Page 43 - Advanced engineering mathematics

1.3 Exact Equations 23

EXAMPLE 1.13
−x
The equation y + y =0 is separable and linear, and the general solution is y(x)=ce . However,

to make a point, try to find a potential function. In differential form,
ydx + dy = 0.
Here M(x, y) = y and N(x, y) = 1. A potential function ϕ would have to satisfy
∂ϕ ∂ϕ
= y and = 1.
∂x ∂y
If we integrate the first of these equations with respect to x, we get

ϕ(x, y) = ydx = xy + g(y).
Then we need
∂ϕ

= 1 = x + g (y).
∂y

But then g (y) = 1 − x, which is impossible if g is a function of y only. There is no potential
function for this differential equation.

We call a differential equation M + Ny = 0 exact if it has a potential function. Otherwise
it is not exact.

There is a simple test to determine whether M + Ny = 0 is exact for (x, y) in a rectangle R

of the plane.

THEOREM 1.1 Test for Exactness

Suppose M, N, ∂N/∂x and ∂M/∂y are continuous for all (x, y) in some rectangle R in the
(x, y)− plane. Then M + Ny = 0 is exact on R if and only if

∂N ∂M
=
∂x ∂y
for (x,y) in R.

Proof If M + Ny = 0 is exact, then there is a potential function ϕ and

∂ϕ ∂ϕ
= M(x, y) and = N(x, y).
∂x ∂y
Then, for (x, y) in R,
2
2
∂M ∂ ∂ϕ ∂ ϕ ∂ ϕ ∂ ∂ϕ ∂N
= = = = = .
∂y ∂y ∂x ∂y∂x ∂x∂y ∂x ∂y ∂x
Conversely, suppose ∂M/∂y and ∂N/∂x are continuous on R, and that
∂N ∂M
= .
∂x ∂y
Choose any (x 0 , y 0 ) in R, and define for (x, y) in R
x y
ϕ(x, y) = M(ξ, y 0 )dξ + N(x,η)dη.
x 0 y 0

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October 14, 2010 14:9 THM/NEIL Page-23 27410_01_ch01_p01-42