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 $\frac{dc}{dt}=0,\text{or }P(c)-D(c)=0.$ Thus we need to solve

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

Substitute values of $P_m=5$, $k=10$ and $r=1/5$ and simplify the expression, we obtain

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

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