Problem 13 - 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.) \(\frac{d y}{d x}=(2 x+1)^{4} ; y=6\) at \(x=0\)

Step-by-Step Solution

Answer

The particular solution is \( y = \frac{(2x+1)^5}{10} + \frac{59}{10} \).

Step 1: Recognize the Differential Equation Type

The given differential equation \( \frac{dy}{dx} = (2x+1)^4 \) is a first-order and separable differential equation. This means we can separate variables to integrate both sides independently.

Step 2: Separate Variables and Integrate

To solve for \( y \), integrate both sides:\[\int dy = \int (2x+1)^4 dx.\]Integrating the left side gives \( y = \), and the right side requires the substitution method or applying the formula for integrating a power function.

Step 3: Calculate the Right Side Integral

Let \( u = 2x + 1 \), hence \( du = 2 dx \) or \( dx = \frac{1}{2} du \). Substitute to get:\[\int (2x+1)^4 dx = \frac{1}{2} \int u^4 du.\]Integrate to obtain \( \frac{1}{2} \times \frac{u^5}{5} + C \). Substitute \( u = 2x + 1 \) back to get:\[\frac{(2x+1)^5}{10} + C.\]

Step 4: State the General Solution

The general solution to the differential equation is:\[y = \frac{(2x+1)^5}{10} + C.\]

Step 5: Apply Initial Condition to Find C

Use the initial condition \( y = 6 \) when \( x = 0 \). Substitute into the general solution:\[ 6 = \frac{(2(0)+1)^5}{10} + C \].This simplifies to \( 6 = \frac{1}{10} + C \). Solve for \( C \) to find \( C = 6 - \frac{1}{10} = \frac{60}{10} - \frac{1}{10} = \frac{59}{10} \).

Step 6: Write the Particular Solution

Substitute \( C = \frac{59}{10} \) back into the general solution to find the particular solution:\[y = \frac{(2x+1)^5}{10} + \frac{59}{10}.\]

Key Concepts

Separable Differential Equations

Separable differential equations are a type of ordinary differential equation (ODE) where the variables can be separated on opposite sides of the equation, allowing each side to be integrated independently. In our exercise, the equation is \( \frac{dy}{dx} = (2x+1)^4 \), which is a classic example of a separable differential equation.

To separate the variables, we aim to express the differential equation in a form where each side involves only one of the variables \( y \) or \( x \). We do this by manipulating the algebraic structure of the equation.

• Move terms involving \( y \) to one side: \( dy = \).
• Move terms involving \( x \) to the other: \( (2x+1)^4 dx \).

After separating, we are prepared to integrate both sides, which is a crucial step in solving the equation.

Initial Conditions in Differential Equations

Initial conditions specify a particular value of the solution to a differential equation at a given point, which helps to determine the constant of integration. In our exercise, we have the initial condition \( y = 6 \) at \( x = 0 \).

This condition transforms a general solution into a particular solution to fit the specific scenario described. The general solution includes a constant \( C \). When we apply the initial condition, we substitute \( x = 0 \) and \( y = 6 \) into the general solution:

\[ 6 = \frac{(2 \cdot 0+1)^5}{10} + C \] This allows us to solve for \( C \), ensuring our solution meets the initial conditions provided.

Initial conditions are vital in real-world applications where starting values are known, such as population models or physical systems at a specific time.

Integration Techniques in Calculus

Integration plays a critical role in solving differential equations, particularly in determining solutions from separated variables. In our example, integrating both sides of \( dy = (2x+1)^4 dx \) is necessary to find the solution for \( y \).

The integration process often involves specific techniques and substitutions to simplify complex expressions:

• Substitution Method: Useful when the integrand involves a composite function. For example, setting \( u = 2x+1 \) simplifies \( (2x+1)^4 \), making the integration process more manageable.
• Power Rule for Integration: Applied to integrate terms like \( u^4 \). This is structured as \( \int u^n \,du = \frac{u^{n+1}}{n+1} + C \).

In our step-by-step solution, these techniques yield the integrated function \( y = \frac{(2x+1)^5}{10} + C \). Integration allows us to express the solution more precisely, accommodating for constants based on initial conditions.