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

\( y \) drops to 1 at \( t = \frac{\ln 10}{2} \); it never increases to 100.

Step-by-step solution

Write Down the Initial-Value Problem

The given differential equation is \( \frac{d y}{d t} = -2 y \) with the initial condition \( y(0) = 10 \). We need to find the time \( t \) when \( y(t) = 100 \) or \( y(t) = 1 \).

Solve the Differential Equation

This is a first-order linear differential equation, which can be solved by separating variables. We separate variables and integrate:\[ \int \frac{1}{y} \, dy = \int -2 \, dt \]The integrals give us:\[ \ln |y| = -2t + C \]where \( C \) is the integration constant.

Solve for the Constant Using Initial Values

Using the initial condition \( y(0) = 10 \), substitute \( t = 0 \) and \( y = 10 \) into the equation:\[ \ln |10| = C \]Thus, \( C = \ln 10 \). The equation becomes:\[ \ln |y| = -2t + \ln 10 \].

Solve for \( y \) as a Function of \( t \)

Exponentiate both sides to solve for \( y \):\[ |y| = e^{-2t + \ln 10} = 10e^{-2t} \]Since \( y > 0 \), we have:\[ y = 10e^{-2t} \].

Find \( t \) When \( y = 100 \)

Set \( y = 100 \) and solve for \( t \):\[ 100 = 10e^{-2t} \]\[ 10 = e^{-2t} \]Taking natural logarithms, we get:\[ \ln 10 = -2t \]Thus,\[ t = -\frac{\ln 10}{2} \].

Find \( t \) When \( y = 1 \)

Set \( y = 1 \) and solve for \( t \):\[ 1 = 10e^{-2t} \]\[ e^{-2t} = \frac{1}{10} \]Taking natural logarithms gives:\[ -2t = \ln \left(\frac{1}{10}\right) = -\ln 10 \]Thus,\[ t = \frac{\ln 10}{2} \].

Key concepts

Separation of Variables

In order to solve the differential equation \(\frac{dy}{dt} = -2y\), we used a technique called "separation of variables." This method is commonly used to solve simple differential equations where you can manipulate and rearrange the equation so that each variable ends up on opposite sides.

These are the basic steps:

• Move all terms involving \(y\) to one side of the equation and all terms involving \(t\) (or time, in this case) to the other side.
• Integrate both sides separately to find a general solution.

If we apply this to our equation:\[ \int \frac{1}{y} \, dy = \int -2 \, dt \]Integrating gives us:\[ \ln |y| = -2t + C \]where \(C\) is an integration constant. Separation of variables is a powerful method because it breaks down the problem into smaller, more manageable pieces through integration. This technique is versatile for initial-value problems as long as the variables can be separated.

Initial Value Problems

An initial value problem is a type of differential equation that, besides the equation itself, includes initial conditions. These conditions are used to find particular solutions, as opposed to general solutions. The initial condition specifies the state of the system at \(t = 0\) or another initial time.

In our exercise, the initial condition is \(y(0) = 10\). It's like saying, "At time zero, the value of \(y\) is 10."

Here's why initial conditions are important:

• They allow us to solve for the constant \(C\) after we integrate, giving us a specific solution tailored to the problem.
• With the initial values, the solution is no longer just any curve but a specific curve that passes through the initial point.

In our work:\[ \ln |y| = -2t + C \]we use the initial condition \(\ln|10| = C\), resulting in \(C = \ln 10\). This makes the solution specific by anchoring it at the point \(t = 0, y = 10\). With this known, we can determine behavior at other points in time.

Exponential Growth and Decay

The equation \(y = 10e^{-2t}\) obtained after solving the differential equation illustrates the concept of exponential decay. Exponential growth and decay are processes where the rate of change of a quantity is proportional to the quantity itself.

With exponential decay, things decrease rapidly at first and then at a slower rate over time.

• In our example, the term \(e^{-2t}\) causes the initial quantity \(10\) to decrease over time as \(t\) increases.
• The negative sign in the exponent \(-2t\) specifically indicates decay rather than growth.

Such behavior is found in numerous natural and economic phenomena, like radioactive decay and cooling processes.

To find the time \(t\) for specific \(y\) values (\(y = 1\) or \(y = 100\)), we solve the equation for \(t\). For \(y=1\), setting \(1 = 10e^{-2t}\) gives \(t = \frac{\ln 10}{2}\), showing how quickly decay reaches low values. Similarly, solving for \(y = 100\) also involves understanding that it is theoretically not possible with decay unless there is an error, highlighting that exponential decay leads to reduction and not growth.