Homework Sourse6000 was invested 2 years ago and over time grew according t
$6000 was invested 2 years ago and over time grew according to the differential equation: dP/dt=kt. Current value of the investment is $7328. How much will the investment be worth in another 5 years from today?
Solution
dP/dt = kt
=> dP = ktdt => P = kt2/2 + c , c is the unknown const of integration
At t=0 , P = c => c = 6000
At t=2, P = 7328 => 7238 = k(4)/2 + 6000 => k = 619
P = 619t2/2 + 6000
Investment value 5 years from now, OR 7 years from beginning i.e. t=7
=> P = 619(49/2) + 6000
=> P = 21165.5 (Ans)
