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1.3 Exact Equations 21

1.3 Exact Equations

A differential equation M(x, y) + N(x, y)y = 0 can be written in differential form as

M(x, y)dx + N(x, y)dy = 0. (1.4)
Sometimes this differential form is the key to writing a general solution. Recall that the
differential of a function ϕ(x, y) of two variables is
∂ϕ ∂ϕ
dϕ = dx + dy. (1.5)
∂x ∂y
If we can find a function ϕ(x, y) such that
∂ϕ ∂ϕ
= M(x, y) and = N(x, y), (1.6)
∂x ∂y
then the differential equation Mdx + Ndy = 0isjust
M(x, y)dx + N(x, y)dy = dϕ = 0.

But if dϕ = 0, then ϕ(x, y) = constant. The equation
ϕ(x, y) = c,
with c an arbitrary constant, implicitly defines the general solution of Mdx + Ndy = 0.

EXAMPLE 1.11
We will use these ideas to solve
x
dy 2x − e sin(y)
= .
x
dx e cos(y) + 1
This equation is neither separable nor linear. Write it in the form of equation (1.4) as
x
x
M(x, y)dx + N(x, y)dy = (e sin(y) − 2x)dx + (e cos(y) + 1)dy = 0.
2
x
Now let ϕ(x, y) = e sin(y) + y − x . Then
∂ϕ x ∂ϕ x
= e sin(y) − 2x = M(x, y) and = e cos(y) + 1 = N(x, y),
∂x ∂y
so equations (1.6) are satisfied. The differential equation becomes just dϕ = 0, with general
solution defined implicitly by
x
2
ϕ(x, y) = e sin(y) + y − x = c.
To verify that this equation does indeed implicitly define the solution of the differential equation,
differentiate it implicitly with respect to x, thinking of y as y(x), to get
x
x
e sin(y) + e cos(y)y + y − 2x = 0

and solve this for y to get

x
2x − e sin(y)

y = ,
x
e cos(y) + 1
which is the original differential equation.
Example 1.11 suggests a method. The difficult part in applying it is finding the function
ϕ(x, y). One magically appeared in Example 1.11, but usually we have to do some work to find
a function satisfying equations (1.6).

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October 14, 2010 14:9 THM/NEIL Page-21 27410_01_ch01_p01-42