Question

Verify that the given function \(y\) is a solution of the differential equation that follows it. Assume that \(C\) is an arbitrary constant. $$y(t)=C_{1} \sin 4 t+C_{2} \cos 4 t ; y^{\prime \prime}(t)+16 y(t)=0$$

Short answer

Question: Verify that the given function \(y(t)=C_{1}\sin 4t+C_{2}\cos 4t\) is a solution to the given differential equation \(y''(t)+16y(t)=0\). Answer: By finding the first and second derivatives of the given function and substituting them into the given differential equation, we have shown that the equation holds true. Therefore, the given function \(y(t)=C_{1}\sin 4t+C_{2}\cos 4t\) is indeed a solution to the given differential equation \(y''(t)+16y(t)=0\).

Step-by-step solution

Find the first derivative of \(y(t)\)

To find the first derivative of the function \(y(t)=C_{1}\sin 4t+C_{2}\cos 4t\), we apply the chain rule: $$y'(t)=\frac{d}{dt}(C_{1}\sin 4t+C_{2}\cos 4t)=4C_{1}\cos 4t-4C_{2}\sin 4t$$

Find the second derivative of \(y(t)\)

Now we find the second derivative of the function \(y(t)\) using the first derivative found in step 1: $$y''(t)=\frac{d}{dt}(4C_{1}\cos 4t-4C_{2}\sin 4t)=-16C_{1}\sin 4t-16C_{2}\cos 4t$$

Substitute the function and its derivatives into the given differential equation

We will now substitute the function \(y(t)\), the first derivative \(y'(t)\), and the second derivative \(y''(t)\) into the given differential equation \(y''(t)+16y(t)=0\): $$(-16C_{1}\sin 4t-16C_{2}\cos 4t)+16(C_{1}\sin 4t+C_{2}\cos 4t)=0$$

Simplify and verify the equation

By simplifying the expression, $$-16C_{1}\sin 4t-16C_{2}\cos 4t+16C_{1}\sin 4t+16C_{2}\cos 4t=0$$ We find that all the terms cancel out, resulting in $$0=0$$ Since the equality holds true after the substitution, the given function \(y(t)=C_{1}\sin 4t+C_{2}\cos 4t\) is indeed a solution to the given differential equation \(y''(t)+16y(t)=0\).

Key concepts

Second Derivative

The second derivative is a concept from calculus that helps us understand the acceleration or concavity of a function. When we take the first derivative of a function, we are measuring how the function is changing at each point, which we often associate with velocity in physics.
However, the second derivative goes a step further. It tells us how the rate of change itself is changing, similar to acceleration. In the context of differential equations, like the one in the original exercise, the second derivative plays a crucial role.
For the function \(y(t) = C_1 \sin 4t + C_2 \cos 4t\), the first derivative \(y'(t)\) is calculated as \(4C_1 \cos 4t - 4C_2 \sin 4t\).
To find the second derivative \(y''(t)\), you differentiate \(y'(t)\) again, yielding \(-16C_1 \sin 4t - 16C_2 \cos 4t\).

• It captures how quickly the initial velocity is changing.
• It also provides critical information required to solve differential equations.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate compositions of functions. When a function is wrapped inside another function, the chain rule helps us find derivatives with respect to a particular variable.
In the exercise, \(y(t) = C_1 \sin 4t + C_2 \cos 4t\) is a composition of trigonometric functions with the inner function being \(4t\).
The chain rule tells us that while differentiating \(\sin 4t\) or \(\cos 4t\), we must multiply the derivative of the outer function by the derivative of the inner function \(4t\).
So, when differentiating \(\sin 4t\), you first differentiate to get \(\cos 4t\) and then multiply by the derivative of \(4t\), which is \(4\).

• This results in a term like \(4 \cos 4t\) for \(y'(t)\).
• The chain rule simplifies the process of differentiating complex functions.

Trigonometric Functions

Trigonometric functions describe relationships between angles and sides of triangles. They form the backbone of many differential equations, particularly in mechanical and electrical engineering.
The original function \(y(t) = C_1 \sin 4t + C_2 \cos 4t\) contains two primary trigonometric functions: \(\sin\) and \(\cos\).

• The \(\sin\) function relates to the vertical component of an angle on the unit circle.
• The \(\cos\) function relates to the horizontal component.

These functions cyclically oscillate between -1 and 1, creating periodic waves. Such properties make them ideal for modeling vibrations and waves in differential equations.
When differentiated, these functions preserve the periodic nature which is crucial to the solution of differential equations. The derivative of \(\sin 4t\) is \(4 \cos 4t\), and the derivative of \(\cos 4t\) is \(-4 \sin 4t\).
These transformations play essential roles when verifying solutions for differential equations, like substituting back into \(y''(t)+16y(t)=0\) to confirm the function as a valid solution.