Question

Determining a Solution In Exercises \(13-20\) , determine whether the function is a solution of the differential equation \(y^{(4)}-16 y=0\) . $$ y=5 \ln x $$

Short answer

The function \(y = 5\ln x\) is not a solution to the differential equation \(y^{(4)}-16 y = 0\).

Step-by-step solution

Understand the Function and the Differential Equation

The function given is \(y=5 \ln x\). The differential equation is \(y^{(4)}-16 y=0\), meaning the fourth derivative of our function subtracted by 16 times the original function equals to zero.

Calculate the First Derivative

Take the derivative of the function \(y=5 \ln x\) with respect to \(x\) using the rule of logarithmic differentiation. The first derivative is \(y'=\frac{5}{x}\).

Calculate the Second Derivative

Take the derivative of the first derivative \(y'=\frac{5}{x}\) with respect to \(x\). The second derivative is \(y''=-\frac{5}{x^2}\).

Calculate the Third Derivative

Take the derivative of the second derivative \(y''=-\frac{5}{x^2}\) with respect to \(x\). The third derivative is \(y'''=\frac{10}{x^3}\).

Calculate the Fourth Derivative

Take the derivative of the third derivative \(y'''=\frac{10}{x^3}\) with respect to \(x\). The fourth derivative is \(y^{(4)}=-\frac{30}{x^4}\).

Substitute into the Differential Equation

Now that we have the fourth derivative, substitute it into the differential equation \(y^{(4)}-16 y=0\) where \(y^{(4)}=-\frac{30}{x^4}\) and \(y=5 \ln x\). After substitution, we have: \(-\frac{30}{x^4}-16(5 \ln x) \neq 0\)

Verify Equation

Check if after substitution the equation holds true. In our case it does not hold true, indicating that the provided function is not a solution for the given differential equation.

Key concepts

Fourth Derivative

When working with differential equations, calculating higher-order derivatives, like the fourth derivative, can reveal important aspects of a function's behavior. The notation \( y^{(4)} \) represents the fourth derivative of the function \( y \) with respect to \( x \). The process involves taking the derivative of a derivative multiple times, which can seem complex at first.

To arrive at the fourth derivative for a function \( y = 5 \ln x \), you start with the first derivative. Each subsequent derivative is calculated using the derivative from the previous step. Here is the sequence:

• First Derivative: \( y' = \frac{5}{x} \)
• Second Derivative: \( y'' = -\frac{5}{x^2} \)
• Third Derivative: \( y''' = \frac{10}{x^3} \)
• Fourth Derivative: \( y^{(4)} = -\frac{30}{x^4} \)

Each step involves applying basic differentiation rules. Understanding and practicing these steps are crucial for mastering calculus and differential equations.

Logarithmic Differentiation

Logarithmic differentiation is a helpful technique in calculus that simplifies the process of finding the derivatives of complicated functions, especially those involving logarithms. This method is particularly useful when the function is a product or quotient of expressions with variable exponents.

In the given exercise, we're dealing with the function \( y = 5 \ln x \). Since logarithms ease the process of differentiation, taking the derivative of \( \ln x \) is straightforward:

• Base formula: \( \frac{d}{dx} \ln x = \frac{1}{x} \)

Thus, applying this rule to the function \( y \), the first derivative is found quickly as \( y' = \frac{5}{x} \).

Logarithmic differentiation makes these calculations simpler and is an essential tool when facing complex differential equations, ensuring we accurately determine derivatives even when initial expressions seem daunting.

Function Substitution

Substituting a function into a differential equation is an important process for verifying solutions. It's about seeing if the calculated derivatives of a function fit the given equation. For our exercise, we are given the equation \( y^{(4)} - 16y = 0 \) and need to check if it holds for the function \( y = 5 \ln x \).

To do this, follow these steps:

• Calculate the relevant derivatives.
• Substitute both the function and its derivatives back into the original equation.

In the given problem:

• Function: \( y = 5 \ln x \)
• Fourth Derivative: \( y^{(4)} = -\frac{30}{x^4} \)
• Substituted equation: \(-\frac{30}{x^4} - 16(5 \ln x) eq 0 \)

Since the equation does not equal zero, this shows \( y = 5 \ln x \) is not a solution to the differential equation. Substitution helps confirm or refute whether a potential solution is valid.

Verification of Solutions

Verifying solutions to a differential equation involves proving if a given function satisfies the equation after performing necessary calculations, like deriving and substitution. This process ensures mathematical accuracy and confirms or denies the suitability of a function as a solution.

Here’s how to perform a verification:

• Compute all necessary derivatives of the given function.
• Substitute these derivatives and the function back into the differential equation.
• Check if the result holds true within the equation, i.e., whether the left side equals the right side.

In the exercise discussed, we substituted the calculated fourth derivative and the original function back into the equation \( y^{(4)} - 16y = 0 \).
The result was \( -\frac{30}{x^4} - 16(5 \ln x) eq 0 \), confirming that \( y = 5 \ln x \) doesn't satisfy the equation. Verification identifies errors in solution assumptions or calculations and indicates whether additional attempts are needed to find a suitable function.