Lecture Notes in Differential Equations - Bruce E. Shapiro

70 LESSON 9. EXACT EQUATIONS
where
Then we conclude that
is a solution where
M y (t, y) = N t (t, y) (9.50)
φ(t, y) = C (9.51)
φ t (t, y) = M(t, y) and φ y (t, y) = N(t, y) (9.52)
We start with the first equation in (9.52), multiply by dt and integrate:
∫
∫
φ = φ t (t, y)dt = M(t, y)dt + g(y) (9.53)
Differentiate with respect to y:
φ y = ∂ ∫
∫
M(t, y)dt + g ′ (y) =
∂y
M y (t, y)dt + g ′ (y) (9.54)
Hence, using the second equation in (9.52)
∫
g ′ (y) = φ y (t, y) − M y (t, y)dt (9.55)
∫
= N(t, y) − M y (t, y)dt (9.56)
Now multiply by dy and integrate:
∫
∫ ( ∫
g(y) = g ′ (y)dy = N(t, y) −
)
M y (t, y)dt dy (9.57)
From equation (9.53)
∫
φ(t, y) = M(t, y)dt +
∫ (
N(t, y) −
∫
)
M y (t, y)dt dy (9.58)
Alternatively, we can start with the second of equations (9.52), multiply by
dy (because the derivative is with respect to y), and integrate:
∫
∫
φ = φ y (t, y)dy = N(t, y)dy + h(t) (9.59)
where the constant of integration depends, possibly, on t, because we only
integrated over y. Differentiate with respect to t:
φ t (t, y) = ∂ ∫
∫
N(t, y)dy + h ′ (t) = N t (t, y)dy + h ′ (t) (9.60)
∂t