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A differential equation is an equation involving an unknown function and its derivatives. Consider the differential equation \(y^{\prime \prime}(t)+y(t)=0\) a. Show that \(y=A \sin t\) satisfies the equation for any constant \(A\) b. Show that \(y=B \cos t\) satisfies the equation for any constant \(B\) c. Show that \(y=A \sin t+B \cos t\) satisfies the equation for any constants \(A\) and \(B\).

Short Answer

Expert verified
Question: Show that the functions a. \(y = A\sin t\), b. \(y = B\cos t\), and c. \(y = A\sin t + B\cos t\) satisfy the differential equation \(y^{\prime\prime}(t)+y(t)=0\). Answer: We found the first and second derivatives of each function and substituted them into the given differential equation. In each case, the equation held true, which implies that all three functions satisfy the differential equation \(y^{\prime\prime}(t)+y(t)=0\) for any constants \(A\) and \(B\).

Step by step solution

01

Find the derivatives of the given functions#a. \(y = A\sin t\)

To find the first and second derivatives of the function \(y = A\sin t\), we can apply the rules of differentiation. First derivative: \(y^{\prime}(t) = A\cos t\) Second derivative: \(y^{\prime\prime}(t) = -A\sin t\) Now, we need to substitute these derivatives into the differential equation and check whether it is satisfied.
02

Substitute the derivatives into the differential equation#a. \(y = A\sin t\)

Substituting the derivatives: \(y^{\prime\prime}(t)+y(t) = -A\sin t + A\sin t = 0\) Since the equation holds true, the function \(y=A\sin t\) satisfies the given differential equation for any constant \(A\). Now we will follow the same process for the function \(y = B\cos t\).
03

Find the derivatives of the given functions#b. \(y = B\cos t\)

First derivative: \(y^{\prime}(t) = -B\sin t\) Second derivative: \(y^{\prime\prime}(t) = -B\cos t\) Now, we need to substitute these derivatives into the differential equation and check whether it is satisfied.
04

Substitute the derivatives into the differential equation#b. \(y = B\cos t\)

Substituting the derivatives: \(y^{\prime\prime}(t)+y(t) = -B\cos t + B\cos t = 0\) Since the equation holds true, the function \(y=B\cos t\) satisfies the given differential equation for any constant \(B\). Finally, we will find the derivatives of the function \(y = A\sin t + B\cos t\) and substitute them into the differential equation.
05

Find the derivatives of the given functions#c. \(y = A\sin t + B\cos t\)

First derivative: \(y^{\prime}(t) = A\cos t - B\sin t\) Second derivative: \(y^{\prime\prime}(t) = -A\sin t - B\cos t\) Now, we need to substitute these derivatives into the differential equation and check whether it is satisfied.
06

Substitute the derivatives into the differential equation#c. \(y = A\sin t + B\cos t\)

Substituting the derivatives: \(y^{\prime\prime}(t) + y(t) = (-A\sin t - B\cos t) + (A\sin t + B\cos t) = 0\) Since the equation holds true, the function \(y=A\sin t + B\cos t\) satisfies the given differential equation for any constants \(A\) and \(B\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
The second derivative in mathematics is a powerful tool used to understand the curvature or concavity of a function. When you differentiate a function twice, you get what's called the "second derivative," \(y''(t)\). For the problem at hand, we are dealing with the differential equation \(y''(t) + y(t) = 0\). This involves calculating the second derivatives of potential solutions and substituting them back into the equation to check if they satisfy it.

In the given problem, two functions, \(y = A\sin t\) and \(y = B\cos t\), are tested against this equation. When differentiating \(y = A\sin t\), the first derivative is \(y'(t) = A\cos t\) and the second derivative is \(y''(t) = -A\sin t\). Similarly, for \(y = B\cos t\), the first derivative is \(y'(t) = -B\sin t\) and the second derivative is \(y''(t) = -B\cos t\).

When these second derivatives are plugged back into the equation \(y''(t) + y(t) = 0\), they simplify to zero, demonstrating that both functions satisfy the equation. Recognizing this approach helps reduce complex differential equations into simpler algebraic verifications.
Sine and Cosine Functions
Sine and cosine functions are fundamental in trigonometry and appear frequently in differential equations. They are periodic functions, meaning they repeat values in regular intervals, and are often used to model wave-like phenomena.

These functions have specific properties that make them extremely useful in solving differential equations:
  • Derivatives are cyclical: The sine and cosine functions have derivatives that also cycle back to sine and cosine. This means their forms are preserved after differentiation. For example, differentiating \(A\sin t\) gives \(A\cos t\), and differentiating \(B\cos t\) results in \(-B\sin t\).
  • Solutions to homogeneous equations: Both functions satisfy the homogeneous linear differential equation of the form \(y''(t) + y(t) = 0\).
Understanding this cyclical nature and their role in differential equations allows us to use sine and cosine functions as a basis for many solutions, notably in problems involving oscillations.
Homogeneous Solutions
A homogeneous differential equation is one that has a form where all terms involve the unknown function and its derivatives. These types of equations hold particular interest because they offer insights into the natural modes of a system — solutions that do not change when scaled by a factor, represented here as constants.

For the given equation \(y''(t) + y(t) = 0\), the solutions \(y = A\sin t\), \(y = B\cos t\), and the combination \(y = A\sin t + B\cos t\) are homogeneous solutions. They represent the general solution because:
  • Linear combination: As the problem illustrates, any linear combination of \(\sin t\) and \(\cos t\) will satisfy the equation, where A and B are arbitrary constants.
  • Independence: \(\sin t\) and \(\cos t\) are linearly independent, meaning they span a two-dimensional solution space of the differential equation.
Homogeneous solutions are essential when modeling systems inherently stable or without external forcing functions, as they naturally arise from the system's inherent dynamics.

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