(3)
Now
P = 1000,   when t = 0
Substituting the values of P and t in (3), we get C = 1000 Therefore, equation (3),
gives
P = 1000 
t20
e
Let t years be the time required to double the principal Then
2000 = 1000
et20
    ⇒  t = 20 loge2
EXERCISE 9 4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1