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xartaan Group Title

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).

  • one year ago
  • one year ago

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  1. xartaan Group Title
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    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

    • one year ago
  2. xartaan Group Title
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    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?

    • one year ago
  3. xartaan Group Title
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    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.

    • one year ago
  4. hartnn Group Title
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    225=-e^c , and take logs of both sides to get ln(225)=-c <----NO

    • one year ago
  5. hartnn Group Title
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    if 225 = e^(-c) then -c = ln 225

    • one year ago
  6. hartnn Group Title
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    y=300+e^(.4t+c). y=300+e^(.4t)e^c. y=300-225e^(.4t)

    • one year ago
  7. hartnn Group Title
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    i put e^c = -225

    • one year ago
  8. xartaan Group Title
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    ah i see

    • one year ago
  9. hartnn Group Title
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    any more doubts ? all clear ?

    • one year ago
  10. xartaan Group Title
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    I think so, I am not so quick, I will work it out!

    • one year ago
  11. xartaan Group Title
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    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.

    • one year ago
  12. hartnn Group Title
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    welcome :)

    • one year ago
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