Question

Solve the following initial-value problems starting from \(y_{0}=10 .\) At what time does \(y\) increase to 100 or drop to \(1 ?$$\frac{d y}{d t}=e^{-4 t}\)

Short answer

\(y\) reaches neither 100 nor 1; it never changes to these values physically.

Step-by-step solution

Understand the Differential Equation

The given differential equation is \(\frac{d y}{d t}=e^{-4 t}\). This is a first-order ordinary differential equation (ODE). Our task is to solve this ODE with the initial condition \(y_{0}=10\) and determine the time \(t\) when \(y=100\) or \(y=1\).

Separate Variables

Separate the variables in the differential equation. We have: \[dy = e^{-4t} \, dt\] Integrate both sides to solve for \(y\).

Integrate Both Sides

Integrate the left side, \(\int dy\), which gives \(y\). Integrate the right side, \(\int e^{-4t} \, dt\), which results in:\[y = -\frac{1}{4}e^{-4t} + C\] where \(C\) is the constant of integration.

Apply Initial Condition

Apply the initial condition \(y(0) = 10\) to solve for \(C\): \[10 = -\frac{1}{4}e^{0} + C\]\[10 = -\frac{1}{4} + C\]\[C = 10 + \frac{1}{4} = \frac{40}{4} + \frac{1}{4} = \frac{41}{4}\]Substitute \(C\) back into the equation.

Solve for Specific Times

The equation for \(y(t)\) is now:\[y = -\frac{1}{4}e^{-4t} + \frac{41}{4}\]For \(y = 100\):\[100 = -\frac{1}{4}e^{-4t} + \frac{41}{4}\]Solve for \(t\).For \(y = 1\):\[1 = -\frac{1}{4}e^{-4t} + \frac{41}{4}\]Solve for \(t\).

Solve for Time when \(y=100\)

Rearrange and solve for \(t\):\[100 = -\frac{1}{4}e^{-4t} + \frac{41}{4}\]\[\frac{400}{1}= -e^{-4t} + 41\]\[ e^{-4t} = 41 - 400 \]Since \(41 - 400 \) is negative, this situation is not possible. So, \(y\) never reaches 100.

Solve for Time when \(y=1\)

Rearrange and solve for \(t\):\[1 = -\frac{1}{4}e^{-4t} + \frac{41}{4}\]\[\frac{4}{1}= -e^{-4t} + 41\]\[ e^{-4t} = 41 - 4 \]\[ e^{-4t} = 37 \]Solving gives,\[-4t = \ln(37)\]\[t = -\frac{\ln(37)}{4}\]

Determine Feasibility and Conclusion

Evaluate \(t = -\frac{\ln(37)}{4}\) using a calculator:\[t \approx -0.8935\]This result doesn't make sense physically since time cannot be negative. The function doesn't change to \(y=1\) again.

Key concepts

Initial Value Problems

In mathematics, an Initial Value Problem (IVP) is a differential equation coupled with a specific starting point or initial condition. This problem seeks to find a function that satisfies the differential equation and also passes through a given point in the space. In the context of the differential equation given by \( \frac{dy}{dt} = e^{-4t} \), the initial value \( y_0 = 10 \) specifies the starting point for the solution curve at time \( t=0 \).
To solve an IVP, you will first solve the differential equation in a general form and then apply the initial condition to solve for any constants of integration. This ensures the solution fits the specific conditions of the problem. In essence, initial value problems allow you to determine a unique solution from a family of potential solutions that satisfy both the differential equation and the initial condition.

Separation of Variables

Separation of variables is a popular method used to solve ordinary differential equations, particularly when the variables can be divided on either side of the equation.
The method involves three main steps:

• Rewrite the differential equation in the form \( f(y) \, dy = g(t) \, dt \).
• Integrate both sides separately, resulting in the antiderivatives: \( \int f(y) \, dy \) and \( \int g(t) \, dt \).
• Simplify and solve for the function of interest, often \( y \), by incorporating the constant of integration.

In our problem, the equation \( \frac{dy}{dt} = e^{-4t} \) is separated into \( dy = e^{-4t} \, dt \). This method is efficient and works best when the equation, after separation, can be integrated directly.

Integration

Integration is a fundamental concept in calculus that involves finding the antiderivative of a given function. It is crucial in solving differential equations, as it allows you to reverse the process of differentiation. In this context, integration turns a complex problem involving rates of change into one where you can find the actual function itself.
In our exercise, we integrate both sides of the equation \( dy = e^{-4t} \, dt \). On the left side, \( \int dy \) results in \( y \), the antiderivative of \( dy \). On the right side, integration of \( \int e^{-4t} \, dt \) leads to \( -\frac{1}{4} e^{-4t} + C \), thanks to the chain rule, where \( C \) is the integration constant.
The integration constant is determined by applying the initial condition, ensuring that the particular solution satisfies \( y(0) = 10 \). This integration process ultimately helps us express \( y \) in terms of \( t \), making it possible to further analyze the problem.

Exponential Functions

Exponential functions are a core component in various mathematical concepts. In differential equations, they often represent growth or decay processes. An exponential function has the form \( f(t) = ae^{bt} \), where \( a \) is a constant, \( e \) is the base of natural logarithms, and \( b \) determines the rate of growth or decay.
In our exercise, \( e^{-4t} \) signifies a decay process, as evidenced by the negative exponent. The function decreases rapidly as \( t \) increases. When integrating \( e^{-4t} \), you incorporate the constant factor from the exponent resulting in \( -\frac{1}{4}e^{-4t} \). The triple role of understanding the nature, integration, and application of exponential functions is essential in solving initial value problems involving growth and decay. Here, recognizing \( e^{-4t} \) assists in forming the solution curve by providing the basis for the rate at which \( y \) changes over time.