Problem 10 - 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.) \(\frac{d y}{d t}=y^{4} ; y=1\) at \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \( y = \left( \frac{1}{3t+1} \right)^{1/3} \).

Step 1: Separate Variables

The given differential equation is \( \frac{d y}{d t}=y^{4} \). Since this is a separable differential equation, we can rewrite it as: \( \frac{1}{y^{4}} \, dy = dt \). This separates the variables \( y \) and \( t \).

Step 2: Integrate Both Sides

Next, integrate both sides of the equation from Step 1. The left side becomes \( \int \frac{1}{y^{4}} \, dy = \int y^{-4} \, dy \), and the right side becomes \( \int dt \).

Step 3: Solve the Integrals

Integrate both expressions: \( \int y^{-4} \, dy = \frac{y^{-3}}{-3} + C_1 = -\frac{1}{3y^3} + C_1 \). The right side is \( \int dt = t + C_2 \).

Step 4: Combine and Solve for y

Equating both integrated parts gives us: \( -\frac{1}{3y^3} + C_1 = t + C_2 \). Combine constants: \( -\frac{1}{3y^3} = t + C \) (where \( C = C_2 - C_1 \)).

Step 5: Apply Initial Condition

Use the initial condition \( y = 1 \) at \( t = 0 \) to find \( C \). Substitute into the equation: \( -\frac{1}{3} = 0 + C \). Thus, \( C = -\frac{1}{3} \).

Step 6: Solve for the Particular Solution

Substitute \( C = -\frac{1}{3} \) back into the equation: \( -\frac{1}{3y^3} = t - \frac{1}{3} \). Rearrange this to find \( y \) in terms of \( t \): \[ y^3 = \frac{1}{3(t+\frac{1}{3})} = \frac{1}{3t + 1} \]Therefore, \( y = \left( \frac{1}{3t+1} \right)^{1/3} \).

Key Concepts

Separable Differential Equations

Separable differential equations are a class of differential equations that can be easily solved by separating the variables. Essentially, you manipulate the equation until the variables are grouped with their respective differentials. For example, a separable differential equation can be expressed as:\[ \frac{dy}{dt} = f(y)g(t) \]In this type of equation, you aim to separate \( y \) and \( t \) on different sides. This allows independent integration over each variable. In the given exercise, starting with \( \frac{dy}{dt} = y^4 \), we separate variables to get \( \frac{1}{y^4} dy = dt \). This clear separation is crucial, as it sets the stage for solving the equation through integration.

• Identify separable structure.
• Rewrite to isolate variables.
• Prepare for integration.

By reorganizing the terms, separable differential equations become solvable using basic calculus tools.

Initial Condition

An initial condition in the context of differential equations is a specified value that helps to determine a particular solution from a general one. It gives the solution a specific starting point. In our exercise, the initial condition is given as \( y = 1 \) when \( t = 0 \). Initial conditions are vital because:

• They ensure uniqueness of the solution.
• They help incorporate real-world scenarios into mathematical solutions.

When an initial condition is applied, it helps in calculating the integration constant \( C \). For this problem, after integrating, we obtain:\[ -\frac{1}{3y^3} + C = t \]Plugging in \( y = 1 \) and \( t = 0 \), we solve for \( C \), which results in the particular value \( C = -\frac{1}{3} \). This makes the solution applicable to the given scenario.

Integration

Integration is a fundamental technique used to solve differential equations, particularly separable ones. Once variables are separated, we integrate both sides of the equation. For this problem, the goal is to integrate\[ \int \frac{1}{y^4} \, dy = \int dt \]The left side integrates to \(-\frac{1}{3y^3} + C_1\) and the right side becomes \(t + C_2\), where \(C_1\) and \(C_2\) are constants of integration.

• Set up integral for each separated side.
• Calculate the integrals independently.
• Combine to form a unified equation.

Integration not only helps find the relationship between \( y \) and \( t \) but also leads to solving for the unknown constant \( C \), which is adjusted with the initial condition.

Particular Solution

A particular solution of a differential equation is obtained by incorporating the initial condition into the general solution. It makes the solution specific to the given conditions. The general solution is initially expressed with an arbitrary constant \( C \), reflecting different possible solutions.Incorporating the initial condition, which is \( y = 1 \) when \( t = 0 \), adjusts the general solution\[ -\frac{1}{3y^3} = t + C \]to a particular one, by solving for \( C = -\frac{1}{3} \). Plugging \( C \) back gives:\[ -\frac{1}{3y^3} = t - \frac{1}{3} \]Rearranging gives \( y = \left(\frac{1}{3t+1}\right)^{1/3} \), a solution tailored to meet the initial condition provided. The particular solution is essential for practical application because it represents one real-world situation modeled by the differential equation.