Problem 42 - Calculus for Biology and Medicine

Question

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d y}{d x}=(y-1)(y+3)-h $$

Step-by-Step Solution

Answer

Equilibria exist for \(h \geq -1\). Stable equilibria depend on \(y\) near zero and \(-3\).

Step 1: Set the Derivative to Zero

To find equilibrium points, we set \(\frac{d y}{d x}\) to zero. This gives us the equation \((y-1)(y+3)-h=0\).

Step 2: Solve for y in Terms of h

Rewrite the equilibrium condition as \((y-1)(y+3) = h\). Simplify and rewrite: \(y^2 + 2y - 3 = h\). This is a quadratic equation in \(y\).

Step 3: Analyze the Quadratic Equation

To find the values of \(h\) for which equilibria exist, the quadratic equation \(y^2 + 2y - (3 + h) = 0\) must have real solutions. The determinant \(D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-(3+h)) = 4 + 12 + 4h\) should be non-negative for real solutions.

Step 4: Determine the Range of h

Solve \(4 + 4h \geq 0\) which simplifies to \(h \geq -1\). Thus, for \(h \geq -1\), there are real equilibria.

Step 5: Classify Stability of Equilibria

Graphically, the term \((y-1)(y+3)\) describes a parabola opening upwards with roots at \(y=1\) and \(y=-3\). For \(h \geq -1\), the vertex of the parabola gives minimum value. Stable equilibria occur where \((y-1)(y+3) < h\) as \(\frac{d y}{d x} > 0\) on \(0 > y > -3\) and \(\frac{d y}{d x} < 0\) on \(y > -3\) and \(y < 0\).

Key Concepts

Equilibrium Points

In the world of differential equations, equilibrium points are values where the system is at rest or changes are not occurring. For our equation \( \frac{d y}{d x} = (y-1)(y+3)-h \), equilibrium points occur when \( \frac{d y}{d x} = 0 \).
This means setting the expression \( (y-1)(y+3)-h \) equal to zero and solving for \( y \). Basically, these points are potential resting places for a system.

• You set the equation to zero: \( (y-1)(y+3) = h \).
• Rewrite as a quadratic: \( y^2 + 2y - 3 = h \).
• Identify values of \( y \) that solve the equation.

Finding equilibrium points is crucial; they help understand the long-term behavior of dynamic systems.

Stability Analysis

Stability analysis helps us determine whether equilibrium points in a system are stable or unstable. A stable equilibrium, if slightly disturbed, will return to its original state, much like a bowl with a marble at the bottom.
An unstable one will diverge if disturbed, similar to a marble on top of an overturned bowl.

• Graphically analyze using the parabola defined by \( (y-1)(y+3) \) with roots at \( y=1 \) and \( y=-3 \). This gives us insight into where the function increases or decreases.
• For \( h \geq -1 \), the parabola opens upward, showing regions of stability on the graph.
• The graphical behavior around equilibrium points informs their stability status—if the derivative \( \frac{d y}{d x} \) changes sign as you cross an equilibrium point, it is stable; otherwise, it is unstable.

Conducting a stability analysis forms a foundational step to predict how systems react over time.

Quadratic Equations

Quadratic equations form the backbone of many solutions in differential equations. For a standard quadratic equation, you have the form \( ax^2 + bx + c = 0 \). The solution often involves finding values of \( x \) that satisfy this equation, utilizing the discriminant \( D = b^2 - 4ac \).
This discriminant decides the nature of the roots:

• If \( D > 0 \), two distinct real roots exist.
• If \( D = 0 \), a single repeated real root exists.
• If \( D < 0 \), no real roots exists, indicating complex roots.

In our case, analyzing \( y^2 + 2y - (3 + h) = 0 \), we find that roots exist for \( h \geq -1 \).
Thus, to ensure equilibria exist, our quadratic equation's discriminant should be non-negative, highlighting the bridge between solving quadratics and understanding dynamics in differential equations.