anonymous
  • anonymous
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).
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.

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hartnn
  • hartnn
225=-e^c , and take logs of both sides to get ln(225)=-c <----NO
hartnn
  • hartnn
if 225 = e^(-c) then -c = ln 225
hartnn
  • hartnn
y=300+e^(.4t+c). y=300+e^(.4t)e^c. y=300-225e^(.4t)
hartnn
  • hartnn
i put e^c = -225
anonymous
  • anonymous
ah i see
hartnn
  • hartnn
any more doubts ? all clear ?
anonymous
  • anonymous
I think so, I am not so quick, I will work it out!
anonymous
  • anonymous
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.
hartnn
  • hartnn
welcome :)

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