Lecture Notes in Differential Equations - Bruce E. Shapiro

56 LESSON 7. AUTONOMOUS ODES
Hence
∫
dy
y(K − y) = 1 ∫ dy
K y + 1 ∫
K
dy
K − y
(7.12)
Using (7.14) in (7.9)
ln
= 1 (ln y − ln(K − y)) (7.13)
K
= 1 K ln y
K − y
Multiplying through by K and exponentiating,
(7.14)
y
= rt + C (7.15)
K − y
y
K − y = ert+C = e rt e C (7.16)
If we set y(0) = y 0 then
Hence
Multiplying by K − y,
e C = y 0
K − y 0
(7.17)
y
K − y = y 0
K − y 0
e rt (7.18)
y 0
y = K y 0
e rt + y e rt (7.19)
K − y 0 K − y 0
Bringing the second term to the left and factoring a y,
(
y 1 + y )
0
e rt = K y 0
e rt (7.20)
K − y 0 K − y 0
Multiplying both sides by K − y 0 ,
Solving for y,
y ( K − y 0 + y 0 e rt) = Ky 0 e rt (7.21)
Ky 0 e rt
y =
(K − y 0 ) + y 0 e rt (7.22)
Ky 0
=
y 0 + (K − y 0 )e −rt (7.23)
We can see from equation (7.23) that for large t the second term in the
denominator approaches zero, hence y → K.