Science:Math Exam Resources/Courses/MATH102/December 2014/Question C 02 (a)/Solution 1

By the definition of steady state, we look for the value of $c$ such that ${\frac {dc}{dt}}=0,\quad {\text{or }}P(c)-D(c)=0.$ Thus we need to solve

${\begin{aligned}P_{m}{\frac {c^{2}}{k^{2}+c^{2}}}&=rc\\P_{m}\cdot c^{2}&=rc(k^{2}+c^{2})\\P_{m}\cdot c^{2}&=k^{2}rc+rc^{3}\\P_{m}\cdot c^{2}-k^{2}rc-rc^{3}&=0\\c(P_{m}c-k^{2}r-rc^{2})&=0.\end{aligned}}$

Substitute values of $P_{m}=5$, $k=10$ and $r=1/5$ and simplify the expression, we obtain

${\begin{aligned}c(P_{m}c-k^{2}r-rc^{2})&=0\\c(5c-20-c^{2}/5)&=0\\{\frac {-c}{5}}(-25c+100+c^{2})&=0\\{\frac {1}{5}}c(c-5)(c-20)=0\end{aligned}}$

which gives $c=0,\ 5,\ 20$ are the steady states.