Lecture Notes in Differential Equations - Bruce E. Shapiro

78 LESSON 10. INTEGRATING FACTORS
We are first asked to verify that (10.4) is not exact. To do this we must
show that M y ≠ N x , so we calculate the partial derivatives.
∂M
∂y = ∂ ( )
sin y
− 2e −t y cos t − sin y
sin t =
∂y y
y 2 (10.6)
∂N
∂t = ∂ ( cos y + 2e −t )
cos t
= − 2 ∂t y
y e−t (sin t + cos t) (10.7)
Since M y ≠ N t we may conclude that equation (10.4) is not exact.
To show that (10.5) is an integrating factor, we multiply the differential
equation by µ(t, y). This gives
(
e t sin y − 2y sin t ) dt + ( e t cos y + 2 cos t ) dy = 0 (10.8)
which is in the form M(t, y)dt + N(t, y)dy with
The partial derivatives are
M(t, y) = e t sin y − 2y sin t (10.9)
N(t, y) = e t cos y + 2 cos t (10.10)
∂M(t, y)
∂y
∂N(t, y)
∂t
= ∂ ∂y (et sin y − 2y sin t) (10.11)
= e t (cos y − 2 sin t) (10.12)
= ∂ ∂t (et cos y + 2 cos t) (10.13)
= e t (cos y − 2 sin t) (10.14)
Since M y = N t , we may conclude that (10.8) is exact.
Since (10.8) is exact, its solution is φ(t, y) = C for some function φ that
satisifes
∂φ
∂t = M(t, y) = et sin y − 2y sin t (10.15)
∂φ
∂y = N(t, y) = et cos y + 2 cos t (10.16)
From (10.15)