43_pdfsam_math 54 differential equation solutions odd

# 43_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.2 separates if we divide by P and multiply by dt . Z 1 P dP = r 100 Z dt ln P = r 100 t + C P ( t )= Ke rt/ 100 , where K is the initial amount of money in the savings account, K = \$1000, and r %isthe interest rate, r = 5. This results in P ( t ) = 1000 e 5 t/ 100 . (2.7) (a) To determine the amount of money in the account after 2 years we substitute t =2into equation (2.7), which gives P (2) = 1000 e 10 / 100 = \$1105 . 17 . (b) To determine when the account will reach \$4000 we solve equation (2.7) for t with P = \$4000: 4000 = 1000 e 5 t/ 100 e 5 t/ 100 =4 t =20ln4 27 . 73 years . (c) To determine the amount of money in the account after 3 1 2 years we need to determine the value of each \$1000 deposit after 3 1 2 years has passed. This means that the initial \$1000 is in the account for the entire 3 1 2 years and grows to the amount which is given by P 0 = 1000 e 5(3 . 5) / 100 . For the growth of the \$1000 deposited after 12 months, we take
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## This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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