Problem 5 - Calculus

Question

Solve starting from \(y_{0}=10\). At what time does \(y\) increase to 100 or drop to \(1 ?\) $$ \frac{d y}{d t}=4 y $$

Step-by-Step Solution

Answer

\( y \) reaches 100 at \( t = \frac{\ln(10)}{4} \) and drops to 1 at \( t = \frac{\ln(0.1)}{4} \).

Step 1: Understand the differential equation

The given differential equation is \( \frac{dy}{dt} = 4y \). This is a first-order linear differential equation, and it indicates exponential growth since the coefficient of \( y \) is positive.

Step 2: Solve the differential equation

To solve \( \frac{dy}{dt} = 4y \), separate variables by dividing both sides by \( y \) and multiplying by \( dt \) to get \( \frac{1}{y} dy = 4 dt \). Integrate both sides: \( \int \frac{1}{y} dy = \int 4 dt \), resulting in \( \ln|y| = 4t + C \), where \( C \) is the constant of integration.

Step 3: Solve for y in terms of t

Exponentiate both sides to solve for \( y \): \( e^{\ln |y|} = e^{4t + C} \) giving \( |y| = e^C e^{4t} \). Simplifying, we have \( y = Ce^{4t} \), where \( C = e^C \) is another constant.

Step 4: Apply initial condition

Substitute \( y(0) = 10 \) into the equation to find \( C \): \( 10 = Ce^0 \), which means \( C = 10 \). So the solution becomes \( y(t) = 10e^{4t} \).

Step 5: Find time when y increases to 100

Set \( y(t) = 100 \) and solve for \( t \): \( 10e^{4t} = 100 \). Dividing by 10 gives \( e^{4t} = 10 \). Taking natural log on both sides, \( 4t = \ln(10) \). Thus, \( t = \frac{\ln(10)}{4} \).

Step 6: Check if y can drop to 1

Set \( y(t) = 1 \) and solve for \( t \): \( 10e^{4t} = 1 \). Dividing by 10 gives \( e^{4t} = 0.1 \). Taking natural log on both sides, \( 4t = \ln(0.1) \). Thus, \( t = \frac{\ln(0.1)}{4} \).

Key Concepts

Exponential Growth

Exponential growth occurs when the rate of change of a quantity is proportional to its current size. In the context of differential equations, you encounter this when dealing with equations like \( \frac{dy}{dt} = ky \). Here, the function \( y(t) \) shows exponential growth if \( k \), a constant, is positive.
Exponential functions have the form \( y(t) = Ce^{kt} \), where \( C \) is the initial value at \( t=0 \). The constant \( e \) is the base of natural logarithms, approximately 2.71828. This results in an increasingly rapid growth or decay over time, depending on the sign of \( k \).

• If \( k > 0 \), the function models exponential growth, like population increase or investment returns.
• If \( k < 0 \), it models exponential decay, seen in processes like radioactive decay or cooling.

In the provided problem, the positive coefficient \( 4 \) signifies that \( y \) grows exponentially over time.

Initial Condition

When solving differential equations, it's vital to have an initial condition to find a particular solution. An initial condition offers a specific starting point or value for the unknown function. Typically, this is written in the form \( y(t_0) = y_0 \), where \( t_0 \) is the initial time, and \( y_0 \) is the initial quantity.
With an initial condition, the arbitrary constant that appears after integration can be determined. In our problem, the initial condition given is \( y(0) = 10 \), which means initially, at time \( t = 0 \), the value of \( y \) is 10.
This information allows us to solve for the constant \( C \) in the general solution \( y(t) = Ce^{4t} \).

• Plugging in \( t = 0 \) and \( y = 10 \), we find \( C = 10 \).
• The specific solution becomes \( y(t) = 10e^{4t} \).

This step is crucial because it transforms a general solution into one that satisfies the initial condition provided in the problem.

Separation of Variables

Separation of variables is a key method for solving first-order differential equations. It involves rearranging an equation so that each variable and its differential appear on opposite sides of the equation.
For the equation \( \frac{dy}{dt} = 4y \), separating variables entails rearranging it to \( \frac{1}{y} dy = 4 dt \). With this setup, each side can be integrated independently.

• The left side involves integrating \( \frac{1}{y} \), resulting in \( \ln |y| \).
• The right side integrates to \( 4t \) plus an integration constant \( C \).

Separation of variables is particularly useful for solvable forms of equations where one side contains only variables of \( y \) and the other only variables of \( t \).
This method simplifies the solving process immensely, especially for equations exhibiting exponential growth.

Integration of Differential Equations

After successfully separating variables in a differential equation, you move on to integration to find a general solution. This step involves integrating both sides of the equation obtained from separation.
For the form \( \frac{1}{y} dy = 4 dt \), integrate both:

• The left side yields \( \ln |y| \).
• The right side results in \( 4t + C \), when considering the constant of integration.

Once integrated, you solve for \( y \) by exponentiating, as seen in \( e^{\ln |y|} = e^{4t + C} \), simplifying to \( |y| = e^C e^{4t} \). This reduces further to \( y = Ce^{4t} \) by expressing \( e^C \) as a constant \( C \).
Without integration, it would be challenging to determine the functional form of solutions to differential equations, especially for equations demonstrating exponential growth dynamics.