Problem 4 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(\frac{d y}{d x}=y(1-y)\)

Step-by-Step Solution

Answer

The solution is \( y = \frac{Ke^x}{1 + Ke^x} \)

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d y}{d x} = y(1 - y) \). This is a separable differential equation because it can be written as a product of a function of x and a function of y.

Step 2: Separate Variables

Rewrite the equation to separate the variables: \[ \frac{1}{y(1 - y)} dy = dx \]

Step 3: Integrate Both Sides

Integrate both sides of the equation. First, use partial fraction decomposition on the left side: \[ \frac{1}{y(1 - y)} = \frac{A}{y} + \frac{B}{1 - y} \] which gives \[ \int \frac{1}{y(1 - y)} dy = \int \left(\frac{A}{y} + \frac{B}{1 - y}\right) dy \]

Step 4: Solve the Partial Fractions

To find A and B, set up the equation \( \frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1 - y} \). Solving this gives A = 1 and B = -1.

Step 5: Integrate Each Term

Integrate each term separately: \[ \int \frac{1}{y} dy - \int \frac{1}{1 - y} dy = \int dx \] \[-\ln|1-y| + \ln|y| = x + C \]

Step 6: Combine Logarithmic Terms

Combine the logarithms on the left-hand side: \[ \ln\left|\frac{y}{1-y}\right| = x + C \]

Step 7: Exponentiate Both Sides

To solve for y, exponentiate both sides to get rid of the natural log: \[ \frac{y}{1-y} = e^{x+C} \] Let \( e^C = K \), so \( \frac{y}{1-y} = Ke^x \)

Step 8: Solve for y

Solve for y: \[ y = \frac{Ke^x}{1 + Ke^x} \]

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique in algebra, used to break down complex fractions into simpler ones. This is particularly useful in integration because simple fractions are easier to integrate. To do this, we express the fraction as a sum of simpler fractions with unknown coefficients.

For example, to decompose \(\frac{1}{y(1-y)}\), we write it as \(\frac{A}{y} + \frac{B}{1 - y}\). Here, A and B are constants that we need to find.

To find A and B, we compare the numerators of both sides after multiplying by the common denominator. So, we set up the equation and solve for A and B:

• Multiply both sides by \(y(1 - y)\): \(\frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1 - y}\)
• Equate the numerators: \(1 = A(1 - y) + By\)
• The numerator equation then simplifies to: \(1 = A - Ay + By\)

By comparing coefficients, we find that A = 1 and B = -1:

• For the constant term: \(A = 1\)
• For the term with y: \(A - B = 0\), so, \(B = -1\)

This allows us to rewrite the fraction and move forward with the integration step.

Integration

Integration is one of the fundamental operations in calculus, and it is the inverse process of differentiation. It is used to find areas, volumes, central points, and many other useful things. When dealing with differential equations, integration helps to find the general solution.

In the context of our problem, we separated the variables to get an integral form:

• On the left-hand side, \( \frac{1}{y(1 - y)} dy \) becomes \( \frac{A}{y} + \frac{B}{1 - y} \), which we then integrate term-wise.
• On the right-hand side, we simply have \( dx \).

Integrating each term separately, we find:

• \( \frac{1}{y} dy \rightarrow \) results in \( \ln|y| \)
• \( \frac{1}{1 - y} dy \rightarrow \) results in \( -\ln|1 - y| \)
• On the right side, the integral of \( dx \) is \( x + C \)

Summarizing these results, we have:
\[ \ln|y| - \ln|1 - y| = x + C \]
This is why mastering integration techniques is so essential for solving differential equations.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is the logarithm to the base of the mathematical constant \( e \), where \( e \) is approximately equal to 2.71828. It is a very common function in calculus, especially when dealing with growth and decay processes.

In differential equations, the natural logarithm is often seen while integrating functions. In our example, after decomposing and integrating, we end up with natural logarithms:

\[-\ln|1-y| + \ln|y| = x + C\]

Using properties of logarithms, we can combine them. One key property is:

• \(\ln a - \ln b = \ln \( \frac{a}{b}\)\)

Applying this property, we get:

\[\ln \(\frac{y}{1 - y}\) = x + C\]

To solve for y, we exponentiate both sides, reversing the logarithm. This yields:

\[\frac{y}{1-y} = e^{x+C} \]Letting K = e^C, we have:

\[\frac{y}{1-y} = Ke^x\]
From here, solving for y gives us the final solution:
\[y = \frac{Ke^x}{1 + Ke^x}\]This demonstrates the power and utility of the natural logarithm in simplifying and solving complex differential equations.