Question

A system of differential equations is given. Obtain an expression for each equilibrium (it may be a function of the constant a). x=a(x3),y=5y,a0x^{\prime}=a(x-3), \quad y^{\prime}=5-y, \quad a \neq 0

Solution

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

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

a0a \neq 0.

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

Solving the first equation:

a(x^3)=0 Since a0 x^3=0x^=3\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) x^=3,y^=5\hat{x} = 3, \hat{y} = 5.

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