Problem 77
Question
Solve the initial value problems $$\frac{d^{2} s}{d t^{2}}=-4 \sin \left(2 t-\frac{\pi}{2}\right), \quad s^{\prime}(0)=100, \quad s(0)=0$$
Step-by-Step Solution
Verified Answer
The solution is \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t \).
1Step 1: Integrate the Second Derivative
We start by integrating the second derivative to find the first derivative. Given \( \frac{d^2 s}{d t^2} = -4 \sin \left( 2t - \frac{\pi}{2} \right) \), we integrate with respect to \( t \). The integral of \( \sin(2t - \frac{\pi}{2}) \) is \( -\frac{1}{2}\cos(2t - \frac{\pi}{2}) \). Thus, the first derivative is \( s'(t) = 2\cos(2t - \frac{\pi}{2}) + C_1 \), where \( C_1 \) is a constant of integration.
2Step 2: Apply Initial Condition for First Derivative
Use the initial condition \( s'(0) = 100 \) to find \( C_1 \). Plug \( t = 0 \) into the first derivative equation: \( s'(0) = 2\cos\left(-\frac{\pi}{2}\right) + C_1 = 100 \). Since \( \cos(-\frac{\pi}{2}) = 0 \), we have \( C_1 = 100 \). Thus, the equation becomes \( s'(t) = 2\cos(2t - \frac{\pi}{2}) + 100 \).
3Step 3: Integrate the First Derivative
Now integrate \( s'(t) = 2\cos(2t - \frac{\pi}{2}) + 100 \) to find \( s(t) \). The integral of \( 2\cos(2t - \frac{\pi}{2}) \) is \( \sin(2t - \frac{\pi}{2}) \) as \( \frac{d}{dt} \sin(2t - \frac{\pi}{2}) \) is \( 2\cos(2t - \frac{\pi}{2}) \). Thus, \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t + C_2 \), where \( C_2 \) is another constant of integration.
4Step 4: Apply Initial Condition for Initial Value
Use the initial condition \( s(0) = 0 \) to find \( C_2 \). Plug \( t = 0 \) into \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t + C_2 \): \( s(0) = \sin(0) + 0 + C_2 = 0 \). Therefore, \( C_2 = 0 \). Thus, \( s(t) = \sin(2t - \frac{\pi}{2}) + 100t \).
Key Concepts
Initial Value ProblemsSecond Order Differential EquationsIntegration Techniques
Initial Value Problems
Initial value problems (IVPs) involve finding a particular solution to a differential equation that satisfies given initial conditions. These initial conditions typically specify the value of the function and possibly its derivatives at a certain point.
Consider a general second-order differential equation and initial conditions:
Consider a general second-order differential equation and initial conditions:
- Differential equation: \( \frac{d^{2} s}{dt^{2}} = f(t) \)
- Initial conditions: Values are given for the function \( s(t) \) and its first derivative \( s'(t) \) at a particular time \( t = t_0 \).
Second Order Differential Equations
Second order differential equations involve the second derivative of a function.
In our example, we have the equation \( \frac{d^{2}s}{dt^{2}} = -4 \sin(2t - \frac{\pi}{2}) \). This suggests that the second derivative of \( s(t) \) with respect to \( t \) is given in terms of trigonometric functions.
The solution process involves finding the first derivative through integration, and then solving for \( s(t) \), the original function. Each integration introduces an unknown constant, which needs to be determined using initial conditions.
Initial conditions are crucial here for calculating these constants, thereby yielding a unique solution that fits the initial setup. In practice, second order differential equations appear in various physics problems like oscillations and wave dynamics.
In our example, we have the equation \( \frac{d^{2}s}{dt^{2}} = -4 \sin(2t - \frac{\pi}{2}) \). This suggests that the second derivative of \( s(t) \) with respect to \( t \) is given in terms of trigonometric functions.
The solution process involves finding the first derivative through integration, and then solving for \( s(t) \), the original function. Each integration introduces an unknown constant, which needs to be determined using initial conditions.
Initial conditions are crucial here for calculating these constants, thereby yielding a unique solution that fits the initial setup. In practice, second order differential equations appear in various physics problems like oscillations and wave dynamics.
Integration Techniques
Integration is a key technique when solving differential equations. For an equation like \( \frac{d^{2}s}{dt^{2}} = -4 \sin(2t - \frac{\pi}{2}) \), we employ integration to reverse differentiation.
Here are the general steps for solving the example problem:
Here are the general steps for solving the example problem:
- First Integration: Integrating \( -4 \sin(2t - \frac{\pi}{2}) \) with respect to \( t \) to find the first derivative \( s'(t) \). This gives us \( 2\cos(2t - \frac{\pi}{2}) + C_1 \).
- Determine the Constant \( C_1 \): Apply the initial condition \( s'(0) = 100 \) to find \( C_1 \).
- Second Integration: Further integrate the first derivative to find the function \( s(t) \). This results in \( \sin(2t - \frac{\pi}{2}) + 100t + C_2 \).
- Determine Constant \( C_2 \): Use the initial condition \( s(0) = 0 \) to find \( C_2 \).
Other exercises in this chapter
Problem 76
Solve the initial value problems $$\frac{d r}{d \theta}=3 \cos ^{2}\left(\frac{\pi}{4}-\theta\right), \quad r(0)=\frac{\pi}{8}$$
View solution Problem 77
Suppose that \(\int_{1}^{x} f(t) d t=x^{2}-2 x+1 .\) Find \(f(x)\).
View solution Problem 78
Find \(f(4)\) if \(\int_{0}^{x} f(t) d t=x \cos \pi x\).
View solution Problem 78
Integrals of nonpositive functions Show that if \(f\) is integrable then $$ f(x) \leq 0 \quad \text { on } \quad[a, b] \quad \Rightarrow \quad \int_{a}^{b} f(x)
View solution