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Solve the initial value problems in Exercises \(55-60\) $$ \frac{d^{2} s}{d t^{2}}=-4 \sin \left(2 t-\frac{\pi}{2}\right), \quad s^{\prime}(0)=100, \quad s(0)=0 $$

Short Answer

Expert verified
The solution is \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t + 1 \).

Step by step solution

01

Recognize the Type of Problem

This is a second-order linear differential equation with constant coefficients driven by a non-homogeneous trigonometric function. We need to solve it given the initial conditions.
02

Integrate to Find First Derivative

Integrate the equation \( \frac{d^2 s}{dt^2} = -4 \sin(2t - \frac{\pi}{2}) \) with respect to \( t \) to find \( \frac{ds}{dt} \). Use the fact that \( \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C \). Thus, \( \frac{ds}{dt} = 2 \cos(2t - \frac{\pi}{2}) + C_1 \).
03

Apply Initial Condition for Velocity

Use the initial condition \( s'(0) = 100 \) to find \( C_1 \). Substitute \( t = 0 \) into \( \frac{ds}{dt} = 2 \cos(-\frac{\pi}{2}) + C_1 = 100 \). Since \( \cos(-\frac{\pi}{2}) = 0 \), we get \( C_1 = 100 \).
04

Integrate to Find Position Function

Integrate \( \frac{ds}{dt} = 2 \cos(2t - \frac{\pi}{2}) + 100 \) with respect to \( t \) to find \( s(t) \). Use \( \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \). This gives us \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t + C_2 \).
05

Apply Initial Condition for Position

Use the initial condition \( s(0) = 0 \) to find \( C_2 \). Substitute \( t = 0 \) into \( s(t) = \sin(-\frac{\pi}{2}) + 100(0) + C_2 = 0 \). Since \( \sin(-\frac{\pi}{2}) = -1 \), we get \( -1 + C_2 = 0 \) which gives \( C_2 = 1 \).
06

Write the Final Solution

Combine the results to write the solution: \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t + 1 \).

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Key Concepts

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

Initial Value Problems
Initial value problems are a type of differential equation problem where the solution is determined by specific conditions given at a starting point. These conditions are known as initial values.
In our exercise, we dealt with a second-order differential equation with two initial conditions:
  • The initial velocity: \( s'(0) = 100 \)
  • The initial position: \( s(0) = 0 \)
These values define the unique solution to our differential equation by specifying the state of the system at time \( t = 0 \). This is essential because without these conditions, a differential equation may have infinitely many solutions.
By applying the initial conditions, we can solve for the unknown constants which appear during the integration process. This pinpoints the exact path of the system from the multitude of possible solutions.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, are fundamental in differential equations, especially when dealing with oscillatory systems. In our problem, we encountered the function \( \sin(2t - \frac{\pi}{2}) \), which illustrates a wave-like behavior.
Trigonometric functions:
  • Model periodic phenomena like waves and oscillations
  • Switch roles in integration: integrating \( \sin(x) \) gives \( -\cos(x) \) and integrating \( \cos(x) \) gives \( \sin(x) \)
These functions have unique properties, such as phase shifts represented by adding or subtracting angles (e.g., \(-\frac{\pi}{2}\)), which affect their graphs' positions along the time axis.
Such shifts are crucial in accurately representing physical systems in equations, reflecting real-world scenarios like mechanical vibrations or sound waves.
Integration Techniques
Integration is a key technique used in solving differential equations. In this exercise, you may notice integrating twice, which is typical of second-order differential equations.
During integration:
  • Initial integration solved for the velocity function, \( \frac{ds}{dt} = 2 \cos(2t - \frac{\pi}{2}) + C_1 \), where constants arose needing definition through initial conditions.
  • Second integration derived the position function. By integrating the velocity, the position \( s(t) \) becomes an expression that includes another constant.
To perform these integrations, we used trigonometric identities and standard integration formulas:
  • \( \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C \)
  • \( \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \)
Successfully applying these techniques is crucial for finding solutions to differential equations and understanding how different variables interact over time.

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Most popular questions from this chapter

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