Lecture Notes in Differential Equations - Bruce E. Shapiro

85
Since M y ≠ N t , equation (10.74) is not exact.
We proceed to check cases 1 through 5 to see if we can find an integrating
factor.
P (1) (t, y) = M y − N t
N
=
(3t + 2y) − (2t + y)
t 2 + ty
= t + y
t 2 + ty = 1 t
(10.79)
(10.80)
(10.81)
This depends only on t, hence we can use case (1). The integrating factor
is
(∫ ) (∫ ) 1
µ(t) = exp P (t)dt = exp
t dt = exp (ln t) = t (10.82)
Multiplying equation (10.74) by µ(t) = t gives
Equation (10.83) has
This time, since
we have an exact equation.
(3t 2 y + y 2 t)dt + (t 3 + t 2 y)dy = 0 (10.83)
M(t, y) = 3t 2 y + y 2 t (10.84)
N(t, y) = t 3 + t 2 y (10.85)
M y = 3t 2 + 2yt = N t (10.86)
The solution of (10.83) is φ(t, y) = C, where C is an arbitrary constant,
and
∂φ
∂t = M(t, y) = 3t2 y + ty 2 (10.87)
∂φ
∂y = N(t, y) = t3 + t 2 y (10.88)
To find φ(t, y) we begin by integrating (10.87) over t:
∫ ∫ ∂φ
φ(t, y) =
∂t dt = (3t 2 + ty 2 )dt = t 3 y + 1 2 t2 y 2 + h(y) (10.89)
Differentiating with respect to y
∂φ
∂y = t3 + t 2 y + h ′ (y) (10.90)