**Unformatted text preview: **Same as in 02(a), to solve the steady states, we just need to solve the following equation C(TC2 — Pmc + fear) 2 0, Obviously c = 0 is one of the steady states and other 2 — Pmc + k2?" = 0. To get only two steady state, from the quadratic formula we need to have b2 — 4m: : 0 and in
2 P3, 1 _ _ _ l
= = — To obtain posmve steady state c, we get r = —. E 16’ 4 steady states are roots of the quadratic equation TC this case (Pm)2 — 4- r - k2? = 0. Thus with the given values of Pm and k we have 7' ...

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- Spring '19
- Cheryl Fu
- Quadratic equation, steady states