Problem 16 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solution is \( x(t) = \frac{1}{3} - \frac{7}{3}e^{3-3t} \); using the initial condition, determine \( C = -\frac{7}{3}e^3 \).

Step 1: Understand the Differential Equation Problem

The differential equation given is \( \frac{d x}{d t} = 1 - 3x \). This is a first-order linear autonomous differential equation. We also have an initial condition \( x(1) = -2 \). Our goal is to find the particular solution to this differential equation that satisfies this initial condition.

Step 2: Separate Variables

To solve the differential equation, first separate the variables, if possible. Rewrite the equation as \( \frac{d x}{d t} + 3x = 1 \). This is already in a form that can be solved using an integrating factor, but is not perfectly separable in the traditional sense. We will use the integrating factor method here.

Step 3: Use Integrating Factor Method

The standard form for the integrating factor is \( \frac{d x}{d t} + P(t)x = Q(t) \). Here, \( P = 3 \) and \( Q = 1 \). The integrating factor is \( e^{\int P(t) dt} = e^{3t} \). Multiply the entire equation by this integrating factor: \[ e^{3t} \frac{d x}{d t} + 3e^{3t}x = e^{3t} \].

Step 4: Write Left Side as a Derivative

Observe that the left side of the equation is the derivative of the product of the integrating factor and \( x \). Rewrite it as: \[ \frac{d}{dt}(e^{3t} x) = e^{3t} \].

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^{3t} x) \, dt = \int e^{3t} \, dt \].Thus, \[ e^{3t} x = \frac{1}{3} e^{3t} + C \], where \( C \) is a constant.

Step 6: Solve for \( x \)

Solve for \( x \) by dividing every term by \( e^{3t} \):\[ x(t) = \frac{1}{3} + Ce^{-3t} \].

Step 7: Apply Initial Condition

Use the initial condition \( x(1) = -2 \) to find \( C \):\[ -2 = \frac{1}{3} + Ce^{-3 \times 1} \].Simplifying, \( -2 = \frac{1}{3} + Ce^{-3} \), leading to:\[ Ce^{-3} = -2 - \frac{1}{3} = -\frac{6}{3} - \frac{1}{3} = -\frac{7}{3} \].Thus,\[ C = -\frac{7}{3}e^3 \].

Step 8: Write the Particular Solution

Substitute \( C \) back into the general solution:\[ x(t) = \frac{1}{3} - \frac{7}{3}e^{3-3t} \].This is the particular solution that satisfies the given differential equation and initial condition.

Key Concepts

Integrating Factor Method

The integrating factor method is a powerful technique to solve first-order linear differential equations. This method is particularly useful when the separation of variables is not easily applicable. Here's how it works:

• First, consider the standard form of the equation, which is written as \( \frac{dy}{dx} + P(x)y = Q(x) \).
• The integrating factor \( \mu(x) \) is a function you multiply through the entire equation. This factor is given by \( e^{\int P(x)\, dx} \).
• Multiply every term by this integrating factor, transforming the left side of the equation into the derivative of the product \( \mu(x) \cdot y \).
• Now the equation looks like \( \frac{d}{dx}(\mu(x) y) = \mu(x) Q(x) \).

Next, simply integrate both sides to solve for the product \( \mu(x) y \), and then solve for \( y \) by dividing through by the integrating factor. This yields the general solution. The integrating factor method helps make complex equations manageable by transforming them into simpler integrable forms.

First-order Linear Differential Equations

First-order linear differential equations involve derivatives of one order with respect to one variable. They are generally expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Dealing with these types of equations requires specific approaches:

• Identification: Recognize the leading term is the first derivative. There are no higher derivatives present in first-order cases.
• Standard Form: It is crucial to arrange the equation in the form discussed above to utilize solving techniques.
• Applications: These equations frequently appear in real-world scenarios across physics, biology, and engineering, providing models for exponential growth and decay, circuit analysis, and cooling processes.

The beauty of first-order equations lies in their simplicity and wide-ranging applications. Their solutions often involve logarithmic or exponential functions, which are foundational concepts in mathematics.

Initial Condition

When solving differential equations, initial conditions are used to find a particular solution from a family of general solutions. Initial conditions are specific values given for the solution or its derivatives at a particular point. Here's why they matter:

• Initial conditions make an indefinite solution specific by solving for constants that appear after integration.
• The given value usually represents the state of the system at the beginning of observation, like in our exercise where \( x(1) = -2 \).
• These conditions are essential in applications, as real-world problems often require knowing how a situation evolves from a certain starting point.

By incorporating initial conditions, one can find a solution that not only solves the differential equation but also aligns perfectly with the initial scenario or data at hand. This leads to more meaningful and representative results.