Page 42 - Advanced engineering mathematics

22 CHAPTER 1 First-Order Differential Equations

EXAMPLE 1.12
Consider
3
dy 2xy + 2
=− .
dx 3x y + 8e 4y
2
2
This is neither linear nor separable. Write
4y
2
2
3
(2xy + 2)dx + (3x y + 8e )dy = 0
so
3
4y
2
2
M(x, y) = 2xy + 2 and N(x, y) = 3x y + 8e .
From equations (1.6), we want ϕ(x, y) such that
∂ϕ ∂ϕ
2
4y
2
3
= 2xy + 2 and = 3x y + 8e .
∂x ∂y
Choose either of these equations and integrate it. If we choose the first equation, then integrate
with respect to x:
∂ϕ

ϕ(x, y) = dx
∂x

3
2
3
= (2xy + 2)dx = x y + 2x + g(y).
In this integration, we are reversing a partial derivative with respect to x,so y is treated like a
constant. This means that the constant of integration may also involve y; hence it is called g(y).
Now we know ϕ(x, y) to within this unknown function g(y). To determine g(y), use the fact that
we know what ∂ϕ/∂y must be
∂ϕ 2 2 4y
= 3x y + 8e
∂y
2
2
= 3x y + g (y).

4y
4y
This means that g (y) = 8e ,so g(y) = 2e . This fills in the missing piece, and

3
4y
2
ϕ(x, y) = x y + 2x + 2e .
The general solution of the differential equation is implicitly defined by
4y
3
2
x y + 2x + 2e = c,
in which c is an arbitrary constant. In this example, we are not able to solve for y explicitly in
terms of x.
A function ϕ(x, y) satisfying equations (1.6) is called a potential function for the differen-
tial equation M + Ny = 0. If we can find a potential function ϕ(x, y), we have at least the

implicit expression ϕ(x, y) = c for the solution.
The method of Example 1.12 may produce a potential function if the integrations can be
carried out. However, it may also be the case that a potential function does not exist.

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October 14, 2010 14:9 THM/NEIL Page-22 27410_01_ch01_p01-42