Problem 47 - Calculus Early Transcendentals

Question

Show that \(y=x^{3}+3 x+1\) satisfies \(y^{\prime \prime \prime}+x y^{\prime \prime}-2 y^{\prime}=0\)

Step-by-Step Solution

Answer

The function satisfies the differential equation as all terms simplify to zero.

Step 1: Find the First Derivative

Compute the first derivative of the function, \(y = x^3 + 3x + 1\). The first derivative, \(y'\), is obtained by differentiating each term:\[y' = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x) + \frac{d}{dx}(1)\]\[y' = 3x^2 + 3\]

Step 2: Find the Second Derivative

Compute the second derivative of the function, which is the derivative of \(y'\). Differentiate \(y' = 3x^2 + 3\):\[y'' = \frac{d}{dx}(3x^2) + \frac{d}{dx}(3)\]\[y'' = 6x\]

Step 3: Find the Third Derivative

Compute the third derivative of the function, which is the derivative of \(y''\). Differentiate \(y'' = 6x\):\[y''' = \frac{d}{dx}(6x)\]\[y''' = 6\]

Step 4: Substitute Derivatives into Given Equation

Substitute the derived values of \(y''', y''\), and \(y'\) into the given differential equation \(y''' + xy'' - 2y' = 0\):\[6 + x(6x) - 2(3x^2 + 3) = 0\]\[6 + 6x^2 - 6x^2 - 6 = 0\]\[0 = 0\]The equation holds true.

Key Concepts

Second Derivative

The second derivative of a function tells us how the rate of change of a function's slope is varying. Essentially, it provides information about the curvature of the graph of the function. For the function given in the problem, which is \(y = x^3 + 3x + 1\), when we initially differentiate it, we find the first derivative as \(y' = 3x^2 + 3\). To find the second derivative, we take the derivative of \(y'\).

Differentiating \(y' = 3x^2 + 3\), gives us the second derivative:

• Start with \(y' = 3x^2 + 3\).
• Differentiate each term: \(\frac{d}{dx}(3x^2) + \frac{d}{dx}(3)\).
• This simplifies to \(6x\).

The second derivative, \(y'' = 6x\), provides insight into how the graph of the original function curves. A positive second derivative indicates the graph is curving upwards, while a negative value would imply a downward curve.

First Derivative

The first derivative of a function tells us about the slope of the tangent line to the curve at any point, which is also an indicator of the function's rate of change.

For the function \(y = x^3 + 3x + 1\), calculating the first derivative is the first step in our process:

• Simplify the expression to its basic parts: \(x^3\), \(3x\), and \(1\).
• Find the derivative of each part: \(\frac{d}{dx}(x^3) = 3x^2\), \(\frac{d}{dx}(3x) = 3\), and \(\frac{d}{dx}(1) = 0\).
• Combine these to obtain \(y' = 3x^2 + 3\).

The first derivative \(y'\) gives us a function that illustrates how quickly or slowly \(y\) changes as \(x\) varies. This is crucial for determining points of increase, decrease, and potential maximum or minimum values in the function.

Third Derivative

The third derivative of a function, quite less common in daily use compared to the first and second derivatives, provides information on how even the curve's rate of curvature changes.

In our exercise, to compute the third derivative, we begin with the second derivative we found, which is \(y'' = 6x\). Calculating the third derivative involves taking the derivative of this expression:

• Start with \(y'' = 6x\).
• Differentiate it: \(\frac{d}{dx}(6x) = 6\).

The third derivative, \(y''' = 6\), informs us about the jerk of the function—how the acceleration that describes the rate of curvature shifts. In certain applications, especially motion, this can be quite insightful.