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What is the equilibrium solution of \(d v / d t=\) \(32-v\) ? How does this relate to the solution \(v=32+C e^{-t}\) ?

Short Answer

Expert verified
The equilibrium solution is \( v = 32 \). As \( t \) approaches infinity, \( v \) approaches 32, matching the equilibrium solution.

Step by step solution

01

- Understand the Differential Equation

We are given the differential equation \( \frac{dv}{dt} = 32 - v \). This is a first-order linear differential equation.
02

- Find Equilibrium Solution

The equilibrium solution occurs when \( \frac{dv}{dt} = 0 \). Set \( 32 - v = 0 \) and solve for \( v \).
03

- Solve for v

From \( 32 - v = 0 \), we get \( v = 32 \). This is the equilibrium solution.
04

- Relate to General Solution

The general solution given is \( v = 32 + Ce^{-t} \). Notice that as \( t \to \,\infty \), \( Ce^{-t} \to 0 \). Therefore, \( v \to 32 \) as \( t \to \,\infty \), which coincides with the equilibrium solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-order linear differential equation
A first-order linear differential equation is a type of differential equation that involves the derivatives of a function as well as the function itself. In general, it can be written as: \(\frac{dy}{dx} + P(x)y = Q(x)\). Here, \(P(x)\) and \(Q(x)\) are known functions of \(x\). Our specific differential equation is \(\frac{dv}{dt} = 32 - v\). It's called 'first-order' because the highest derivative in the equation is the first derivative, i.e., \(\frac{dv}{dt}\), and 'linear' because it can be written in the standard linear form \( \frac{dy}{dx} + P(x)y = Q(x) \). In our case, it's already in the required linear form with \( P(t) = 1 \) and \( Q(t) = 32 \). Understanding this form is key to solving and analyzing the behavior of the solution.
General solution
The general solution of a differential equation encompasses all possible solutions. For our equation \( \frac{dv}{dt} = 32 - v \), the general solution is given by \( v = 32 + Ce^{-t} \), where \( C \) is a constant determined by initial conditions. To find this solution, you can separate variables and integrate: \( \frac{dv}{32-v} = dt \). Integrating both sides, we obtain: \[ -\text{ln}|32 - v| = t + C' \] where \( C' \) is the integration constant. Solving for \( v\) yields the exponential form, \( v = 32 + Ce^{-t} \). The general solution thus describes not just one specific function, but an entire 'family' of solutions depending on the constant \( C \) determined by initial values.
Long-term behavior in differential equations
Long-term behavior in differential equations refers to how the solution behaves as \(t \to \,\infty \). For the equation \( \frac{dv}{dt} = 32 - v \), we see that the given general solution is \( v = 32 + Ce^{-t} \). As time \( t \) approaches infinity, the exponential term \( Ce^{-t} \) approaches zero because \( e^{-t} \) decays to zero. This means that no matter the initial value of the constant \( C \), the term \( Ce^{-t} \) will become negligible for large \( t \). Thus, the solution \( v \) will approach \( 32 \). This value, \( v = 32 \), is called the equilibrium solution. It represents a stable state where the system will settle over the long term. In real-world terms, it's where the quantity described by \( v \) (like a population, temperature, concentration, etc.) remains constant if unaffected by external changes.

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Most popular questions from this chapter

Solve the Logistic equation, \(d y / d t=\alpha y(1-\) \((1 / K) y\) ), by viewing it as a Bernoulli equation and then solve the resulting linear equation by using an integrating factor rather than the method of undetermined coefficients that is illustrated in the examples.

An object that weighs \(48 \mathrm{lb}\) is released from rest at the top of a plane metal slide that is inclined \(30^{\circ}\) to the horizontal. Air resistance \((\mathrm{lb})\) is numerically equal to one-half the velocity (ft/s), and the coefficient of friction is \(\mu=1 / 4\). Using Newton's second law of motion by summing the forces along the surface of the slide, we find the following forces: (a) the component of the weight parallel to the slide: \(F_{1}=48 \sin 30^{\circ}=24\); (b) the component of the weight perpendicular to the slide: \(N=48 \cos 30^{\circ}=\) \(24 \sqrt{3}\); (c) the frictional force (against the motion of the object): \(F_{2}=-\mu N=-\frac{1}{4} \times\) \(24 \sqrt{3}=-6 \sqrt{3}\); and (d) the force due to air resistance (against the motion of the object): \(F_{3}=-\frac{1}{2} v\). Because the mass of the object is \(m=\) \(48 / 32=3 / 2\), we find that the velocity of the object satisfies the initial value problem \(m d v / d t=F_{1}+F_{2}+F_{3}\) or $$ \frac{3}{2} \frac{d v}{d t}=24-6 \sqrt{3}-\frac{1}{2} v, \quad v(0)=0 . $$ Solve this problem for \(v(t)\). Determine the distance traveled by the object at time \(t\), \(x(t)\), if \(x(0)=0\).

(a) Find the solution to the IVP \(d y / d t=\) \(-r(1-y / A) y, y(0)=y_{0}\), where \(r\) and \(A\) are positive constants. (b) If \(y_{0}A\), show that \(\lim _{t \rightarrow \infty} y(t)=\infty\). (d) Show that if \(y_{0}>A\), then \(y(t)\) has a vertical asymptote at \(t=\) \(\frac{1}{r} \ln \left(\frac{y_{0}}{y_{0}-A}\right)\).

(See Exercise 21) Determine the minimum initial velocity needed to launch the lunar module (used on early space missions) from the surface of the Earth's moon given that the moon's radius is \(R \approx 1080\) miles and the acceleration of gravity of the moon is \(16.5 \%\) of that of the Earth.

Determine the time of death if the temperature of a corpse is \(79^{\circ} \mathrm{F}\) when discovered at \(3: 00\) p.m. and \(68^{\circ} \mathrm{F} 3\) h later. Assume that the temperature of the surroundings is \(60^{\circ} \mathrm{F}\) (normal body temperature is \(98.6^{\circ} \mathrm{F}\) ).

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