Problem 18 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$\frac{d x}{d t}+\frac{2}{4-t} x=5, x(0)=4$$

Step-by-Step Solution

Answer

The short answer based on the given solution is: \( x(t) = -4(t-4)^2 - 5(t-4) \)

Step 1: Identify the integrating factor

Given the equation \( \frac{dx}{dt} + \frac{2}{4-t}x = 5 \), let's first write it in the standard form: \( \frac{dx}{dt} + P(t)x = Q(t) \), where in this case, \( P(t) = \frac{2}{4-t} \) and \( Q(t) = 5 \). The integrating factor (IF) is given by \( e^{\int P(t) dt} \). So we will find the integrating factor: \( IF = e^{\int \frac{2}{4-t} dt} \)

Step 2: Calculate the integrating factor

Now, calculate the integral: \( IF = e^{-2\int \frac{1}{t-4} dt} \) Use the substitution \( u = t-4, du = dt \). Then the integral becomes: \( IF = e^{-2\int \frac{1}{u} du} \) \( IF = e^{-2\ln|u|} \), where \( |u| = |t-4| \) Apply exponent rules and simplify: \( IF = \frac{1}{(t-4)^2} \)

Step 3: Multiply both sides of the equation by the integrating factor

Now, we multiply both sides of the given equation by the integrating factor: \( \frac{1}{(t-4)^2} \frac{dx}{dt} + \frac{2}{4-t} \frac{1}{(t-4)^2} x = \frac{1}{(t-4)^2} \cdot 5 \) Now, observe that the left side is the derivative of \( x(t) \cdot (t-4)^{-2} \). This is because: \( \frac{d}{dt}[x(t) \cdot (t-4)^{-2}] = x'(t) \cdot (t-4)^{-2} - 2x(t) \cdot (t-4)^{-3} \) Therefore, rewrite the equation as: \( \frac{d}{dt} [x(t) \cdot (t-4)^{-2}] = 5(t-4)^{-2} \)

Step 4: Integrate both sides of the equation

Integrate both sides of the equation: \( \int \frac{d}{dt} [x(t)(t-4)^{-2}] dt = 5\int (t-4)^{-2} dt \) We obtain: \( x(t)(t-4)^{-2}= -5(t-4)^{-1} + C \)

Step 5: Apply the initial condition and solve for x(t)

Now we apply the initial condition \( x(0) = 4 \) in the above equation: \( 4(0-4)^{-2} = -5(0-4)^{-1} + C \) \( 16 = 20 + C \) So, \( C = -4 \). Now, plug the value of C back into the equation: \( x(t)(t-4)^{-2} = -5(t-4)^{-1} - 4 \) Multiply both sides by \((t-4)^2\) to solve for x(t): \( x(t) = -5 (t-4) - 4(t-4)^2 \) Finally, we have found the function x(t): \( x(t) = -4(t-4)^2 - 5(t-4) \)

Key Concepts

Initial-Value Problem

In differential equations, an **Initial-Value Problem (IVP)** involves not just solving a differential equation, but also finding a specific solution that meets given conditions. These conditions are typically expressed as the value of the unknown function at a particular point (often with respect to time or another independent variable).

Consider the equation you've encountered: \[\frac{dx}{dt} + \frac{2}{4-t} x = 5, \quad x(0) = 4\]Here, the function \(x(t)\) is what we're solving for, and it's specified that \(x(0) = 4\). This tells us that when \(t = 0\), the value of \(x\) is 4. This initial condition is crucial.

Without it, you might find a general solution containing arbitrary constants. The initial condition allows you to pinpoint the exact solution that matches the condition you have. It essentially "anchors" the solution in the context of the problem, providing a specific, tailored answer rather than a broad family of solutions.

Integrating Factor

An **Integrating Factor** is a clever technique used in solving linear first-order differential equations. It essentially helps to simplify the equation to make it more directly solvable. The integrative step usually turns the problem into an equation involving the derivative of a single function.

For the problem given:\[\frac{dx}{dt} + \frac{2}{4-t}x = 5\]The integrating factor, denoted as \( IF \), is calculated by \( e^{\int P(t) dt} \), where \( P(t) = \frac{2}{4-t} \) here. This calculation usually involves integration, and in this particular problem, involves the substitution method to solve the integral.

Once you have the integrating factor \( IF = \frac{1}{(t-4)^2} \), it is used to multiply through the entire differential equation. This step creates an equation where the left-hand side becomes the derivative of a single expression, simplifying the equation and hence easing the process of finding a solution.

Linear Differential Equation

A **Linear Differential Equation** is called "linear" because it conforms to specific standards where the unknown function and its derivatives appear to the power of one (not squared or higher). This particular class of differential equations is widely studied because of their predictable behavior and solutions.

The general form of a linear first-order differential equation is:\[\frac{dy}{dt} + P(t)y = Q(t)\]Our example fits this form as:\[\frac{dx}{dt} + \frac{2}{4-t}x = 5\]Here, the equation maintains a linear relationship, where \(x\) and its derivative \(\frac{dx}{dt}\) are combined linearly with coefficients being functions of \(t\).

The linearity gives us the leverage to use systematic methods like the integrating factor, which simplifies finding solutions. In contrast, nonlinear differential equations often require more advanced techniques or numerical methods for solutions.

Solving Techniques

**Solving Techniques** for differential equations include a variety of methods, each suited to different types of problems. For linear first-order differential equations especially, tactics like separation of variables and integrating factors are quite common.

In this context, the solution process involved:

• Identifying the form of the equation and writing it in standard linear form.
• Calculating the integrating factor, which effectively transformed the equation into a simpler form.
• Multiplying the entire equation by the integrating factor to aid in easily calculating the derivative of a product of functions.
• Integrating both sides to find the solution.
• Implementing the initial condition to determine the specific solution that fits the problem.

By following these systematic techniques, one can navigate through solving the differential equation to reach the particular solution that fulfills the initial-value condition.