Problem 34 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. $$\frac{d^{2} x}{d t^{2}}+\omega^{2} x=F_{0} \cos \gamma t, \quad x(0)=0, x^{\prime}(0)=0$$

Step-by-Step Solution

Answer

The solution is \( x(t) = -\frac{F_0}{\omega^2 - \gamma^2} \cos(\omega t) + \frac{F_0}{\omega^2 - \gamma^2} \cos(\gamma t) \).

Step 1: Recognize the type of differential equation

The given equation \( \frac{d^{2} x}{d t^{2}} + \omega^2 x = F_0 \cos \gamma t \) is a non-homogeneous linear second-order differential equation with constant coefficients.

Step 2: Write the complementary solution

The complementary solution, \( x_c(t) \), solves the homogeneous equation \( \frac{d^{2} x}{d t^{2}} + \omega^2 x = 0 \). The characteristic equation is \( r^2 + \omega^2 = 0 \), whose roots are \( r = \pm i\omega \). Thus, the complementary solution is \( x_c(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) \).

Step 3: Find a particular solution

To find a particular solution, \( x_p(t) \), for the non-homogeneous part \( F_0 \cos \gamma t \), assume a particular solution of the form \( x_p(t) = A \cos(\gamma t) + B \sin(\gamma t) \). Substitute this into the differential equation, and equate coefficients for \( \cos(\gamma t) \) and \( \sin(\gamma t) \).

Step 4: Substitute and simplify to find coefficients

Substitute \( x_p(t) = A \cos(\gamma t) + B \sin(\gamma t) \) into \( \frac{d^{2} x}{d t^{2}} + \omega^2 x = F_0 \cos \gamma t \). Expand and equate coefficients:\[ -A\gamma^2 \cos(\gamma t) - B\gamma^2 \sin(\gamma t) + A\omega^2 \cos(\gamma t) + B\omega^2 \sin(\gamma t) = F_0 \cos(\gamma t) \]. This results in the system: \[ (A(\omega^2 - \gamma^2)) = F_0 \quad \text{and} \quad B(\omega^2 - \gamma^2) = 0 \].

Step 5: Solve for A and B

From the equations, \( B(\omega^2 - \gamma^2) = 0 \) gives us \( B = 0 \) (assuming \( \omega^2 eq \gamma^2 \)). Solve for \( A \) using \( A(\omega^2 - \gamma^2) = F_0 \):\[ A = \frac{F_0}{\omega^2 - \gamma^2} \]. Thus, the particular solution is \( x_p(t) = \frac{F_0}{\omega^2 - \gamma^2} \cos(\gamma t) \).

Step 6: Form the general solution

Combine the complementary and particular solutions: \[ x(t) = x_c(t) + x_p(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) + \frac{F_0}{\omega^2 - \gamma^2} \cos(\gamma t) \].

Step 7: Apply initial conditions

Given \( x(0) = 0 \) and \( x'(0) = 0 \), substitute \( t = 0 \) into the general solution:For \( x(0) = 0 \), \[ C_1 + \frac{F_0}{\omega^2 - \gamma^2} = 0 \].For \( x'(0) = 0 \), differentiate the general solution:\[ x'(t) = -C_1 \omega \sin(\omega t) + C_2 \omega \cos(\omega t) - \frac{F_0 \gamma}{\omega^2 - \gamma^2} \sin(\gamma t) \].Substitute \( t = 0 \): \[ 0 = C_2 \omega \].Thus, \( C_2 = 0 \) and \( C_1 = -\frac{F_0}{\omega^2 - \gamma^2} \).

Step 8: Write the final solution

Substitute the constants back into the general solution: \[ x(t) = -\frac{F_0}{\omega^2 - \gamma^2} \cos(\omega t) + \frac{F_0}{\omega^2 - \gamma^2} \cos(\gamma t) \].This is the solution to the initial-value problem.

Key Concepts

Second-Order Differential Equations

A second-order differential equation involves a second derivative, such as the equation \( \frac{d^{2} x}{d t^{2}} + \omega^2 x = F_0 \cos \gamma t \) presented in the exercise. These equations represent systems where the change in a rate of change is directly dependent on the state of the system itself.

In the general form of a second-order linear differential equation:

• \( \frac{d^{2} x}{d t^{2}} \) is the second derivative, indicating the acceleration or the curvature of \( x \).
• \( \omega^2 x \) represents a term related to the property of the system (e.g., oscillatory behavior).
• The term \( F_0 \cos \gamma t \) adds a forcing function, making it non-homogeneous.

Second-order differential equations often appear in physics and engineering to model phenomena such as mechanical vibrations, electrical circuits, and wave propagation. The challenge often lies in finding solutions that fit specific boundary or initial conditions, which guide the behavior of the solution.

Initial-Value Problems

An initial-value problem provides specific values at a starting point, which help satisfy a differential equation as it evolves over time. These values might describe the initial position and velocity of an object, for example.

In our exercise, the initial conditions are given as:

• \( x(0) = 0 \): The position at time \( t = 0 \) is zero.
• \( x'(0) = 0 \): The velocity at time \( t = 0 \) is also zero.

Initial-value problems are critical in ensuring that the solution not only satisfies the differential equation itself but also adheres to the real-world scenario described by these conditions.
This is crucial in applications where the starting state of a system influences its future behavior, such as in projectile motion or electrical signals.

Complementary and Particular Solutions

The solution to a non-homogeneous differential equation, like the one in our problem, consists of finding both a complementary and a particular solution.

**Complementary Solution:**
The complementary solution \( x_c(t) \) addresses the homogeneous part of the differential equation (without the forcing term). The equation \( \frac{d^{2} x}{d t^{2}} + \omega^2 x = 0 \) was solved leading to the general form: \[ x_c(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) \] These solutions arise from solving the characteristic equation \( r^2 + \omega^2 = 0 \). The constants \( C_1 \) and \( C_2 \) will be determined later by initial conditions.

**Particular Solution:**
The particular solution \( x_p(t) \) accounts for the forced component. By assuming a solution form \( A \cos(\gamma t) + B \sin(\gamma t) \) and substituting it back, we solve for coefficients matching the non-homogeneous part, resulting in: \[ x_p(t) = \frac{F_0}{\omega^2 - \gamma^2} \cos(\gamma t) \]
The overall solution \( x(t) \) is then the sum of both, adjusted to fit any initial conditions. This gives a complete response of the system to both natural and forced influences, essential in achieving accurate models of real systems like damped harmonic oscillators.