Lecture Notes in Differential Equations - Bruce E. Shapiro

81
Proof. In each of the five cases we are trying to show that µ(t, y) is an
integrating factor of the differential equation
M(t, y)dt + N(t, y)dy = 0 (10.36)
We have already shown in the discussion leading to equation (10.3) that
µ(t, y) will be an integrating factor of (10.36) if µ(t, y) satisfies
∂
∂y (µ(t, y)M(t, y)) = ∂ (µ(t, y)N(t, y)) (10.37)
∂t
By the product rule for derivatives,
µ(t, y)M y (t, y) + µ y (t, y)M(t, y) = µ(t, y)N t (t, y) + µ t (t, y)N(t, y) (10.38)
To prove each theorem, we need to show that under the given assumptions
for that theorem, the formula for µ(t, y) satisfies equation (10.38).
Case 1. If µ(t, y) is only a functions of t then (a) µ y (t, y) = 0 (there is
no y in the equation, hence the corresponding partial is zero); and (b)
µ t (t, y) = µ ′ (t) (µ only depends on a single variable, t, so there is no
distinction between the partial and regular derivative). Hence equation
(10.37) becomes
Rearranging,
µ(t)M y (t, y) = µ(t)N t (t, y) + µ ′ (t)N(t, y) (10.39)
d
dt µ(t) = M y(t, y) − N t (t, y)
µ(t) (10.40)
N(t, y)
Separating variables, and integrating,
∫
∫
1 d
µ(t) dt µ(t)dt = My (t, y) − N t (t, y)
dt (10.41)
N(t, y)
Hence
∫
ln µ(t) =
(∫
P (1) (t)dt =⇒ µ(t) = exp
)
P (1) (t)dt
(10.42)
as required by equation (10.27). (Case 1)
Case 2. If µ(t, y) = µ(y) is only a function of y then (a) µ t (t, y) = 0 (because
µ has no t-dependence); and (b) µ y (t, y) = µ ′ (t) (because µ is only a
function of a single variable y). Hence (10.38) becomes
µ(y)M y (t, y) + µ ′ (y)M(t, y) = µ(y)N t (t, y) (10.43)