Problem 24
Question
Find the critical points and phase portrait of the given autonomous first- order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the \(x y\) -plane determined by the graphs of the equilibrium solutions. $$ \frac{d y}{d x}=10+3 y-y^{2} $$
Step-by-Step Solution
Verified Answer
Critical points are at \( y = 5 \) (asymptotically stable) and \( y = -2 \) (unstable).
1Step 1: Identify the Critical Points
To find the critical points, we set the right-hand side of the differential equation to zero: \[ 10 + 3y - y^2 = 0. \] Rearranging gives: \[ y^2 - 3y - 10 = 0. \] This is a quadratic equation, which we can factor: \[ (y - 5)(y + 2) = 0. \] Solving this gives the critical points: \( y = 5 \) and \( y = -2 \).
2Step 2: Analyze the Stability of Critical Points
To determine the stability of each critical point, we need the derivative \( f'(y) \) of \( f(y) = 10 + 3y - y^2 \).Compute the derivative:\[ f'(y) = 3 - 2y. \]Evaluate at each critical point:1. At \( y = 5 \): \[ f'(5) = 3 - 2(5) = 3 - 10 = -7. \] A negative derivative implies \( y = 5 \) is asymptotically stable.2. At \( y = -2 \): \[ f'(-2) = 3 - 2(-2) = 3 + 4 = 7. \] A positive derivative implies \( y = -2 \) is unstable.
3Step 3: Sketch the Solution Curves
In the \(xy\)-plane, draw horizontal lines at \( y = 5 \) and \( y = -2 \) to denote the critical points. These are equilibrium solutions.- For \( y > 5 \), \( f'(y) < 0 \), so solutions curve towards \( y = 5 \).- Between \(-2 < y < 5\), \( f'(y) > 0 \) when moving down from above, indicating instability moving away from \( y = -2 \). Thus, solutions curve towards \( y = 5 \).- For \( y < -2 \), \( f'(y) > 0 \), indicating solutions move away from \( y = -2 \).The flow towards and away from these critical points suggest the movement of the solutions within different regions.
Key Concepts
Understanding Critical Points in Differential EquationsDrawing the Phase PortraitStability Analysis of Critical PointsThe Role of Solution Curves
Understanding Critical Points in Differential Equations
Critical points, also known as equilibrium points, occur where the rate of change of a function is zero. In the context of differential equations, these points are crucial in understanding the behavior of solutions over time. To find them, you set the right-hand side of an equation to zero, looking for values that fulfill this condition. For the equation \( \frac{dy}{dx} = 10 + 3y - y^2 \), this means solving \( 10 + 3y - y^2 = 0 \). By factoring, we find the critical points at \( y = 5 \) and \( y = -2 \). These critical points represent the state where the function doesn't grow or shrink; essentially they are flat spots in the phase space.
Drawing the Phase Portrait
A phase portrait is a graphical representation that shows the trajectories of differential equations in the phase plane. It provides a visual way to understand how solutions behave around critical points.
For our equation example, the phase portrait begins by marking the critical points \( y = 5 \) and \( y = -2 \) on the vertical axis of a diagram.
From here, we draw lines to represent possible paths (or solutions) of the equation as they move with the rate determined by \( \frac{dy}{dx} = 10 + 3y - y^2 \). By analyzing the direction of these paths, we can infer the long-term behavior of solutions starting at different initial conditions. Visualizing these solutions as arrows helps us understand the flow of the system towards stability or instability.
For our equation example, the phase portrait begins by marking the critical points \( y = 5 \) and \( y = -2 \) on the vertical axis of a diagram.
From here, we draw lines to represent possible paths (or solutions) of the equation as they move with the rate determined by \( \frac{dy}{dx} = 10 + 3y - y^2 \). By analyzing the direction of these paths, we can infer the long-term behavior of solutions starting at different initial conditions. Visualizing these solutions as arrows helps us understand the flow of the system towards stability or instability.
Stability Analysis of Critical Points
Stability analysis involves determining if solutions near a critical point converge to or diverge from it, classifying it as stable, unstable, or semi-stable.
For stability, examine the derivative \( f'(y) \) of \( f(y) = 10 + 3y - y^2 \): \( f'(y) = 3 - 2y \). For \( y = 5 \), we find \( f'(5) = -7 \), a negative value indicating asymptotic stability. This means that small perturbations around \( y = 5 \) will decay over time, with solutions gravitating back to this point.
Conversely, at \( y = -2 \), \( f'(-2) = 7 \), a positive value suggesting instability. Solutions near this point tend to move away, diverging as time progresses.
For stability, examine the derivative \( f'(y) \) of \( f(y) = 10 + 3y - y^2 \): \( f'(y) = 3 - 2y \). For \( y = 5 \), we find \( f'(5) = -7 \), a negative value indicating asymptotic stability. This means that small perturbations around \( y = 5 \) will decay over time, with solutions gravitating back to this point.
Conversely, at \( y = -2 \), \( f'(-2) = 7 \), a positive value suggesting instability. Solutions near this point tend to move away, diverging as time progresses.
The Role of Solution Curves
Solution curves offer insights into the behavior of differential equations over time. They show how the variables evolve from different initial states.Given the equation \( \frac{dy}{dx} = 10 + 3y - y^2 \), sketching the solution curves aids in understanding the dynamics between critical points.
Draw horizontal lines at the critical points \( y = 5 \) and \( y = -2 \). These lines represent equilibrium states; no change happens at these levels. For areas above \( y=5 \) and below \( y=-2 \), analyze the derivative's sign - it indicates the direction solutions will move.- Above \( y = 5 \), solutions curve towards \( y=5 \) as they stabilize.- Between \( -2 < y < 5 \), solutions have complex behavior influenced by both critical points, but ultimately move towards \( y=5 \) due to its stability.- Below \( y = -2 \), solutions veer away from the unstable \( y=-2 \).
Solution curves form a roadmap of potential behaviors, driven by the equation's structure and critical points.
Draw horizontal lines at the critical points \( y = 5 \) and \( y = -2 \). These lines represent equilibrium states; no change happens at these levels. For areas above \( y=5 \) and below \( y=-2 \), analyze the derivative's sign - it indicates the direction solutions will move.- Above \( y = 5 \), solutions curve towards \( y=5 \) as they stabilize.- Between \( -2 < y < 5 \), solutions have complex behavior influenced by both critical points, but ultimately move towards \( y=5 \) due to its stability.- Below \( y = -2 \), solutions veer away from the unstable \( y=-2 \).
Solution curves form a roadmap of potential behaviors, driven by the equation's structure and critical points.
Other exercises in this chapter
Problem 24
Solve the given initial-value problem. $$ \left(\frac{3 y^{2}-t^{2}}{y^{5}}\right) \frac{d y}{d t}+\frac{t}{2 y^{4}}=0, \quad y(1)=1 $$
View solution Problem 24
In Problems, find an implicit and an explicit solution of the given initial- value problem. \(\frac{d y}{d x}=\frac{y^{2}-1}{x^{2}-1}, \quad y(2)=2\)
View solution Problem 24
Each \(D E\) in Problems 23-30 is of the form given in (5). In Problems 23-28, solve the given differential equation by using an appropriate substitution. $$ \f
View solution Problem 24
In Problems 23-28, find an implicit and an explicit solution of the given initial-value problem. $$ \frac{d y}{d x}=\frac{y^{2}-1}{x^{2}-1}, \quad y(2)=2 $$
View solution