Question

Consider the differential equation \(y^{\prime \prime}(t)+k^{2} y(t)=0,\) where \(k\) is a positive real number. a. Verify by substitution that when \(k=1,\) a solution of the equation is \(y(t)=C_{1} \sin t+C_{2} \cos t .\) You may assume that this function is the general solution. b. Verify by substitution that when \(k=2,\) the general solution of the equation is \(y(t)=C_{1} \sin 2 t+C_{2} \cos 2 t\). c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture.

Short answer

2. What are the constants used in the general solution? 3. For k=2, what is the general solution?

Step-by-step solution

Calculate the first and second derivatives of the given function in each case

For the given function, \(y(t)=C_{1} \sin t+C_{2} \cos t\), we derive \(y'(t)\) and \(y''(t)\) as follows: \(y'(t) = C_{1} \cos t - C_{2} \sin t\) \(y''(t) = -C_{1} \sin t - C_{2} \cos t\) For the given function, \(y(t)=C_{1} \sin 2t+C_{2} \cos 2t\), we derive \(y'(t)\) and \(y''(t)\) as follows: \(y'(t) = 2C_{1} \cos 2t - 2C_{2} \sin 2t\) \(y''(t) = -4C_{1} \sin 2t - 4C_{2} \cos 2t\)

Verify the solutions when \(k=1\) and \(k=2\)

Now, we need to verify if the resulting functions satisfy the given differential equation when \(k = 1\) and \(k = 2\). For \(k=1\), we've calculated \(y''(t) = -C_{1} \sin t - C_{2} \cos t\): \(y''(t) + y(t) = (-C_{1} \sin t - C_{2} \cos t) + (C_{1} \sin t + C_{2} \cos t) = 0\) So, it's verified that \(y(t)=C_{1} \sin t+C_{2} \cos t\) is a solution when \(k=1\). For \(k=2\), we've calculated \(y''(t) = -4C_{1} \sin 2t - 4C_{2} \cos 2t\): \(y''(t) + 4y(t) = (-4C_{1} \sin 2t - 4C_{2} \cos 2t) + 4(C_{1} \sin 2t + C_{2} \cos 2t) = 0\) So, it's verified that \(y(t)=C_{1} \sin 2t+C_{2} \cos 2t\) is a solution when \(k=2\).

Determine the general solution for arbitrary values of \(k\)

Using the patterns of our solutions in (a) and (b), we can conjecture the general solution to be in the form: \(y(t) = C_{1} \sin kt + C_{2} \cos kt\) Now, we verify our conjecture by taking the first and second derivatives: \(y'(t) = kC_{1} \cos kt - kC_{2} \sin kt\) \(y''(t) = -k^{2}C_{1} \sin kt - k^{2}C_{2} \cos kt\) Plugging our conjectured solution into the given ordinary differential equation, we get: \(y''(t) + k^{2}y(t) = (-k^{2}C_{1} \sin kt - k^{2}C_{2} \cos kt) + k^{2}(C_{1} \sin kt + C_{2} \cos kt) = 0\) This confirms that \(y(t) = C_{1} \sin kt + C_{2} \cos kt\) is the general solution of the given differential equation for arbitrary \(k>0\).

Key concepts

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, particularly when dealing with oscillatory phenomena like waves. In this exercise, we work with the sine and cosine functions, which are examples of trigonometric functions. These functions are periodic, meaning they repeat their values in regular intervals. Their periodicity and properties make them very useful for solving differential equations.
Specifically, for the equation given in the exercise, the solutions involve the functions \( \sin(t) \) and \( \cos(t) \), indicating a dependency on time. These functions can be adjusted for frequency by multiplying by a constant such as \( k \).

• \( \sin(t) \) is a sine function, which starts at 0, goes up to 1, down to -1, and back to 0 around the unit circle, cycling every \( 2\pi \) radians (or 360 degrees).
• \( \cos(t) \) is a cosine function, which starts at 1, goes down to -1, back to 1, following the same cycle.

In this context, trigonometric functions express the solution to the differential equation, providing insights into the system's oscillatory nature. The constants \( C_1 \) and \( C_2 \) scale these solutions to fit initial conditions or specific scenarios.

General Solution

When solving differential equations, finding the general solution is a key objective. The general solution encompasses all possible solutions of the differential equation. It includes arbitrary constants, whose values can be set based on initial conditions.
In this exercise, the general solution for the differential equation \( y''(t) + k^2 y(t) = 0 \) was proposed and verified as \( y(t) = C_1 \sin kt + C_2 \cos kt \). This is a unified solution that satisfies any particular case with the right values of \( C_1 \) and \( C_2 \).

• The constants \( C_1 \) and \( C_2 \) are determined by initial conditions, such as initial position and velocity in physical applications.
• For different values of \( k \), the frequency of the sine and cosine waves changes, affecting the pace of oscillations in the solution.

This implies that the differential equation supports a family of solutions characterized by these periodic trigonometric forms, showcasing how systems can behave under different parameters.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations that involve a function and its derivatives. The term "ordinary" distinguishes them from partial differential equations, as they involve derivatives with respect to a single variable.
In this exercise, the equation \( y''(t) + k^2 y(t) = 0 \) represents an ODE, where \( y(t) \) is a function of one variable \( t \), typically representing time.

• The second derivative \( y''(t) \) represents the acceleration or curvature of the function \( y(t) \).
• The constant \( k^2 \) ties the rate of change directly to the function's value, indicating harmonic behavior.

This type of ODE is often seen in mechanical and electrical systems, where functions describe oscillations or vibrations. To solve such equations, we usually seek solutions in terms of known functions, often leading us to use trigonometric or exponential functions. The process involves finding a particular form that satisfies the equation when derivatives are plugged back into it, thereby verifying the solution.