Question

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

Question: Verify that the given functions \(y=A\sin t\), \(y=B\cos t\), and \(y=A\sin t + B\cos t\) satisfy the differential equation \(y^{\prime \prime}(t)+y(t)=0\). Answer: All three functions satisfy the given differential equation for any constants \(A\) and \(B\).

Step-by-step solution

Calculate derivatives for each function

We need the first and second derivatives of \(y=A\sin t\), \(y=B\cos t\), and \(y=A\sin t + B\cos t\). For function \(y=A\sin t\): First derivative: \(y^{\prime}(t) = A\cos t\) Second derivative: \(y^{\prime \prime}(t) = -A\sin t\) For function \(y=B\cos t\): First derivative: \(y^{\prime}(t) = -B\sin t\) Second derivative: \(y^{\prime \prime}(t) = -B\cos t\) For function \(y=A\sin t + B\cos t\): First derivative: \(y^{\prime}(t) = A\cos t - B\sin t\) Second derivative: \(y^{\prime \prime}(t) = -A\sin t - B\cos t\)

Substitute derivatives into the differential equation

Now we will substitute the derivatives of each function back into the differential equation \(y^{\prime \prime}(t)+y(t)=0\) and check if the equation holds. a. For \(y=A\sin t\), we have: \(-A\sin t + A\sin t = 0\) \(0 = 0\) (This is true, so \(y=A\sin t\) satisfies the differential equation) b. For \(y=B\cos t\), we have: \(-B\cos t + B\cos t = 0\) \(0 = 0\) (This is true, so \(y=B\cos t\) satisfies the differential equation) c. For \(y=A\sin t + B\cos t\), we have: \((-A\sin t - B\cos t) + (A\sin t + B\cos t) = 0\) \(0 = 0\) (This is true, so \(y=A\sin t + B\cos t\) satisfies the differential equation) Thus, all given functions satisfy the given differential equation for any constants \(A\) and \(B\).

Key concepts

Understanding Derivatives

Derivatives are fundamental in calculus. They measure how a function changes as its input changes. Think of the derivative as the slope of the function at a particular point.
In the context of differential equations, derivatives help us determine how the system behaves over time.

• The first derivative of a function gives us the rate of change or the velocity of the function.
• The second derivative provides the acceleration or how the rate of change is itself changing.

In the solution of the differential equation, derivatives are evaluated for specific functions.For example, if we take the function \(y = A \sin t\), its first derivative is \(y'(t) = A \cos t\), and the second derivative is computed as \(y''(t) = -A \sin t\).
This shows how the function's rate of change (cosine part) and its acceleration (sine part) are connected.

Unpacking Trigonometric Functions

Trigonometric functions like sine and cosine are pivotal in mathematics, especially when dealing with oscillating systems. These functions describe periodic phenomena across various scientific fields.

• \(\sin t\) represents the vertical component of a point moving around a circle.
• \(\cos t\) represents the horizontal component.

In solving differential equations, trigonometric functions often appear because they naturally model repetitive patterns over time.
In our differential equation problem, you see functions like \(y = A \sin t\) and \(y = B \cos t\), which serve as particular solutions.
Because sine and cosine functions are related by their derivatives, they elegantly solve equations that require a balance of forces, like \(y''(t) + y(t) = 0\).This specific equation models simple harmonic motion where forces such as tension and gravity are in equilibrium.

Introduction to Initial Value Problems

Initial value problems (IVPs) are a class of differential equations with known values at the start. For example, if you know the position and velocity of a particle at time \(t=0\), you can determine its future behavior.
When working with differential equations, once you've found a general solution, adding initial conditions transforms it into an initial value problem.
This makes the solution specific by plugging in the initial values of the function and its derivatives.

• For instance, a spring system might have a starting point where its displacement and velocity are specified.
• The general solution, such as \(y = A \sin t + B \cos t\), includes these constants \(A\) and \(B\).
• These constants are determined using the initial values provided.

Essentially, IVPs allow the creation of a unique solution tailored to the given initial conditions, mimicking real-world scenarios such as predicting a projectile's path.