Lecture Notes in Differential Equations - Bruce E. Shapiro

71
From the first of equations (9.52),
∫
M(t, y) =
N t (t, y)dy + h ′ (t) (9.61)
hence
∫
h ′ (t) = M(t, y) −
N t (t, y)dy (9.62)
Multiply by dt and integrate,
∫
h(t) = h ′ (t)dt =
∫ ( ∫
M(t, y) −
)
N t (t, y)dy dt (9.63)
From (9.59)
∫
φ(t, y) =
∫ ( ∫
N(t, y)dy + M(t, y) −
)
N t (t, y)dy dt (9.64)
The following theorem summarizes the derivation.
Theorem 9.2. If M(t, y)dt + N(t, y)dy = 0 is exact, i.e., if
∂M(t, y)
∂y
=
∂N(t, y)
∂t
(9.65)
The φ(t, y) = C is a solution of the ODE, where either of the following
formulas can be used for φ:
∫
∫ ( ∫
)
φ(t, y) = M(t, y)dt + N(t, y) − M y (t, y)dt dy (9.66)
∫
∫ ( ∫ )
φ(t, y) = N(t, y)dy + M(t, y) − N t (t, y)dy dt (9.67)
In practice, however, it is generally easier to repeat the derivation rather
than memorizing the formula.
Example 9.3. Solve
This has the form M(t, y)dt + N(t, y)dy where
Checking that it is exact,
(t + 2y)dt + (2t − y)dy = 0 (9.68)
M(t, y) = t + 2y (9.69)
N(t, y) = 2t − y (9.70)
M y (t, y) = 2 = N t (t, y) (9.71)