Question

Solve each of the following initial value problems and plot the solutions for several values of \(y_{0}\). Then describe in a few words how the solutions resemble, and differ from, each other. $$ \begin{array}{l}{\text { (a) } d y / d t=-y+5, \quad y(0)=y_{0}} \\ {\text { (c) } d y / d t=-2 y+10, \quad y(0)=y_{0}}\end{array} \quad \text { (b) } d y / d t=-2 y+5, \quad y(0)=y_{0} $$

Short answer

A) All three solutions approach a horizontal asymptote at \(y=5\) as \(t \to \infty\), but the second ODE solution approaches this asymptote at a faster rate. The first and third ODEs have solutions with the same rate of decay, but may start from different initial positions. B) All three solutions approach a horizontal asymptote at \(y=5\) as \(t \to \infty\), but the first and third ODE solutions approach this asymptote at a faster rate. The second ODE has solutions with the same rate of decay, but may start from different initial positions. C) The solutions of the three ODEs have different horizontal asymptotes, with the second ODE having a faster decay rate than the first and third ODEs. D) All three solutions approach a horizontal asymptote at \(y=5\) as \(t \to \infty\), and all of them have the same rate of decay and follow an identical path to their common asymptotic value. Answer: A) All three solutions approach a horizontal asymptote at \(y=5\) as \(t \to \infty\), but the second ODE solution approaches this asymptote at a faster rate. The first and third ODEs have solutions with the same rate of decay, but may start from different initial positions.

Step-by-step solution

(a) Solve the first ODE: \(\frac{dy}{dt} = -y + 5\), \(y(0) = y_{0}\)

Separate the variables: $$\frac{dy}{y - 5} = -dt$$ Integrate both sides: $$\int \frac{dy}{y - 5} = -\int dt$$ $$\ln |y - 5| = -t + C$$ Exponentiate both sides: $$y - 5 = e^{-t + C}$$ $$y(t) = 5 + e^{-t} C_1$$ Use the initial condition to find \(C_{1}\): $$y(0) = y_{0}$$ $$y_{0} = 5 + e^0 C_1$$ $$C_1 = y_0 - 5$$ So the solution is: $$y(t) = 5 + (y_0 - 5)e^{-t}$$

(b) Solve the second ODE: \(\frac{dy}{dt} = -2y + 5\), \(y(0) = y_{0}\)

Separate the variables: $$\frac{dy}{2y - 5} = -dt$$ Integrate both sides: $$\int \frac{dy}{2y - 5} = -\int dt$$ $$\frac{1}{2}\ln |2y - 5| = -t + C$$ Exponentiate both sides: $$2y - 5 = e^{-2t + C}$$ $$y(t) = \frac{5 + e^{-2t} C_1}{2}$$ Use the initial condition to find \(C_{1}\): $$y(0) = y_{0}$$ $$y_{0} = \frac{5 + e^0 C_1}{2}$$ $$C_1 = 2y_0 - 5$$ So the solution is: $$y(t) = \frac{5 + (2y_0 - 5)e^{-2t}}{2}$$

(c) Solve the third ODE: \(\frac{dy}{dt} = -2y + 10\), \(y(0) = y_{0}\)

Separate the variables: $$\frac{dy}{2y - 10} = -dt$$ Integrate both sides: $$\int \frac{dy}{2y - 10} = -\int dt$$ $$\frac{1}{2}\ln |2y - 10| = -t + C$$ Exponentiate both sides: $$2y - 10 = e^{-t + C}$$ $$y(t) = 5 + \frac{e^{-t} C_1}{2}$$ Use the initial condition to find \(C_{1}\): $$y(0) = y_{0}$$ $$y_{0} = 5 + \frac{e^0 C_1}{2}$$ $$C_1 = 2(y_0 - 5)$$ So the solution is: $$y(t) = 5 + (y_0 - 5)e^{-t}$$

Plot the solutions for various values of \(y_{0}\).

To plot the solutions, use a graphing software or calculator to generate the following equations for different values of \(y_{0}\) (e.g., \(y_{0}=1,2,3,4\)): 1. \(y(t) = 5 + (y_0 - 5)e^{-t}\), first ODE 2. \(y(t) = \frac{5 + (2y_0 - 5)e^{-2t}}{2}\), second ODE 3. \(y(t) = 5 + (y_0 - 5)e^{-t}\), third ODE

Describe how the solutions resemble and differ from each other.

1. All three ODE solutions approach a horizontal asymptote at \(y = 5\) as \(t \to \infty\). This means that the long-term behavior of the solutions is the same. 2. In the second ODE solution, the solutions approach the asymptote at a faster rate due to the factor \(e^{-2t}\), which causes a steeper decline in the solution values. 3. In the first and third ODEs, the solutions have the same rate of decay (\(e^{-t}\)), but for different values of y_0, they will start from different initial positions. This means that, while their rate of decay is the same, they do not follow an identical path to their common asymptotic value.

Key concepts

Initial Value Problems

Initial value problems are a fundamental aspect of differential equations where we seek a function that satisfies an ordinary differential equation (ODE) and meets specific conditions at the initial point.
Here, an initial condition, such as \(y(0) = y_0\), defines the starting value for the variable, allowing us to solve for integration constants and find a unique solution specific to that starting point.

This initial condition is crucial because it ensures the differential equation has a single solution for the given starting point, directing the trajectory of the solution through the differential equation's landscape.

• It directly influences the solution and allows a tailored solution fitting the problem's context.
• Each unique initial condition can significantly alter how the solution behaves, even within the same differential equation.
• Consider for example, the ODE \(\frac{dy}{dt} = -y + 5\). With different initial values \(y_0\), the trajectory of the solution’s graph can vary distinctly, though all trajectories share long-term characteristics.

Separable Equations

A separable equation is a type of differential equation where variables can be separated onto opposite sides of the equation.
This form makes solving straightforward, as it allows easy integration of both sides.

For example, in the equation \(\frac{dy}{dt} = -y + 5\), we can rewrite it by moving all \(y\)-dependent terms to one side and the \(t\)-dependent terms to the other:
\(\frac{dy}{y - 5} = -dt\).
This separation facilitates the integration step, producing logarithmic functions that can be solved more easily.

• The technique involves moving all terms in \(y\) to one side of the equation and all terms in \(t\) to the other.
• Both sides of the equation can then be integrated, leading to a solution involving the natural logarithm.
• This method is crucial for solving many first-order ODEs because of its simplicity in integration.

Exponential Decay

Exponential decay describes a process where quantities decrease at a rate proportional to their current value.
It is a common model in disciplines like physics, chemistry, and even economics.

In differential equations like \(\frac{dy}{dt} = -y + 5\), the term \(e^{-t}\) emerges after solving as a component of the solution, indicating how rapidly the variable \(y\) decreases over time towards a stable value.

• Exponential decay is typically characterized by a term like \(e^{-\lambda t}\), where \(\lambda > 0\).
• In the provided problem, \(\lambda = 1\) and \(\lambda = 2\), explain the rate of decay which is directly tied to how quickly the function approaches its asymptotic value.
• The faster rate of decay is seen through steeper curves on the graph, as demonstrated in equations like \(\frac{dy}{dt} = -2y + 5\).

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as it approaches an infinitely large or small value (often time in dynamic systems).
In the context of the given differential equations, this behavior reveals much about the long-term trends of the function.

For equations like \(\frac{dy}{dt} = -y + 5\), all solutions have an asymptote at \(y = 5\) as \(t \rightarrow \infty\).

• This means, regardless of the initial value, \(y\) will settle at 5, reflecting stability or equilibrium in the system.
• The asymptote’s presence is due to the diminishing effect of the exponential decay terms over time, leading the function to converge to a single value.
• Understanding asymptotic behavior is vital as it provides insights into the stability and long-term predictions of the system, useful in fields like engineering and environmental studies.