Problem 20 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Solve the given initial value problem using either integration by parts or a formula from Table 6.1. Note that Exercises 19 and 20 involve separable differential equations. $$ \frac{d y}{d x}=\frac{x y}{3+x}, \text { where } y=1 \text { when } x=1 $$

Step-by-Step Solution

Answer

The solution to the differential equation is \( y = 64 e^{x-1} (3 + x)^{-3} \).

Step 1: Recognize the Differential Equation

Identify that the given differential equation is a separable differential equation: \( \frac{dy}{dx} = \frac{xy}{3+x} \)

Step 2: Separate the Variables

Rearrange the terms to separate the variables on each side: \( \frac{1}{y} dy = \frac{x}{3+x} dx \)

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{1}{y} dy = \int \frac{x}{3+x} dx \) The left side integrates directly to \( \ln |y| \).

Step 4: Simplify Right Side using Substitution

To integrate the right side, use the substitution \( u = 3 + x \), then \( du = dx \). The integral becomes \( \int \frac{x}{u} du = \int \frac{u-3}{u} du = \int 1 - \frac{3}{u} du \).

Step 5: Evaluate the Integral

Integrate to get: \( u - 3 \ln|u| = x - 3 \ln|3+x| \). So: \( \ln |y| = x - 3 \ln|3+x| + C \)

Step 6: Solve for y

Exponentiate both sides to solve for y: \( |y| = e^{x - 3 \ln|3+x| + C} \). Thus: \( y = Ce^{x} (3 + x)^{-3} \)

Step 7: Apply Initial Condition

Use the initial condition \( y(1) = 1 \) to find C. \( 1 = C e^{1} (4)^{-3} \), solving for C gives \( C = 64 e^{-1} \).

Step 8: Final Solution

Substitute C back into the equation: \( y = 64 e^{-1} e^{x} (3 + x)^{-3} \). Simplify to: \( y = 64 e^{x-1} (3 + x)^{-3} \).

Key Concepts

integration by parts

Integration by parts is a key technique in calculus used to integrate the product of two functions. It is based on the product rule for differentiation and is expressed as: ewline ewline \( \ \ \int u dv = uv - \int v du \ \ \ \) ewline Here are the steps to use integration by parts effectively: ewline ewline

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• Identify parts of the integrand you will set as u and dv.
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• Differentiate u to get du, and integrate dv to get v.
• Substitute into the integration by parts formula.
ewline ewline This method is particularly useful when integrating products of polynomial, exponential, logarithmic, or trigonometric functions. ewline ewline In our specific problem, however, we made use of substitution for simplification instead of directly employing integration by parts.

substitution method

The substitution method, also known as u-substitution, simplifies integrals by making a substitution that transforms the integral into a basic form. It's particularly useful here when working with the function involving composite expressions. ewline ewline The general steps in substitution are: ewline ewline

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• Choose a substitution: Set u equal to an expression in the integral.
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• Differentiate u to find du: This will provide a new variable, du, to replace the original differential.
• Substitute all instances of the original variable in the integral.
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• Integrate with respect to the new variable.
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• Replace u with the original expression.
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ewline In our problem, we used the substitution \(u = 3 + x\) which helped simplify the integral \(\frac{x}{3+x}\) by transforming it into \(\frac{u-3}{u}\). This made the integration process straightforward.

initial value problem

An initial value problem (IVP) specifies values of a function and its derivatives at a point. The goal is to find a function that satisfies both the differential equation and the initial conditions. ewline ewline Steps to solve an IVP: ewline

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• Solve the differential equation to get the general solution.
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• Use the initial conditions to find the specific solution by substituting into the general solution.

ewline In our problem, we were given \(\frac{dy}{dx} = \frac{xy}{3+x} \) and \ y=1\ \ when \(x=1\). We solved the equation, then used the initial condition to find the integration constant (C). This pinpointed the specific solution out of the family of potential solutions.

integration techniques

Integration techniques are varied methods used in evaluating integrals. Several techniques include: ewline ewline

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• Direct Integration: Directly apply integration rules.
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• Substitution (u-substitution): Replace complex expressions with simpler ones.
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• Integration by parts: Useful for products of functions.
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• Partial Fractions: Decompose rational functions into simpler fractions.
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ewline Each problem may require a different approach. In our exercise, we primarily used the separation of variables technique along with substitution to find the integral of a differential equation. Understanding these techniques enables solving a wide range of integrals efficiently.

calculus for business

Calculus has important applications in business, particularly in optimizing functions and modeling growth. Some key applications include: ewline ewline

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• Cost Functions: Analyzing cost functions and finding cost minimization points.
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• Revenue and Profit Functions: Analyzing revenue functions for maximizing profit.
• Marginal Analysis: Using derivatives to analyze marginal cost, revenue, and profit.

ewline By understanding differential equations and integration, one can model and predict changes over time, helping businesses make informed decisions. Our problem with initial value and solving differential equations is also closely related to real-world business scenarios, such as predicting product demand over time or changes in investment values.