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Variables separable differential equations

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Please see picture attached - thank you in advance for any help
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Zacken
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Original post by jonathan14


Please see picture attached - thank you in advance for any help


What part are you having trouble with?

What specific issue do you have?

What have you tried so far?
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jonathan14
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Original post by Zacken
What part are you having trouble with?

What specific issue do you have?

What have you tried so far?


I formed an integral of dp/p(100-p) = kdt

I wasn't sure whether to form partial fractions or not? my original attempt did not work for me
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Zacken
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Original post by jonathan14
I formed an integral of dp/p(100-p) = kdt

I wasn't sure whether to form partial fractions or not? my original attempt did not work for me


Yes, you need partial fractions:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{\,\mathrm{d}p}{p(100-p)} = kt + c \iff \frac{1}{100} \int \frac{1}{p} - \frac{1}{p-100} \, \mathrm{d}p = kt + c\end{equation*}



Which integrates to logarithms.
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jonathan14
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Original post by Zacken
Yes, you need partial fractions:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\int \frac{\,\mathrm{d}p}{p(100-p)} = kt + c \iff \frac{1}{100} \int \frac{1}{p} - \frac{1}{p-100} \, \mathrm{d}p = kt + c\end{equation*}



Which integrates to logarithms.


I hate to be a pain but if I'm being honest I wasn't how to proceed after integrating.
Original post by jonathan14
I hate to be a pain but if I'm being honest I wasn't how to proceed after integrating.


After integrating, you'll have something of the form f(p)=kt+c. Then for part (a), just rearrange that formula to find p explicitly in terms of t. (At some point you may need to divide the numerator and denominator of a fraction by e^(100kt) in order to get the e^(-100kt) in the denominator, if my intuition is right). You will then get some combination of k and c as the coefficient of e^(-100kt) (or possibly, there'll be a k and/or c in the exponent, so use the rule e^(a+b)=e^a*e^b to make it the coefficient,) so to get to the given answer, relabel that coefficient as A, since it's just another constant.

For part (b), put in the given numbers and you should get two equations in terms of A and k. You can solve these simultaneously to find the value of k as requested.
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Zacken
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Original post by jonathan14
I hate to be a pain but if I'm being honest I wasn't how to proceed after integrating.


What do you get after integrating? Why don't you post a picture of your working?

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