Question

Determining a Solution In Exercises \(13-20\) , determine whether the function is a solution of the differential equation \(y^{(4)}-16 y=0\) . $$ y=2 \sin x $$

Short answer

No, \(y = 2 \sin x\) is not a solution to the differential equation \(y^{(4)} - 16y = 0\).

Step-by-step solution

Compute the Fourth Derivative of the Function

The fourth derivative of \(y = 2 \sin x\) is \(y^{(4)} = 2 \sin x\). This is established by understanding that the derivative of \(\sin x\) is \(\cos x\), the derivative of \(\cos x\) is \(-\sin x\), the derivative of \(-\sin x\) is \(-\cos x\), and finally the derivative of \(-\cos x\) is \(\sin x\).

Substituting into the Differential Equation

Substitute the value of \(y^{(4)}\) and \(y\) into the differential equation \(y^{(4)} - 16y = 0\). This gives \(2 \sin x - 16(2 \sin x) = 0\). Simplifying, we get \(-30 \sin x = 0\).

Verify the Result

The equation \(-30 \sin x = 0\) is only zero if \(\sin x = 0\), which is not always the case. Therefore, \(y = 2 \sin x\) is not a solution to the differential equation \(y^{(4)} - 16y = 0\).

Key concepts

Fourth Derivative

The fourth derivative of a function is obtained by taking the derivative of a function four times. In the context of the given exercise, we start with the simple function \( y = 2 \sin x \). Understanding the process of differentiation is essential to solve this problem correctly, especially when dealing with trigonometric functions.

• The first derivative of \( \sin x \) is \( \cos x \).
• The second derivative, which is the derivative of \( \cos x \), is \(-\sin x \).
• The third derivative is the derivative of \(-\sin x \), resulting in \(-\cos x \).
• Finally, the fourth derivative is found by differentiating \(-\cos x \) to obtain \( \sin x \).

The result is that the fourth derivative of \( y = 2 \sin x \) is simply \( y^{(4)} = 2 \sin x \), which shows how the cycle of derivatives repeats every four steps with sine and cosine.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in calculus, especially in problems involving periodic functions or oscillations. These functions have distinct properties that make their derivatives cycle and repeat every four derivatives.

• \( \sin x \) becomes \( \cos x \) after one differentiation.
• \( \cos x \) turns into \(-\sin x \).
• \(-\sin x \) becomes \(-\cos x \).
• \(-\cos x \) cycles back to \( \sin x \).

This cyclical nature can be immensely helpful in solving differential equations, as it allows for recognizing pattern repetition.
Since the original function \( y = 2 \sin x \) and its fourth derivative are the same, this periodicity naturally aligns with functions that involve differential equations such as the one posed in this exercise.

Solution Verification

Solution verification involves substituting your computed derivatives back into the original differential equation to confirm if it satisfies the equation.
Here, the differential equation is \( y^{(4)} - 16y = 0 \). After computing that \( y^{(4)} = 2 \sin x \), we substitute \(2 \sin x\) for \( y^{(4)} \) and \( 2 \sin x \) for \( y \) itself.

• The equation transforms into \(2 \sin x - 16(2 \sin x) = -30 \sin x \).
• For the original equation to hold true, this expression must equal zero.
• However, \(-30 \sin x = 0\) only holds true when \( \sin x = 0\), which is not sufficient for all \( x \).

Thus, through this substitution, we see that \( y = 2 \sin x \) does not satisfy the original differential equation for every value of \( x \), showing it is not a solution. Verifying solutions in this way ensures that any assumptions made in differentiation are correct and complete.