Lecture Notes in Differential Equations - Bruce E. Shapiro

22 LESSON 3. SEPARABLE EQUATIONS
In other words, the equation y ′ = f(t, y) is separable if and only if
f(t, y) ∂2 f
∂t∂y = ∂f ∂f
∂t ∂y
(3.48)
Example 3.9. Determine if
is separable.
dy
dt = 1 + t2 + y 3 + t 2 y 3 (3.49)
From the right hand side of the differential equation we see that
f(t, y) = 1 + t 2 + y 3 + t 2 y 3 (3.50)
Calculating all the necessary partial derivatives,
Hence
∂f
∂t = 2t + 2ty3 (3.51)
∂f
∂y = 3y2 + 3t 2 y 2 (3.52)
and
∂f ∂f
∂t ∂y = ( 2t + 2ty 3) ( 3y 2 + 3t 2 y 2) (3.53)
= 6ty 2 + 6t 3 y 2 + 6ty 5 + 6t 3 y 5 (3.54)
f(t, y) ∂2 f
∂t∂y = ( 1 + t 2 + y 3 + t 2 y 3) 6ty 2 (3.55)
= 6ty 2 + 6t 3 y 2 + 6ty 5 + 6t 3 y 5 (3.56)
= ∂f ∂f
∂t ∂y
Consequently the differential equation is separable.
(3.57)
Of course, knowing that the equation is separable does not tell us how to
solve it. It does, however, tell us that looking for a factorization of
f(t, y) = a(t)b(y) (3.58)
is not a waste of time. In the previous example, the correct factorization is
dy
dt = (1 + y3 )(1 + t 2 ) (3.59)