Lecture Notes in Differential Equations - Bruce E. Shapiro

74 LESSON 9. EXACT EQUATIONS
Rearranging the ODE,
(by − at)dt + (bt + cy)dy = 0 (9.93)
which is of the form M(t, y)dt + N(t, y)dy = 0 with
Since
M(t, y) = by − at (9.94)
N(t, y) = bt + cy (9.95)
M y = b = N t (9.96)
equation (9.93) is exact. Therefore the solution is φ(t, y) = K, where
∂φ(t, y)
= M(t, y) = by − at
∂t
(9.97)
∂φ(t, y)
= N(t, y) = bt + cy
∂y
(9.98)
and K is any arbitrary constant (we don’t use C because it is confusing
with c already in the problem). Integrating equation in (9.97) over t gives
∫
φ(t, y) = (by − at)dt = byt − a 2 t2 + h(y) (9.99)
Differentiating with respect to y and setting the result equal to the (9.98),
bt + h ′ (y) = bt + cy (9.100)
h ′ (y) = cy =⇒ h(y) = c 2 y2 (9.101)
From equation (9.99)
∫
φ(t, y) =
(by − at)dt = byt − a 2 t2 + c 2 y2 (9.102)
Therefore the solution of (9.92) is
∫
(by − at)dt = byt − a 2 t2 + c 2 y2 = K (9.103)
for any value of the constant K.
Example 9.6. Solve the initial value problem
}
(2ye 2t + 2t cos y)dt + (e 2t − t 2 sin y)dy = 0
y(0) = 1
(9.104)