In: Advanced Math
Initially, $2000 is deposited into a retirement account which pays 5% interest per year, compounded continuously. New money is added to the account at the rate of $1000 per year. How much will be in the account after 10 years?
Please use the method of ordinary differential equations to setup and complete the problem. Also please show all steps and work clearly so I can follow your logic and learn to solve similar ones myself. I will rate your answer for you and leave feedback. Thank you kindly!
From here we can do
I there mate of inteuest constant is 5% which is Here we know the formula. if it Compoun ded a times in a year. then amount aften it is A= P(a + eht Where a be the rate of interest Nove ded In our case we compoun dedo Continuously, that is Here n is infinity. So tin Continion our Annaul value
nyo nyo ntd Is himP (+ jot = p lopenTe + 4) 3) tu p ph man (4+*%) da 1. Redres - Peta since limon ( 1 * * * = e that is for continious Case X40 we have Annaul value A= Pent Mone Heue P = 2000 es= 5% = 5 A= 10 year. so for principal 2000 we have matweity value A = 2000 e 360x10 - 2000 e ½
ca. l None also 1000 dollan added aften nu so for this 1000 dollen we get [ Compounded & year] - B 1000 e 1000 Similary for second yeary 1000 dollar. Ann aul value 1000 x 8 so tolato so Conturing this process for 1000 extera o invest ment we have annoul value. +(63200) 1000 6400) 9+ le Shoo)et -- +68/100) 07 10th year added value
= 1ooo = 1000 at a tuh & ... +48 = 88/900 1- .no 2 1000 1 - 1 11- 107 [ 1 - 08/100 so total amount in the account att en genting to adding 1000 dollar in o 10th year if 2000 €9/2 + 1000 2- 1-0 120 1- 04/27 1ooo zerut 1 - 04/20 x 16950.21