Problem 47 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

The function satisfies the differential equation as the expression simplifies to 0.

Step 1: Identify the Function and Its Derivatives

We are given the function \( y = x^3 + 3x + 1 \). To show that this function satisfies the given differential equation \( y''' + xy'' - 2y' = 0 \), we first need to find its derivatives: the first, second, and third derivatives.

Step 2: Calculate the First Derivative

Calculate the first derivative \( y' \) of the function \( y = x^3 + 3x + 1 \). Using the power rule of differentiation, we have:\[ y' = \frac{d}{dx}(x^3 + 3x + 1) = 3x^2 + 3 \].

Step 3: Calculate the Second Derivative

Calculate the second derivative \( y'' \) by differentiating the first derivative \( y' = 3x^2 + 3 \). Again, using the power rule:\[ y'' = \frac{d}{dx}(3x^2 + 3) = 6x \].

Step 4: Calculate the Third Derivative

Calculate the third derivative \( y''' \) by differentiating the second derivative \( y'' = 6x \). Applying the power rule:\[ y''' = \frac{d}{dx}(6x) = 6 \].

Step 5: Substitute into the Differential Equation

Now substitute \( y''', \ y'', \) and \( y' \) into the left side of the differential equation:- \( y''' = 6 \)- \( xy'' = x(6x) = 6x^2 \)- \( -2y' = -2(3x^2 + 3) = -6x^2 - 6 \).

Step 6: Simplify the Expression

Combine these results:\[ y''' + xy'' - 2y' = 6 + 6x^2 - 6x^2 - 6 \].Simplify:\[ 6 + 6x^2 - 6x^2 - 6 = 0 \].

Step 7: Conclusion

Since the expression simplifies to zero, the original function \( y = x^3 + 3x + 1 \) satisfies the differential equation \( y''' + xy'' - 2y' = 0 \).

Key Concepts

Derivatives

Understanding derivatives is crucial when working with differential equations. A derivative represents the rate of change of a function concerning its variable, typically x.

In the context of our original problem, we need derivatives up to the third level. Here is how derivatives can be broken down:

• First Derivative: This measures the slope or rate of change of the original function. For the function \( y = x^3 + 3x + 1 \), the first derivative is \( y' = 3x^2 + 3 \).
• Second Derivative: This derivative gives us information about the curvature or concavity of the original function. It is the derivative of the first derivative. In our example, \( y'' = 6x \).
• Third Derivative: This derivative is the rate of change of the second derivative, often related to how the curvature itself changes. Here, it simplifies to \( y''' = 6 \).

Mastering derivatives allows for solving complex equations, like the one in this problem.

Power Rule

The power rule is a fundamental technique in calculus used to find derivatives of polynomial functions quickly. It's especially useful when you're dealing with functions presented as a sum of power terms.

The power rule states that if you have a term of the form \( x^n \), its derivative is \( nx^{n-1} \). Let's apply this to gain deeper insight into the exercise:

• For \( y = x^3 + 3x + 1 \), applying the power rule to the term \( x^3 \) gives \( 3x^2 \).
• Similarly, for \( 3x \), which is \( 3x^1 \), the derivative becomes \( 3 \cdot 1x^{1-1} \), simplifying to \( 3 \).
• Constant terms like \( +1 \) disappear when differentiated because their rate of change is zero.

With this rule, finding derivatives becomes a straightforward process. Learning to apply it efficiently is key for handling more intricate mathematical problems.

Differential Equation Solution Steps

Solving differential equations involves finding a function that satisfies a given equation involving derivatives.

Our task was to demonstrate that \( y = x^3 + 3x + 1 \) meets a specific differential equation, \( y''' + xy'' - 2y' = 0 \). Here's a deeper dive into the necessary steps:

• Identify and Differentiate: First, identify the function and its relevant derivatives up to the third derivative, as shown in previous sections.
• Substitute: Insert the derivatives into the given differential equation. This action tests if, when combined, they result in the equation simplifying to zero.
• Simplify: Combining the derivatives results in verifying the equation. Specifically, you notice terms like \( 6x^2 \) cancel each other out, leading to \( 0 \).
• Conclusion: When the equation holds true (simplifies to zero), it confirms that the initial function is indeed a solution.

Following these structured steps ensures a thorough understanding and solution of differential equations.