y'=ky, you are told when t=3 y is 2 and the rate of change of y is 5. so k is 5/2.. what is y(6)?

Answer 1

Ok. so nobody yet has correctly helped you.

dy/dt = ky
dy/y = k dt , integrating both sides....
ln y = kt + B , put the exponential function raised to these...

e^ ln y = e^(kt + B) = {e^kt} {e^B}
y = Ce^kt ............. here C = e^B.

To solve for C: (you already have k = 5/2)
2 = C e^(2.5 * 3)
C = 2 e^(-15/2)

y(t) = 2 e^(-15/2) e^(5/2 t) = 2 e^( [-15+5t] / 2 )
y(6) = 2 e^(15/2).


d:

Answer 2

y' = ky
intergrating y',
u get
y = kyt + c

since t=3, y=2, k = 2.5
c = -15

y(6) = (2.5)(6)y - 15
= 15y - 15

Answer 3

Hello

Where is t?