Lecture Notes in Differential Equations - Bruce E. Shapiro

72 LESSON 9. EXACT EQUATIONS
Hence (9.68) is exact and the solution is given by φ(t, y) = C where
Integrating (9.73) over t,
∂φ
∂t
∂φ
∂y
= M(x, y) = t + 2y (9.72)
= N(t, y) = 2t − y (9.73)
∫ ∫
∂φ(t, y)
φ(t, y) = dt = (t + 2y)dt = 1 ∂t
2 t2 + 2ty + h(y) (9.74)
because the constant of integration may be a function of y. Taking the
partial with respect to y and setting the result equal to (9.73)
Thus
∂φ
∂y = 2t + h′ (y) = 2t − y (9.75)
h ′ (y) = −y =⇒ h(y) = − y2
2
we conclude that h ′ (y) = −y or h(y) = −y 2 /2. From equation (9.74)
Therefore the solution is
for any constant C.
(9.76)
φ(t, y) = 1 2 t2 + 2ty − 1 2 y2 (9.77)
1
2 t2 + 2ty − 1 2 y2 = C (9.78)
Example 9.4. Find the one-parameter family of solutions to
y ′ y cos t + 2tey
= −
sin t + t 2 e y + 2
We can rewrite the equation (9.79) as
(9.79)
(y cos t + 2te y )dt + (sin t + t 2 e y + 2)dy = 0 (9.80)
which has the form M(t, y)dt + N(t, y)dy = 0 where
M(t, y) = y cos t + 2te y (9.81)
N(t, y) = sin t + t 2 e y + 2 (9.82)