Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Integrate the Differential Equation

To find the general solution, we need to integrate the differential equation \( \frac{ds}{dt} = 16t^2 + 4t - 1 \). This yields:\[s(t) = \int (16t^2 + 4t - 1)\, dt = \frac{16}{3}t^3 + \frac{4}{2}t^2 - t + C = \frac{16}{3}t^3 + 2t^2 - t + C.\]

Step 2: Apply the Initial Condition

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

Step 3: Write the Particular Solution

Substitute \( C = 100 \) back into the general solution. The particular solution is:\[s(t) = \frac{16}{3}t^3 + 2t^2 - t + 100.\]

Key Concepts

Differential Equations

Differential equations are mathematical equations that describe how a quantity changes in relation to another. In this exercise, we work with the differential equation \( \frac{ds}{dt} = 16t^2 + 4t - 1 \). This tells us how the function \( s(t) \), which depends on \( t \), changes as \( t \) changes.
These equations are crucial in many fields, like physics and engineering, as they model real-world systems:

• Motion of objects
• Heat transfer
• Population growth

Understanding these equations allows us to predict how systems behave over time.

Integration

Integration is a fundamental concept in calculus used to reverse-differentiate a function. You can think of it as finding the area under a curve. In the context of our differential equation, integration helps us transition from \( \frac{ds}{dt} \) back to \( s(t) \).
When we integrate \( 16t^2 + 4t - 1 \), it involves finding antiderivatives:

• \( \frac{16}{3}t^3 \) from \( 16t^2 \)
• \( 2t^2 \) from \( 4t \)
• \(-t \) from \(-1 \)

This integration results in the general solution \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + C \), where \( C \) is an integration constant. This constant appears because integration can produce multiple families of solutions that differ by a constant term.

Initial Conditions

Initial conditions are specific conditions or values provided to solve for constant terms in differential equations. They help us find a unique solution for our problem.
For this exercise, we're given that \( s(0) = 100 \). This means when \( t = 0 \), the function \( s(t) \) should equal 100. By substituting these values into our general solution \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + C \), we can solve for \( C \).
This process simplifies our results to a single function, aligning with the specific criteria from the initial conditions. Initial conditions are pivotal in real-life scenarios, ensuring models align with known data.

Particular Solution

The particular solution is the result we get when we apply specific initial conditions to the general solution of a differential equation. Our general solution is \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + C \), and from our exercise, we found \( C = 100 \) by using the initial condition. Thus, the particular solution is \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + 100 \).
In essence, the particular solution tailors the general equation to a specific scenario and ensures the model reflects real-world situations accurately.
This individualized solution holds the power to accurately capture the dynamics and behaviors specified by the original problem, offering precise predictions.