\[\frac{dn}{dt}=0.1n(1-\frac{n}{50})\]
\[n]
\[\frac{dn}{dt}=0.1n-\frac{0.1n^{2}}{50}\]
\[500\int]
\[10\left [ ln (n) + ln (1-n) \right ] = t + c\]
\[\frac{dn}{dt}=0.1n-\frac{0.1n^{2}}{50}\]
\[500\int]
\[10\left [ ln (n) + ln (1-n) \right ] = t + c\]
\[\frac{dn}{dt}=0.1n-\frac{0.1n^{2}}{50}\]
\[500\int]
\[10\left [ ln (n) + ln (1-n) \right ] = t + c\]
\[\frac{1}{n(50-n)}=\frac{A}{n}+\frac{B}{50-n}\]
\[n]
\[n]
\[\frac{1}{n(50-n)}=\frac{1}{50n}+\frac{1}{50(50-n)}\]
\[\frac{1}{n(50-n)}=\frac{A}{n}+\frac{B}{50-n}\]
\[n]
\[n]
\[\frac{1}{n(50-n)}=\frac{1}{50n}+\frac{1}{50(50-n)}\]
\[e^{t}+c]
\[(e^{t})^{0.1}+c]
\[t="ln(n)^{10}-ln(50-n)^{10}\" +="+" c=""][br][br][tex]\[e^{t}+c][/tex][br][br][tex]\[(e^{t})^{0.1}+c][/tex][br][br]I still cant get the answer -_-' Something else I am missing?
\[ln(\frac{n}{50-n})]
\[\frac{n}{50-n}=e^{\frac{t+c}{10}}\]
\[ln(\frac{n}{50-n})]
\[\frac{n}{50-n}=e^{\frac{t+c}{10}}\]
\[n]
\[n(1+e^{\frac{1}{10}}e^{\frac{c}{10}})]
\[n]
\[n]
\[n(1+e^{\frac{1}{10}}e^{\frac{c}{10}})]
\[n]