Question

Another differential equation \[\frac{ dy }{ dt }=0.4(y-300)\] and y(0)=75.

So, I separate the variables into dy/(y-300 = .4dt

integrate both side for ln(y-300)=.4t+c

exponentiate then add 300 to the right for y=300+e^(.4t+c).

Answer 1

Now I think I should use my initial value, so 75=300+e^c

=> -225=e^c => 225=-e^c , and take logs of both sides to get ln(225)=-c

Answer 2

Next I should plug -ln(225) for c in the equation ln(y-300)=.4t+c, giving me ln(y-300)=.4t-ln(225) right? Or and I doing something wrong?

Answer 3

I added the log from the right to the left, and using property of logs combined them to a quotient, but the answer after simplification was wrong, so I must be missing something.

Answer 4

225=-e^c , and take logs of both sides to get ln(225)=-c <----NO

Answer 5

if 225 = e^(-c) then -c = ln 225

Answer 6

y=300+e^(.4t+c).
y=300+e^(.4t)e^c.
y=300-225e^(.4t)

Answer 7

i put e^c = -225

Answer 8

ah i see

Answer 9

any more doubts ? all clear ?

Answer 10

I think so, I am not so quick, I will work it out!

Answer 11

Ah, there really isn't any working out to do because e^c is a constant and its all in terms of y. So simple, thanks again.

Answer 12

welcome :)