ODE question part 2 Watch

Prokaryotic_crap
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can someone please check if this is right? thanks

Q. a new company is being formed. A market analysis predicts that the rate of growth of the company's income at any future timme t will be proportional to the difference between the actual income at that time and an upper limit of $10 million. It is also estimated that the income after 4 years will be $4 million. The income is $0 initially. Find and expression for the income at time t.
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working:

\frac{dI}{st} = k(10^6-I)

\int \frac{1}{10^6-I} dI = k\int dt

ln(10^6-I)^{-1} = k[t + c]

(10^6-I)^{-1} = e^{kt}e^c, now plugging in I=0 when t=0

(10^6-0)^{-1} = e^{0k}e^c

c = ln(10^6)^{-1}

how do i find k?
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Chewwy
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it is also estimated that the income after 4 years will be 4 million.
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Prokaryotic_crap
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yes but what formula do i plug the values in, 3rd line of working?
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Hancock orbital
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(Original post by Chewwy)
it is also estimated that the income after 4 years will be 4 million.
PC you should consider substituting this to find k.
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Prokaryotic_crap
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yes, i know but in which equation, thats my problem, is it in ln(10^6-I)^{-1} = k[t + c] as i know c, t and I? or is it another one, please tell me
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dramaminedreams
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(Original post by Prokaryotic_crap)
yes, i know but in which equation, thats my problem, is it in ln(10^6-I)^{-1} = k[t + c] as i know c, t and I? or is it another one, please tell me
That would seem reasonable
You should also, consider, what would happen if you substituted a line after, and ask why you still get the same answer. (my point is either line is fine)
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Hancock orbital
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(Original post by Prokaryotic_crap)
yes, i know but in which equation, thats my problem, is it in ln(10^6-I)^{-1} = k[t + c] as i know c, t and I? or is it another one, please tell me
Any - the equations will give k regardless which one you use.
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Prokaryotic_crap
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so plugiing t=4, I=4^6 and c=ln(10^6)^{-1}...

ln(10^6-4^6)^{-1} = k[4+ln(10^6)^{-1}]

ln(6^6)^{-1} = k[4+ln(10^6)^{-1}]

 \frac{ln(6^6)^{-1}}{4+ln(10^6)^{-1}} = k

k=0.023, something is wrong
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Prokaryotic_crap
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Finally!

im pretty confident thatk=\frac{ln(10^6)}{4}, i done it by plugging the value and also done it by using logs and the same value came up. Here is my first method:

(10^6-I)^{-1}=e^{4k}e^c

(10^6-I)^{-1}=e^{4k}e^{-ln(10^6)}, plugging t=4 and I=4x10^6

(6000000)^{-1} = e^{4k}-(1000000)

10^6=e^{4k}

k =\frac{ln(10^6)}{4}
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Hancock orbital
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(Original post by Prokaryotic_crap)
Finally!

im pretty confident that
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
k=\frac{ln(1^6}{4)
, i done it by plugging the value and also done it by using logs and the same value came up. Here is my first method:

(10^6-I)^{-1}=e^{4k}e^c

(10^6-I)^{-1}=e^{4k}e^{-ln(10^6)}, plugging t=4 and I=4x10^6

(6000000)^{-1} = e^{4k}-(1000000)

1^6=e^{4k}

k =\frac{ln(1^6)}{4}
Check you valu for when you take way I and note that the log(1) = 0 always so k is not the same value.
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Prokaryotic_crap
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k=\frac{ln(10^6)}{4}, corrected now
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Prokaryotic_crap
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right?
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Prokaryotic_crap
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yep, pretty confident
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