general integral

x2 1 -3-− --= C y y

of the given differential equation.

"exact differential equation" is owned by pahio.
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