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 1
cos
1
cos
dy
x
dx
x
= +−
2