Problem 44 - Calculus

Question

Decide whether the statement is true or false. Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=2 x-y .\) Justify your answer. If \(g(x)\) is a different solution to the differential equation \(d y / d x=2 x-y,\) then \(\lim _{x \rightarrow \infty}(g(x)-f(x))=0\) [Hint: Show that \(w=g(x)-f(x)\) satisfies the differential equation \(d w / d x=-w .]\)

Step-by-Step Solution

Answer

True, \( \lim_{x \rightarrow \infty} (g(x) - f(x)) = 0 \).

Step 1: Understand the Problem

We have a differential equation \( \frac{dy}{dx} = 2x - y \). We assume \( y = f(x) \) is a solution to this equation. We need to determine if a different solution \( g(x) \) to the equation satisfies \( \lim_{x \rightarrow \infty} (g(x) - f(x)) = 0 \).

Step 2: Define the Difference Function \(w\)

Let \( w(x) = g(x) - f(x) \). We are analyzing \( g(x) \) and \( f(x) \) in terms of their difference \( w(x) \). We need to find the behavior of \( w(x) \) as \( x \to \infty \).

Step 3: Find the Differential Equation for \(w(x)\)

Differentiate \( w(x) \) with respect to \( x \): \[ \frac{d}{dx}[g(x) - f(x)] = \frac{dg}{dx} - \frac{df}{dx} \]Substituting the original differential equation for both \( g(x) \) and \( f(x) \):\[ \frac{dg}{dx} = 2x - g(x), \quad \frac{df}{dx} = 2x - f(x) \]Thus,\[ \frac{dw}{dx} = (2x - g(x)) - (2x - f(x)) = -w(x) \]

Step 4: Solve the Differential Equation for \(w(x)\)

The differential equation \( \frac{dw}{dx} = -w(x) \) is separable. We rearrange it as\[ \frac{dw}{w} = -dx \]Integrate both sides: \[ \int \frac{1}{w} dw = \int -1 dx \] This gives \[ \ln |w| = -x + C \] where \( C \) is the constant of integration. Exponentiating both sides, we obtain \[ |w(x)| = e^{-x + C} = Ce^{-x} \] Thus, \[ w(x) = Ce^{-x} \] where \( C \) is the constant.

Step 5: Evaluate \(\lim_{x \rightarrow \infty} w(x)\)

As \( x \to \infty \), the exponential term \( e^{-x} \to 0 \). Therefore, \[ \lim_{x \rightarrow \infty} w(x) = \lim_{x \rightarrow \infty} Ce^{-x} = 0 \] which means \[ \lim_{x \rightarrow \infty} (g(x) - f(x)) = 0 \].

Key Concepts

Limit of a Function

The concept of limits is fundamental in calculus and helps us understand the behavior of functions as they approach a certain point, or as they tend to infinity. In our exercise, we are considering whether the difference between two solutions to a differential equation tends to zero as the independent variable approaches infinity.

This involves examining the limit \(\lim_{x \rightarrow \infty}(g(x) - f(x)) = 0.\)

In simple terms, as we let \(x\) increase indefinitely, we are interested in what happens to the expression \(g(x) - f(x)\). If it approaches zero, it suggests that the solutions \(g(x)\) and \(f(x)\) become increasingly similar as \(x\) grows.

• The behavior at infinity is crucial in many applications, for instance in physics and engineering, providing insights into asymptotic behavior.
• Finding the limit often involves simplifying expressions and identifying dominant terms at large values of \(x\).

Separable Differential Equations

Separable differential equations are a special class that can be easily solved using a straightforward technique. Once an equation is identified as such, it can be manipulated into two expressions that involve only one variable each on the different sides of the equation.

Take the equation: \(\frac{dw}{dx} = -w.\)

Here, the variables \(w\) and \(x\) can be separated:

• Move all terms involving \(w\) to one side: \(\frac{1}{w} dw = -dx.\)
• Integrate both sides: \(\int \frac{1}{w} dw = \int -1 dx.\)

The separability allows for such manipulation and integration, showing the power of this method. It results in a general solution which includes an arbitrary constant, often derived from an initial condition.

• It simplifies complex problems and is applicable in various scientific fields.
• The technique is a foundation for solving more advanced differential equations.

Exponential Functions

Exponential functions occur frequently in mathematics, characterized by the constant \(e\), approximately 2.718, raised to the power of a variable. They exhibit unique growth or decay properties, depending on whether the exponent is positive or negative.

In our solution, the function \(|w(x)| = Ce^{-x}\)utilizes an exponential function to describe the decay of \(w\) over time as \(x\) increases. This term \(Ce^{-x}\) approaches zero as \(x\) goes to infinity.

• Exponential decay, as seen here, is typically very rapid. This is why \(\lim_{x \rightarrow \infty} Ce^{-x} = 0.\)
• Such functions are pivotal in natural sciences, modeling processes like radioactive decay and population dynamics.
• The constant \(C\) determines the initial value or condition of the exponential function.