Exercise · ZW-EQUILIBRIUM-POINTS-005 · ★★☆ Intermediate
How many real equilibria does \(y'=y^2+1\) have?
Choose one option.
Show answer
- B. 0
Explanation
Since \(y^2+1\ge1>0\) for every \(y\), the right-hand side never vanishes: there are no real equilibria.
Review the concept: Equilibrium points
An equilibrium point of an autonomous equation \(y'=f(y)\) is a constant value \(y^*\) of the unknown at which \(f\) vanishes.
\[f(y^*)=0\;\Longleftrightarrow\; y(t)\equiv y^*\text{ is a solution}\]
More practice problems: equilibrium points
- ZW-EQUILIBRIUM-POINTS-001 Find the equilibrium point of y'=3y-6.
- ZW-EQUILIBRIUM-POINTS-002 What are the equilibria of y'=y^2-4y+3?
- ZW-EQUILIBRIUM-POINTS-003 How many equilibria does y'=y(y-2)(y+1) have?
- ZW-EQUILIBRIUM-POINTS-004 Which of these equations is autonomous?
- ZW-EQUILIBRIUM-POINTS-006 How many equilibria does y'=sin y have in the interval [0,2π]?
- ZW-EQUILIBRIUM-POINTS-007 Find the equilibria of y'=2y(1-tfracy10).
- ZW-EQUILIBRIUM-POINTS-008 A tank satisfies y'=12-4y. At what value does it stay constant?
- ZW-EQUILIBRIUM-POINTS-009 If y^* is an equilibrium and y(0)=y^*, what is the solution of the initial value problem?