Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d s}{d t}=16 t^{2}+4 t-1 ; s=100 \text { at } t=0 $$

Step-by-Step Solution

Answer

The particular solution is \( s = \frac{16}{3}t^3 + 2t^2 - t + 100 \).

Step 1: Integrate the Differential Equation

The differential equation is \( \frac{ds}{dt} = 16t^2 + 4t - 1 \). To find the general solution, integrate both sides with respect to \( t \). \[\int \frac{ds}{dt} \, dt = \int (16t^2 + 4t - 1) \, dt\]This gives:\[s = \int (16t^2 + 4t - 1) \, dt = \frac{16}{3}t^3 + \frac{4}{2}t^2 - t + C = \frac{16}{3}t^3 + 2t^2 - t + C\]

Step 2: Apply Initial Condition to Find C

We are given that \( s = 100 \) when \( t = 0 \). Substitute \( t = 0 \) and \( s = 100 \) into the general solution to find \( C \).\[100 = \frac{16}{3}(0)^3 + 2(0)^2 - (0) + C\]Which simplifies to:\[100 = C\]Thus, \( C = 100 \).

Step 3: Write the Particular Solution

Substitute \( C = 100 \) back into the general solution to find the particular solution:\[s = \frac{16}{3}t^3 + 2t^2 - t + 100\]This is the solution to the differential equation with the given initial condition.

Key Concepts

Initial Value Problem

An initial value problem in differential equations involves finding a solution to a differential equation that not only satisfies the equation itself but also meets a specific initial condition. The initial condition is typically given as a point in time where the solution takes a specific value. For example, in our problem, the initial condition is

• \( s = 100 \)
• when \( t = 0 \)

This means that when the variable \( t \) is zero, the solution \( s \) must equal 100. The purpose of applying this condition is to determine any unknown constants in the general solution, allowing us to find what is known as a 'particular solution'. This kind of problem is foundational in understanding how initial conditions influence the behavior of solutions in dynamic systems.

Integration Techniques

Solving differential equations often requires using integration, a fundamental technique in calculus. To solve our exercise, we integrated the right-hand side of the differential equation to find the general solution. The equation \( \frac{ds}{dt} = 16t^2 + 4t - 1 \)was integrated with respect to \( t \), resulting in:

• \( s = \int (16t^2 + 4t - 1) \, dt \)
• This becomes \( s = \frac{16}{3}t^3 + 2t^2 - t + C \)

Here's a brief rundown of the integration process:- Break down the differential equation into simpler components.- Integrate each term separately.- Combine coefficients and terms to form the general solution.This technique is essential for solving many types of problems where functions describe changing quantities.

Particular Solution

After finding the general solution to a differential equation, a particular solution is one where we determine the constant(s) such that specific conditions are met. These conditions often relate to initial values provided in the problem statement. In our example, the general solution was

• \( s = \frac{16}{3}t^3 + 2t^2 - t + C \)

To find the particular solution, we applied the initial condition \( s = 100 \) at \( t = 0 \). Solving for \( C \), we found:

• \( C = 100 \)
• Thus the particular solution is \( s = \frac{16}{3}t^3 + 2t^2 - t + 100 \)

With the particular solution, we have a complete description of the system's behavior under the specified initial condition. This is especially useful in real-world applications, where knowing exact behavior at certain points is crucial.