Lecture Notes in Differential Equations - Bruce E. Shapiro

69
Hence
dg
dy = 0 =⇒ g = C′ (9.39)
for some constant C ′ . Substituting (9.39) into (9.35) gives
But since φ = C is a solution, we conclude that
is a solution for some constant C ′′ .
φ = t 2 + y 2 t + C ′ (9.40)
t 2 + y 2 t = C ′′ (9.41)
Method 2 Start with equation (9.32) and integrate over y first, treating t
as a constant, and treating the constant of integration as a function h(t) :
∂φ
= 2ty (9.42)
∂y
∫ ∫ ∂φ
∂y dy = 2tydy (9.43)
φ = ty 2 + h(t) (9.44)
Differentiate with respect to t and use equation (9.31)
∂φ
∂t = ∂ ∂t (ty2 + h(t)) = y 2 + dh
dt = 2t + y2 (9.45)
where the last equality follows from equation (9.31). Thus
dh
dt = 2t =⇒ h = t2 (9.46)
We can ignore the constant of integration because we will pick it up at the
end. From equation (9.44),
Since φ = C is the solution of (9.25), we obtain
is the solution.
φ = ty 2 + h(t) = ty 2 + t 2 (9.47)
ty 2 + t 2 = C (9.48)
Now we can derive the method in general. Suppose that
M(t, y)dt + N(t, y)dy = 0 (9.49)