Question

# A system of differential equations is given. Obtain an expression for each equilibrium (it may be a function of the constant a). $x^{\prime}=a(x-3), \quad y^{\prime}=5-y, \quad a \neq 0$

Solution

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Equilibria are pairs of values $(\hat{x},\hat{y})$ that simultaneously satisfy the pair of equations.

$a ( \hat{x} - 3 )=0 \text{ \ and \ } 5 - \hat{y}=0$

$a \neq 0$.

We can calculate the equilibria by solving the second equation for $\hat{y}$, and then solving the first equation for $\hat{x}$. There is one solution to the second equation: $\hat{y}=5$.

Solving the first equation:

$\begin{gather*} a ( \hat{x} - 3 )=0\\ \text{ Since a \neq 0 } \Rightarrow \hat{x} - 3=0\\ \Rightarrow \hat{x} = 3 \end{gather*}$

Therefore the equilibrium is (i) $\hat{x} = 3, \hat{y} = 5$.

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