What is a slope field or direction field for a first-order ODE y’ = f(x,y)?
- A grid of points (x_k,y_k)
- Tiny line segments drawn at each point with slope f(x_k,y_k)
The process involves choosing a point, computing the slope, and drawing a segment centered at that point.
In the slope field for y’ = x, what is the slope when x = -2?
-2
The slope is constant along the vertical line at x = -2.
In the slope field for y’ = y, what is the slope when y = 0?
0
This indicates a flat slope along the x-axis.
Define isoclines.
Curves where the slope is constant: f(x,y)=c
Along an isocline, every segment has the same slope c.
What are equilibrium solutions in the context of ODEs?
Constant solutions y≡C
Found by setting y’=0, which leads to f(x,y)=0.
Differentiate between stable and unstable equilibrium.
- Stable: Nearby solutions move toward C
- Unstable: Nearby solutions move away
Stability indicates how solutions behave near equilibrium points.
What is a stable attractor?
A solution y_p(x) that nearby solutions approach as x→∞
This concept is crucial in understanding long-term behavior of solutions.
What is the terminal velocity in the context of the equation dv/dt = 32 - 0.16v?
200 ft/s
Terminal velocity occurs when acceleration equals zero.
What is the general form of the solution for the ODE y’ = x - y?
y = x - 1 + Ce^{-x}
This solution describes the behavior of the system as x approaches infinity.
What is the phase line in the context of autonomous ODEs?
A vertical line marking equilibria where f(y)=0 with arrows indicating direction of y’
This helps visualize stability and behavior of solutions.
What is the classification of the ODE y’ = y^2 - 4?
- Order: 1
- Autonomous: yes
- Linear: no
- Separable: yes
This classification helps determine the methods for solving the ODE.
What is the behavior of solutions for y’ = y^2 - 4 near the equilibria y = ±2?
- y = -2 is stable (arrows point toward it)
- y = 2 is unstable (arrows point away)
Stability analysis is crucial for understanding the dynamics of the system.
True or false: The equation y’ = 2y + 1 is separable.
FALSE
It is a sum, not a product, and requires a different method.
What is the method to determine if an ODE is separable?
If the right side can be written as g(x)h(y)
This allows for the separation of variables for integration.
In the equation y’ = -x/y, how do you separate the variables?
y,dy = -x,dx
This allows for integration of both sides.
What is the solution to the separable ODE y’ = -x/y?
y^2 + x^2 = C
This represents a family of circles in the xy-plane.
What is the equation for the rate of change of y in the context of the problem?
y’ = dfrac{dy}{dt} = 2y + 1
This equation is separable because it only depends on y.
Why is the equation y’ = 2y + 1 considered separable?
It can be rearranged to dfrac{dy}{2y+1} = dt
This separation allows for integration.
What is the integral of dfrac{1}{1+y^2} with respect to y?
arctan(y)
This integral is used when solving the equation y’ = 2x(1 + y^2).
In the equation y’ = 2xy^2, what does it indicate about the rate of change of y?
It depends on:
* how big x is
* how big y^2 is
This relationship shows the interaction between x and y.
What is the reason for the interval of validity in the solution of y’ = 2xy^2?
The solution has a denominator: 1 - y_0x^2
When this denominator equals 0, the solution becomes invalid.
Why does the equation y’ = dfrac{t^2}{1-y^2} have no equilibria?
An equilibrium requires y’ = 0 for all t, but t^2 is not always 0
Therefore, no constant solution exists.
What does the equation dfrac{dy}{dt} = dfrac{3t^2+1}{1+2y} represent?
Numerator: time pushing change
Denominator: y resisting change
This reflects the dynamics of the system.
What happens after integrating the equation dfrac{dy}{dt} = dfrac{3t^2+1}{1+2y}?
You get: y^2 + y = stuff
Solving for y requires using the quadratic formula.
