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}=-2 y\)

Short answer

The function decreases to 1 at time \( t = \frac{\ln(10)}{2} \), but never increases to 100.

Step-by-step solution

Set Up the Differential Equation

We start with the given differential equation \( \frac{dy}{dt} = -2y \). This is a separable first-order differential equation, which means we can separate the variables \( y \) and \( t \).

Separate Variables

To separate the variables, rearrange the equation to isolate \( dy \) and \( dt \): \( \frac{dy}{y} = -2 dt \).

Integrate Both Sides

Integrate both sides to find the general solution: \( \int \frac{dy}{y} = \int -2 dt \). This gives \( \ln |y| = -2t + C \), where \( C \) is the integration constant.

Solve for \( y \)

To find \( y \), solve the equation \( \ln |y| = -2t + C \) by exponentiating both sides: \( |y| = e^{C}e^{-2t} \). We can express this as \( y = Ce^{-2t} \), where \( C = e^{C} \).

Apply Initial Condition

Use the initial condition \( y_0 = 10 \) to solve for \( C \). Substitute \( y(0) = 10 \) into \( y = Ce^{-2t} \): \( 10 = Ce^{0} \), so \( C = 10 \). Therefore, the specific solution is \( y = 10e^{-2t} \).

Find Time for \( y = 100 \) or \( y = 1 \)

We need to find \( t \) when \( y = 100 \) or \( y = 1 \). First, for \( y = 100 \), \( 100 = 10e^{-2t} \) gives \( e^{-2t} = 10 \). Taking the natural logarithm: \( -2t = \ln(10) \), so \( t = -\frac{\ln(10)}{2} \), which is negative. Since time cannot be negative in this context, this situation does not occur. Now for \( y = 1 \), \( 1 = 10e^{-2t} \) gives \( e^{-2t} = \frac{1}{10} \). Thus, \( -2t = \ln\left(\frac{1}{10}\right) \) leading to \( t = \frac{\ln(10)}{2} \).

Key concepts

First-Order Differential Equations

A first-order differential equation is one in which the highest order derivative is the first derivative. These equations often describe a variety of processes such as growth and decay phenomena, including population models and chemical reactions. In the context of our exercise, we encounter the differential equation \(\frac{dy}{dt} = -2y\). Here, \(dy/dt\) is the rate of change of \(y\) over time \(t\). This equation is first-order because it involves only the first derivative of \(y\). It also suggests a relationship between the change in \(y\) and the current amount of \(y\). Understanding how these relationships form is crucial for setting up the correct mathematical models that describe real-world processes.

Initial Value Problems

An initial value problem often involves finding a function that goes through a specific starting point. These problems are widely used in physics and engineering to predict future behavior from known initial conditions. In our problem, the initial condition is given as \(y_0 = 10\). This means that at time \(t=0\), the value of \(y\) is 10. By incorporating this information into our solution, we can solve for any unknown constants and find specific solutions related to the initial conditions. This is a critical step, as it allows solutions to be tailored to particular situations rather than just providing a general solution. Initial conditions provide the anchor point from which changes over time are calculated.

Exponential Decay

Exponential decay describes a process where the quantity decreases at a rate proportional to its current value. This mathematical phenomenon is frequently seen in contexts such as radioactive decay, cooling of hot objects, and depreciation of asset value over time. In our problem, \(y = 10e^{-2t}\), the function describes an exponential decay, where the value of \(y\) decreases as time \(t\) increases. The term \(e^{-2t}\) indicates that as \(t\) becomes larger, the exponential function diminishes quickly, leading \(y\) closer to zero. This is significant when predicting when \(y\) will reach certain threshold values or decay significantly, as illustrated by solving for \(t\) when \(y = 1\). Understanding exponential decay is essential when modeling real-world behaviors that decrease over time.

Integration Technique

Integration is a fundamental tool in calculus, used for finding solutions to differential equations. In solving our separable differential equation, integration allows us to combine anti-derivatives to find a family of solutions. By separating variables first to \(\frac{dy}{y} = -2 dt\), and then integrating both sides, we obtain \(\ln |y| = -2t + C\). This process helps to transition from a differential equation to a solvable algebraic equation. The constant \(C\) is particularly important because it encapsulates the integration constants and initial conditions, allowing our solutions to fit the specific scenarios described in the initial value problem. Mastering integration techniques not only solves equations but also plays a pivotal role in analyzing dynamic systems.