The solution of (1) is
given by
40
40
200
C
t
t
ye
e
=
+
 or y (t) = 200 + C 
40
t
e
− (2)
where, c is the constant of integration Note that when t = 0, y = 250 Therefore, 250 = 200 + C
or
C = 50
Then (2) reduces to
y = 200 + 50 
40
t
e
−