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xartaan
 3 years ago
Another differential equation \[\frac{ dy }{ dt }=0.4(y300)\] and y(0)=75.
So, I separate the variables into dy/(y300 = .4dt
integrate both side for ln(y300)=.4t+c
exponentiate then add 300 to the right for y=300+e^(.4t+c).
xartaan
 3 years ago
Another differential equation \[\frac{ dy }{ dt }=0.4(y300)\] and y(0)=75. So, I separate the variables into dy/(y300 = .4dt integrate both side for ln(y300)=.4t+c exponentiate then add 300 to the right for y=300+e^(.4t+c).

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xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0Now 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

xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0Next I should plug ln(225) for c in the equation ln(y300)=.4t+c, giving me ln(y300)=.4tln(225) right? Or and I doing something wrong?

xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1225=e^c , and take logs of both sides to get ln(225)=c <NO

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1if 225 = e^(c) then c = ln 225

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1y=300+e^(.4t+c). y=300+e^(.4t)e^c. y=300225e^(.4t)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1any more doubts ? all clear ?

xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0I think so, I am not so quick, I will work it out!

xartaan
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, 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.
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