Lecture Notes in Differential Equations - Bruce E. Shapiro

66 LESSON 9. EXACT EQUATIONS
This is not the only way we could have solved the problem. We could have
stated by dividing by the right-hand side,
To give
which is also in the desired form, but with
dy
= dt (9.8)
3t + y
− dt + 1 dy = 0 (9.9)
3t + y
M(t, y) = −1 (9.10)
N(t, y) = 1
3t + y
(9.11)
In general there are infinitely many ways to do this conversion.
Now recall from calculus that the derivative of a function φ(t, y) with respect
to a third variable, say u, is given by
d ∂φ dt
φ(t, y) =
du ∂t du + ∂φ dy
∂y du
(9.12)
and that the exact differential of φ(t, y) is obtained from (9.12) by multiplication
by du:
dφ(t, y) = ∂φ
∂t
∂φ
dt + dy (9.13)
∂y
We can integrate the left hand side:
∫
dφ = φ + C (9.14)
If, however, the right hand side is equal to zero, so that dφ = 0, then
∫
0 = dφ = φ + C =⇒ φ = C ′ (9.15)
where C ′ = −C is still a constant.
Let us summarize:
for some constant C.
dφ = 0 =⇒ φ = C (9.16)
If dφ = 0 then the right hand side of (9.12) is also zero, hence
∂φ
∂t
∂φ
dt + dy = 0 =⇒ φ = C (9.17)
∂y