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Mathematics 18 Online
OpenStudy (anonymous):

p'=3p-p^2 just need a start. Thanks

OpenStudy (anonymous):

what does \(p'\) represent?

OpenStudy (anonymous):

dp/dt

OpenStudy (anonymous):

i don't remember any of this stuff, but i think you can divide, get \[\frac{p'}{3p-p^2}=1\] integrate both sides get \[\int\frac{p'}{3p-p^2}=t\]

OpenStudy (anonymous):

hm i will try.thanx

OpenStudy (anonymous):

partial fractions should do the integral

OpenStudy (anonymous):

what i have figured so far is that p(t)=0 and p(t)=3

OpenStudy (anonymous):

partial fractions did it! thanx:)

OpenStudy (anonymous):

\[\frac{1}{3y-y^2}=\frac{1}{3}\left(\frac{1}{y}-\frac{1}{y-3}\right)\]

OpenStudy (anonymous):

yw

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