Source: PS3 worksheet
Tags: equilibrium solutions, systems of ODEs, first derivative test, local maximum, population model, differential equations, Calc 2
Difficulty: Intermediate | Prerequisites: Calc I derivatives, the First Derivative Test, basic ODE concepts (what it means for a function to solve a differential equation)
This material sits at the intersection of differential equations and Calc I analysis tools. You are asked to find steady states of a two-species population model (where nothing changes over time) and then to use the First Derivative Test to determine whether a given ODE can produce solutions with local maxima. Both problems require connecting earlier derivative reasoning to new ODE contexts. If you are comfortable setting derivatives to zero and interpreting sign changes, you have the core toolkit already.
Equilibrium solutions are constant solutions where every derivative equals zero. Finding them means solving a plain algebraic system. The First Derivative Test, applied directly to an ODE, lets you rule out local maxima by showing the derivative can never change sign from positive to negative.
Equilibrium solution (steady state)
A constant solution to a system of differential equations where all time derivatives equal zero. In simple terms, nothing is changing: the system has settled.
System of ODEs
Two or more ordinary differential equations that share the same unknown functions. Think of it as multiple interacting quantities whose rates of change depend on each other.
First Derivative Test
A Calc I technique for classifying critical points: if f'(x) changes from positive to negative at a point, the function has a local maximum there; from negative to positive, a local minimum. In the ODE context, you apply this to dy/dx directly.
Local maximum
A point where y(x) is larger than at all nearby x-values. Formally, y(c) is a local max if y(c) >= y(x) for all x in some open interval around c.
Population model
A system of differential equations describing how the sizes of interacting species change over time. The coefficients encode birth, death, and interaction rates.
Setting up the problem
The system is:
dx/dt = -500x + xy
dy/dt = -600y + 3xy
An equilibrium solution means dx/dt = 0 and dy/dt = 0 simultaneously.
You are looking for constant values x and y (both positive) that satisfy both equations at once.
Solving the system algebraically
From dx/dt = 0: -500x + xy = 0, which factors as x(-500 + y) = 0.
Either x = 0 or y = 500.
From dy/dt = 0: -600y + 3xy = 0, which factors as y(-600 + 3x) = 0.
Either y = 0 or x = 200.
The trivial equilibrium is x = 0, y = 0 (both species extinct, given in the problem).
For both species to have positive population, take y = 500 (from the first equation) and x = 200 (from the second equation).
Result
The positive equilibrium is x(t) = 200, y(t) = 500.
Why it works
Plug back in: dx/dt = -500(200) + (200)(500) = -100,000 + 100,000 = 0. Similarly dy/dt = -600(500) + 3(200)(500) = -300,000 + 300,000 = 0. Both derivatives vanish, confirming equilibrium.
The problem
The model is dy/dx = x² + y⁴.
The question: can y(x) ever have a local maximum?
Applying the First Derivative Test
For y(x) to have a local maximum at some point x = c, by the First Derivative Test, y'(x) must change sign from positive to negative as x passes through c.
At the point c itself, a local max requires y'(c) = 0.
But y'(c) = c² + y(c)⁴. Both c² and y(c)⁴ are non-negative for all real values, and c² + y(c)⁴ = 0 only when c = 0 and y(c) = 0 simultaneously.
Checking the critical point
The only candidate is c = 0 with y(0) = 0.
Near x = 0, for any x not equal to 0: y'(x) = x² + y(x)⁴ >= x² > 0.
So y'(x) is strictly positive on both sides of x = 0 (except at the single point x = 0 itself where it equals y(0)⁴ = 0).
The derivative does not change from positive to negative. It goes from positive, touches zero, and returns to positive.
Conclusion
The model dy/dx = x² + y⁴ cannot produce a local maximum. The derivative y'(x) = x² + y⁴ is always non-negative, and at the only point where it vanishes, it does not change sign from positive to negative. The model is consistent with the experimental requirement.
Two-species population system
\frac{dx}{dt} = -500x + xy \qquad \frac{dy}{dt} = -600y + 3xyEquilibrium condition: set both equations to zero and solve for x, y > 0.
ODE with non-negative derivative
\frac{dy}{dx} = x^2 + y^4Since x² >= 0 and y⁴ >= 0 for all real x and y, the right-hand side is always non-negative.
Equilibrium analysis is how ecologists predict stable population levels in predator-prey or competing-species models (Lotka-Volterra systems). Engineers use the same algebra to find steady-state operating points of circuits, chemical reactors, and control systems.
The First Derivative Test on an ODE is used to verify qualitative behaviour before committing to a model. If your physical system cannot produce a peak in output, but your candidate ODE allows one, the model is wrong.
Students often forget to factor out the common variable (x or y) and jump straight to dividing both sides by x or y. Dividing by x loses the x = 0 solution. Always factor first, then consider each factor separately.
A common error is to assume y'(c) = 0 is enough to declare a local maximum. It is not. The First Derivative Test requires a sign change from positive to negative, not merely a zero.
Some students try to solve dy/dx = x² + y⁴ explicitly. You do not need to. The question is about the sign of dy/dx, which you can read straight off the right-hand side.
Students sometimes confuse equilibrium with stability. Equilibrium means the derivatives are zero. Stability (whether small perturbations grow or shrink) is a separate question not asked here.
⚠️ "Find the equilibrium" is a standard exam question for systems of ODEs. Practise the factor-then-solve pattern.
⚠️ Exam problems often ask whether a solution can have a local max or min. The technique is always the same: check the sign of dy/dx using the ODE itself.
⚠️ Verification by substitution (plugging the equilibrium back in) is quick marks on an exam and a good habit.
True or false: An equilibrium solution to a system of ODEs is one where all unknown functions are constant.
True.
Fill in the blank: To find equilibrium points, set all ______ equal to zero.
Time derivatives (dx/dt, dy/dt, etc.).
True or false: If dy/dx >= 0 for all x, then y(x) can still have a local maximum.
False. A local max requires dy/dx to change from positive to negative.
Fill in the blank: Factoring x(-500 + y) = 0 gives two cases: x = 0 or y = ______.
Q: Given the system dx/dt = -500x + xy and dy/dt = -600y + 3xy, find all equilibrium solutions.
A: Set both derivatives to zero. Factor: x(-500 + y) = 0 gives x = 0 or y = 500. Factor: y(-600 + 3x) = 0 gives y = 0 or x = 200. The equilibria are (0, 0) and (200, 500).
Q: Suppose dy/dx = x⁶ + y². Can y(x) have a local maximum? Explain.
A: No. x⁶ + y² >= 0 for all x and y, so dy/dx is never negative. By the First Derivative Test, dy/dx cannot change from positive to negative, so no local maximum is possible.
Q: A student solves -600y + 3xy = 0 by dividing both sides by y, getting x = 200. What solution did they miss?
A: They missed y = 0. Dividing by y is only valid when y is not equal to zero. The correct approach is to factor as y(-600 + 3x) = 0 and consider both y = 0 and x = 200.
Q: For the system above, verify that (200, 500) is an equilibrium by substitution.
A: dx/dt = -500(200) + (200)(500) = -100,000 + 100,000 = 0. dy/dt = -600(500) + 3(200)(500) = -300,000 + 300,000 = 0. Both zero, confirmed.
This connects to Lotka-Volterra models in ecology and to phase-plane analysis, which you may encounter later in the course or in a differential equations module. The factoring technique used here is the same algebraic method from solving polynomial equations in precalculus.
The First Derivative Test argument connects to later work on concavity and the Second Derivative Test, and to qualitative analysis of direction fields.
equilibrium point, steady state, fixed point, constant solution, systems of differential equations, population dynamics, Lotka-Volterra, predator-prey, first derivative test ODE, local maximum test, sign of derivative, non-negative derivative, qualitative analysis, Calc 2 NYU, Calculus II for Engineers