Chapter 4: Problem 7
Given that \(x(t)=c_{1} \cos \omega t+c_{2} \sin \omega t\) is the general solution of \(x^{\prime \prime}+\omega^{2} x=0\) on the interval \((-\infty, \infty),\) show that a solution satisfying the initial conditions \(x(0)=x_{0}, x^{\prime}(0)=x_{1}\) is given by \(x(t)=x_{0} \cos \omega t+\frac{x_{1}}{\omega} \sin \omega t\).
Short Answer
Step by step solution
Understanding the problem
Apply initial conditions
Differentiate the general solution
Substitute the constants
Verify the solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
General Solution
The general solution for this specific differential equation is given as \(x(t) = c_1 \cos \omega t + c_2 \sin \omega t\). This solution uses the properties of sine and cosine functions, which are periodic in nature to encompass all possible oscillatory motions that the system might possess.
- \(c_1\) and \(c_2\) are constants that will be determined by specific initial conditions.
- \(\omega\) is a constant that often represents angular frequency in physical systems.
Initial Conditions
For our problem, the initial conditions provided are \(x(0) = x_0\) and \(x'(0) = x_1\). By knowing the behavior of the system at time \(t=0\), we can solve for the constants within our general solution.
- The first condition, \(x(0) = x_0\), allows us to find \(c_1\).
- The second condition, \(x'(0) = x_1\), helps us determine \(c_2\).
Harmonic Oscillator
For the differential equation \(x'' + \omega^2 x = 0\), the motion described is characteristic of such an oscillator. The solutions given by \(\cos\) and \(\sin\) functions represent simple harmonic motions.
- The frequency of oscillation is determined by \(\omega\), a constant within the equation.
- The combination of sine and cosine ensures that all possible phase shifts in motion are covered.
Second Derivative
For the function \(x(t) = c_1 \cos \omega t + c_2 \sin \omega t\), the second derivative, \(x''(t)\), is pivotal. It aids in verifying that our function satisfies the original differential equation, \(x'' + \omega^2 x = 0\).
- Computing \(x''(t)\) results in \(-c_1 \omega^2 \cos \omega t - c_2 \omega^2 \sin \omega t\).
- Adding \(\omega^2 x(t)\) to \(x''(t)\) should equal zero, confirming that the general solution is correct.