Page 44 - Advanced engineering mathematics

24 CHAPTER 1 First-Order Differential Equations

In these integrals, x and y are thought of as fixed, and the integration variables are ξ and η
respectively. Now, y appears on the right side only in the second integral, so the fundamental
theorem of calculus gives us immediately
∂ϕ
= N(x, y).
∂y
Computing ∂ϕ/∂x is less straightforward, since x occurs in both integrals defining ϕ(x, y).For
∂ϕ/∂x, use the condition that ∂M/∂y = ∂N/∂x to write

∂ϕ ∂ x ∂ y
= M(ξ, y 0 )dξ + N(x,η)dη
∂x ∂x ∂x
x 0 y 0
∂N
y
= M(x, y 0 ) + (x,η)dη
∂x
y 0
∂M
y
= M(x, y 0 ) + (x,η)dη
∂y
y 0
= M(x, y 0 ) + M(x, y) − M(x, y 0 ) = M(x, y).
This completes the proof.
In the case of ydx + dy = 0, M(x, y) = y and N(x, y) = 1, so
∂M ∂N
= 1 and = 0.
∂y ∂x
Theorem 1.1 tells us that this differential equation is not exact on any rectangle in the plane. We
saw this in Example 1.13.

EXAMPLE 1.14
We will solve the initial value problem
2
y

(cos(x) − 2xy) + (e − x )y = 0; y(1) = 4.
In differential form,
y
2
(cos(x) − 2xy)dx + (e − x )dy = 0 = Mdx + Ndy
with
2
y
M(x, y) = cos(x) − 2xy and N(x, y) = e − x .
Compute
∂M ∂N
=−2x =
∂y ∂x
for all (x, y). By Theorem 1.1, the differential equation is exact over every rectangle, hence over
the entire plane. A potential function ϕ(x, y) must satisfy
∂ϕ ∂ϕ y 2
= cos(x) − 2xy and = e − x .
∂x ∂y
Choose one of these to integrate. If we begin with the second, then integrate with respect to y:

y
2
y
2
ϕ(x, y) = (e − x )dy = e − x y + h(x).

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October 14, 2010 14:9 THM/NEIL Page-24 27410_01_ch01_p01-42