Problem 15 - Calculus for Biology and Medicine

Question

In Problems , solve each autonomous differential equc tion. $$ \frac{d x}{d t}=-2 x, \text { where } x(1)=3 $$

Step-by-Step Solution

Answer

The solution is \( x(t) = 3 e^{2 - 2t} \).

Step 1: Understanding the Problem

We need to solve the differential equation \( \frac{dx}{dt} = -2x \) given the initial condition \( x(1) = 3 \). This is a first-order linear autonomous differential equation.

Step 2: Separate Variables

In this equation, the variables can be separated. We rewrite it as \( \frac{1}{x} dx = -2 dt \). This allows us to integrate both sides separately.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( x \) and the right side with respect to \( t \). This yields \( \ln |x| = -2t + C \), where \( C \) is the integration constant.

Step 4: Solve for x

Exponentiate both sides to solve for \( x \): \( |x| = e^{-2t + C} = e^C e^{-2t} \). Define a new constant \( A = e^C \), so \( x = A e^{-2t} \).

Step 5: Apply the Initial Condition

Use the initial condition \( x(1) = 3 \) to find \( A \). Plugging in these values gives \( 3 = A e^{-2 \cdot 1} = A e^{-2} \). Therefore, \( A = 3 e^{2} \).

Step 6: Write the Final Solution

Substitute \( A \) back into the solution for \( x(t) \): \( x(t) = 3 e^{2} e^{-2t} = 3 e^{2 - 2t} \). This is the general solution satisfying the given initial condition.

Key Concepts

Integration

Integration is a fundamental concept in calculus that helps in solving differential equations. When tackling a differential equation like \( \frac{dx}{dt} = -2x \), we often use integration to find a general solution. This involves finding an antiderivative of the function or expression, allowing us to move from the derivative back to the original function. In the problem, we first separated the variables to write the equation as \( \frac{1}{x} dx = -2 dt \). By integrating both sides separately, we find the antiderivatives:

• Left side: Integrating \( \frac{1}{x} \), we get \( \ln |x| \).
• Right side: Integrating \(-2 \), we get \(-2t \).

After integrating, we introduce the integration constant \( C \) because integration is reverse differentiation and always includes an arbitrary constant. Thus, the integration result is \( \ln |x| = -2t + C \). This essential step converts the differential equation into a more workable form.

First-Order Differential Equation

A first-order differential equation involves derivatives of the first order, meaning the highest derivative in the equation is the first derivative. In this exercise, \( \frac{dx}{dt} = -2x \) is a classic example of a first-order differential equation. These types of equations are particularly prevalent because they describe a wide range of natural phenomena. They involve rates of change and can model various processes like population growth, radioactive decay, or in this case, exponential decay.For first-order autonomous differential equations, where the right-hand side depends only on \( x \), techniques like separation of variables and integration are crucial for solving them. The steps include isolating the variables and integrating each side with respect to its variable. The solution obtained represents an infinite family of solutions, distinguished by an added constant \( C \), until an initial condition is applied for specificity.

Initial Value Problem

Solving an initial value problem means finding a specific solution to a differential equation that satisfies an initial condition. In this problem, the initial condition \( x(1) = 3 \) provides a starting point for the solution. Why is this important? Initial conditions allow us to determine the unique solution from a family of potential solutions. The general solution obtained from the integration step was \( x = A e^{-2t} \). However, the integration constant \( A \) needs to be evaluated.Applying the initial condition, substitute \( t = 1 \) and \( x = 3 \) into the equation:

• The equation becomes \( 3 = A e^{-2} \).
• Solving for \( A \), we find \( A = 3 e^{2} \).

Substitute \( A \) back into the general solution, providing a specific solution \( x(t) = 3 e^{2 - 2t} \) that meets the initial condition. This process of using the initial value narrows down the infinite possibilities to one precise solution, tailoring it to the conditions given in the problem.