Lecture Notes in Differential Equations - Bruce E. Shapiro

79
Differentiating with respect to y,
From equation (10.16),
∫ ∂φ(t, y)
φ(t, y) = dt + h(y) (10.17)
∂t
∫
= M(t, y)dt + h(y) (10.18)
∫
= (e t sin y − 2y sin t)dt + h(y) (10.19)
∂φ(t, y)
∂y
= e t sin y + 2y cos t + h(y) (10.20)
= e t cos y + 2 cos t + h ′ (y) (10.21)
e t cos y + 2 cos t + h ′ (y) = e t cos y + 2 cos t (10.22)
h ′ (y) = 0 =⇒ h(y) = constant (10.23)
Recall the that solution is φ(t, y) = C where from (10.20), and using h(y) =
K,
φ(t, y) = e t sin y + 2 cos t + K (10.24)
hence
is the general solution of (10.8).
e t sin y + 2y cos t = C (10.25)
There is no general method for solving (10.3) to find an integrating factor;
however, sometimes we can find an integrating factor that works under
certain simplifying assumptions. We will present five of these possible simplifications
here as theorems. Only the first case is actually discussed in
Boyce & DiPrima (6th Edition).
Theorem 10.2. Integrating Factors, Case 1. If
P (1) (t, y) = M y(t, y) − N t (t, y)
N(t, y)
≡ P (1) (t) (10.26)
is only a function of t, but does not depend on y, then
(∫ ) (∫ )
µ (1) (t, y) = exp P (1) My (t, y) − N t (t, y)
(t)dt = exp
dt
N(t, y)
is an integrating factor.
(10.27)