Question

Involve equations of the form \(d y / d t=f(y)\). In each problem sketch the graph of \(f(y)\) versus \(y,\) determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. $$ d y / d t=a y+b y^{2}, \quad a>0, \quad b>0, \quad y_{0} \geq 0 $$

Short answer

Answer: The equilibrium points are \(y_1=0\) (unstable) and \(y_2=-\frac{a}{b}\) (asymptotically stable).

Step-by-step solution

Write the given differential equation in the required form

The given first-order differential equation is \(d y / d t=a y+b y^{2}\), where \(a>0\) and \(b>0\).

Identify the equilibrium points

Equilibrium points occur when the derivative of \(y\) with respect to \(t\) equals zero: \(d y / d t = 0\). So, our task is to solve the following equation for \(y\): $$ a y + b y^{2} = 0 $$ By factoring \(y\) from the equation, we get: $$ y(a+by)= 0 $$ This gives us two equilibrium points, \(y_1=0\) and \(y_2=-\frac{a}{b}\).

Determine the stability of each equilibrium point

To determine the stability, we need to check whether the graph of \(f(y)\) is increasing or decreasing around each equilibrium point. For \(y_1=0\): \(f'(y)=(a+by)y=ay+by^2\) Since \(f'(0)=a\cdot0+b\cdot0^2=0\), signs of \(f'(y)\) on left and right of 0 are decided by sign of `a` alone, which is positive by a being greater than 0.. So, the graph of \(f(y)\) is increasing on both sides of \(y_1\), making it an unstable equilibrium point. For \(y_2=-\frac{a}{b}\): Since both \(a>0\) and \(b>0\), \(y_2\) is negative. Now \(f'(y)=(a+by)y\). for \(y=y_2+\epsilon\) on the right side of \(y_2\), \(\epsilon > 0\), In this case, since \(y_2\) is negative, \(a+by_2\) is positive, and \(y_2+\epsilon\) is also negative. Thus, \(f'(y_2+\epsilon)<0\). For \(y=y_2-\epsilon\) on the left side of \(y_2\), \(\epsilon > 0\), Since \(a+by_2-\epsilon b <0\), \(y_2-\epsilon\) is also negative. Thus, \(f'(y_2-\epsilon)>0\). Hence, the graph of \(f(y)\) is decreasing to the right and increasing to the left of \(y_2\), making it an asymptotically stable equilibrium point.

Sketch the graph of \(f(y)\)

To sketch the graph of \(f(y)\) given by \(f(y)=ay+by^2\), we can observe the following: 1. At \(y=0\), \(f(y) = 0\) 2. At \(y=-\frac{a}{b}\), \(f(y) = 0\) 3. Since the coefficient of the quadratic term, \(b\), is positive, the graph is a parabola opening upwards. 4. As \(y\) increases from \(-\infty\) to \(0\), the function increases and reaches \(0\) at the unstable equilibrium point \(y_1=0\). 5. As \(y\) increases further, the function continues to increase, but it will again reach \(0\) at the asymptotically stable equilibrium point \(y_2=-\frac{a}{b}\). 6. For \(y<-a/b\), the function is negative, and for \(y>0\), the function is positive. Now we can sketch the graph of \(f(y)\) considering these points, and show the equilibrium points of \(y_1=0\) as unstable and \(y_2=-\frac{a}{b}\) as asymptotically stable.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They arise whenever a rate of change is involved in a physical, biological, or economic process. The equaition in our example, a first-order differential equation in the form of \( d y / d t=a y+b y^{2} \), perfectly showcases this: it models the rate of change of a quantity, \( y \), over time, \( t \). When we talk about first-order, it means that only the first derivative of \( y \) with respect to \( t \) is considered, which makes it slightly simpler than higher-order differential equations involving second derivatives or above.

Understanding the solutions to these equations can give you insights into the behavior of the modeled system over time, like predicting population growth or the cooling of a hot object. The constants \( a \) and \( b \) in the differential equation adjust the model's sensitivity to changes in \( y \), indicating the strength and direction of these changes.

Stability of Equilibrium Points

In the context of differential equations, equilibrium points, sometimes called critical points, are values at which the rate of change of the function is zero. This means the system is in a sort of balance at that point and, if undisturbed, will remain there indefinitely. In our example with the equation \( d y / d t=a y+b y^{2} \), finding the equilibrium points involves setting the derivative to zero and solving for \( y \).

The stability of these points is crucial as it describes the behavior of the system when it is slightly perturbed from the equilibrium. If it returns to the equilibrium after a disturbance, it is considered stable; if it moves away, it is considered unstable. The stability analysis performed in the step-by-step solution involves looking at the sign of the derivative \( f'(y) \) around these points to predict whether the system will be pushed back or further from equilibrium when perturbed. Understanding the system's stability helps us predict its long-term behavior without having to solve the differential equation explicitly.

Asymptotically Stable

When talking about an equilibrium point, 'asymptotically stable' denotes a condition where, after a small disturbance, the system not only returns to equilibrium but does so in an ever-decreasing oscillation. In essence, the system will approach the equilibrium point as time goes to infinity. It shows a strong form of stability because the system essentially draws itself back to equilibrium over time.

The step-by-step solution determined that in our equation \( y_2=-\frac{a}{b} \) is an asymptotically stable equilibrium. Mathematically, this gets verified by examining the sign of \( f'(y) \) — if negative, the function is decreasing and thus stabilizing back to the equilibrium point. This concept is particularly important in systems where we expect a return to steady state after disturbances such as in ecological systems or engineered feedback controllers.

Unstable Equilibrium

On the flip side, an 'unstable equilibrium' exists when perturbation of the system results in a movement away from the equilibrium point. In such a case, even slight disturbances can cause significant changes over time, and the system may never return to its original state. This concept is highlighted in our example, where \( y_1=0 \) represents an unstable equilibrium point.

Unstable equilibriums play a critical role in phenomena such as tipping points in climate change or market crashes in economics. Even when such a point is determined, as in the step-by-step solution, understanding its implications requires further analysis of the larger system and context as unstable equilibrium points are sensitive to initial conditions and can lead to drastic changes.