# In each of the Problems below, determine the critical point x = x⁰, and then classify its type and examine its stability by making the transformation x = x⁰ + u. 1. dx/ dt = [ 1 1 [ 2 x − 1 −1 ] 0 ] 2. dx/ dt = [ -2 1 [ -2 x − 1 −2 ] 1 ] 3. dx/ dt = [ -1 -1 [ -1 x − 2 −1 ] 5 ] 4. dx/ dt = [ 0 −β [ α x − α, β, γ, δ > 0 δ 0 ] −γ ]

1 answer

###### Question:

In each of the Problems below, determine the critical point x = x⁰, and then classify its type and examine its stability by making the transformation x = x⁰ + u.

1. dx/ dt = [ 1 1 [ 2

x −

1 −1 ] 0 ]

2. dx/ dt = [ -2 1 [ -2

x −

1 −2 ] 1 ]

3. dx/ dt = [ -1 -1 [ -1

x −

2 −1 ] 5 ]

4. dx/ dt = [ 0 −β [ α

x − α, β, γ, δ > 0

δ 0 ] −γ ]

1. dx/ dt = [ 1 1 [ 2

x −

1 −1 ] 0 ]

2. dx/ dt = [ -2 1 [ -2

x −

1 −2 ] 1 ]

3. dx/ dt = [ -1 -1 [ -1

x −

2 −1 ] 5 ]

4. dx/ dt = [ 0 −β [ α

x − α, β, γ, δ > 0

δ 0 ] −γ ]

## Answers

Answer:\bf xy=1\iff y = \cfrac{1}{x}\iff y = x^{-1}\\\\ -----------------------------\\\\ \cfrac{dy}{dt}=-x^{-2}\cdot \cfrac{dx}{dt}\impliedby \textit{using the chain-rule} \\\\\\ \left. \cfrac{dy}{dt}=-\cfrac{\frac{dx}{dt}}{x^2} \right|{ \begin{array}{llll} x=2\\ \frac{dx}{dt}=-2 \end{array}}\implies \cfrac{dy}{dt}=-\cfrac{-2}{2^2}Step-by-step explanation:

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