Problem 39 - 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. $$\text { If } y=f(x), \text { then } d^{2} y / d x^{2}=2-(2 x-y)$$

Step-by-Step Solution

Answer

True, the second derivative matches the expression given.

Step 1: Understand the Given Expression

We need to determine if the statement is true or false. We have a solution to the differential equation \( \frac{dy}{dx} = 2x - y \). The statement asks us to find \( \frac{d^2y}{dx^2} \) and check whether it is equal to \( 2 - (2x - y) \).

Step 2: Differentiate the Given Expression

Differentiate the given differential equation \( \frac{dy}{dx} = 2x - y \) with respect to \( x \) to find \( \frac{d^2y}{dx^2} \). Applying differentiation on both sides, we get \( \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(2x - y) \).

Step 3: Apply the Differentiation

Compute the derivative of each term: The derivative of \( 2x \) with respect to \( x \) is \( 2 \), and the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \). Thus, \( \frac{d^2y}{dx^2} = 2 - \frac{dy}{dx} \).

Step 4: Substitute \( \frac{dy}{dx} = 2x - y \)

Substitute \( \frac{dy}{dx} = 2x - y \) into the expression for \( \frac{d^2y}{dx^2} \): \( \frac{d^2y}{dx^2} = 2 - (2x - y) \).

Step 5: Simplify the Expression

Simplify the expression \( 2 - (2x - y) \) to confirm the match. This results in \( 2 - 2x + y \) which simplifies to \( 2 + y - 2x \).

Step 6: Verify the Result

Since the simplified result is \( 2 - (2x - y) \), which matches \( \frac{d^2y}{dx^2} \), the statement is true.

Key Concepts

Second Derivative

When we talk about a second derivative, we're essentially exploring how the rate of change of a function's slope evolves. To put it simply, if we know how fast a function's value is changing at any point, the second derivative tells us how fast that rate of change itself is changing.
The second derivative, denoted as \( \frac{d^2y}{dx^2} \), measures the curvature or concavity of the graph of a function \( y = f(x) \). Here's what it can tell us:

• If \( \frac{d^2y}{dx^2} > 0 \), the function is curving upwards (concave up).
• If \( \frac{d^2y}{dx^2} < 0 \), the function is curving downwards (concave down).
• If \( \frac{d^2y}{dx^2} = 0 \), the function has an inflection point, where the curvature changes.

In the context of the given problem, we determined \( \frac{d^2y}{dx^2} \) by differentiating \( \frac{dy}{dx} = 2x - y \) with respect to \( x \). This process highlighted how differentiation allows us to grasp deeper layers of how a function behaves.

Differentiation

Differentiation is a mathematical process that involves finding the derivative of a function. It's a tool that helps us understand how a function changes as its input changes. Imagine you're hiking and you'd like to know how steep the hill is at every point. Differentiation provides that slope information.
For any function \( y = f(x) \), the derivative \( \frac{dy}{dx} \) conveys the rate of change of \( y \) concerning \( x \). Performing differentiation gives invaluable insights, such as:

• Determining the slope of the tangent line at a point.
• Identifying critical points where the slope is zero (potential maxima or minima).
• Understanding how quantities accumulate or decay over time.

In the context of our example, we differentiated the equation \( \frac{dy}{dx} = 2x - y \) to find the second derivative \( \frac{d^2y}{dx^2} \). Differentiation is key here, as it allows us to delve into the dynamics within a differential equation, revealing patterns of growth and decay.

Initial Value Problem

An Initial Value Problem (IVP) in mathematics involves finding a function that satisfies a differential equation and also meets given initial conditions. Think of it as solving a mystery where you're providing a starting clue for the detective to figure out the whole story.
When solving an IVP, you're usually given:

• A differential equation that describes a relationship between variables.
• An initial condition, specifying the value of the unknown function at a particular point (e.g., \( y(a) = b \)).

The goal is to find a function that fits the differential equation and passes through the specified initial point. For the exercise we were discussing, even though we don't have a specific initial value, the underlying process is crucial when tackling more complex differential problems.
By connecting an equation with initial conditions, we can provide a unique solution, adapting the abstract to specific scenarios. Thus, understanding IVPs becomes fundamental for solving real-world problems modeled by differential equations.