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\(\frac{dP}{dt}=rP(1-\frac{P}{K})\)

 

Rearrange the differential equation into standard form.

 

I feel like I've attacked this every way possible but I can't seem to get into standard form.

 

Any help is greatly appreciated,

 

Thank you.

vest4R  Mar 26, 2018
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4+0 Answers

 #1
avatar+179 
+1

ultimately I have to solve it, but I should be right once it's in standard form.

 

I was thinking i'm being too picky and try to solve it with the separation methord. But I don't have an expression with t...

vest4R  Mar 26, 2018
edited by vest4R  Mar 26, 2018
 #2
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+2

I don't know what you mean by 'standard form'.

Assuming that the p inside the bracket is the same as the P at the front,  the equation can be written as

\(\displaystyle \frac{dP}{dt}=\frac{rP}{K}(K-P),\)

from which we have, (assuming that the r and K are constants),

\(\displaystyle \int\frac{dP}{P(K-P)}=\frac{r}{K}\int dt.\)

Now it's a partial fractions job on the first integral.

 

Tiggsy

Guest Mar 26, 2018
 #3
avatar+179 
+1

thank you triggsy! just had a look at Partial Fraction Decomposition. Turns out they haven't taught us that yet.

 

thanks!

vest4R  Mar 26, 2018
 #4
avatar+92219 
+1

Thanks Tiggsy :)

Melody  Mar 26, 2018

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