Problem 35 - Calculus for Biology and Medicine

Question

$$ \text { In Problems } 35-38, \text { solve each differential equation. } $$ $$ \frac{d y}{d x}=y(1+y) $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{Ce^x}{1 - Ce^x} \), \( C \) is a constant.

Step 1: Separate Variables

First, we need to rewrite the differential equation \( \frac{dy}{dx} = y(1+y) \) in a form where all \( y \) terms are on one side and all \( x \) terms are on the other side. This can be achieved by rewriting it as \( \frac{1}{y(1+y)} dy = dx \).

Step 2: Integrate Both Sides

Next, we integrate both sides of the equation. The left side becomes \( \int \frac{1}{y(1+y)} \, dy \), and the right side becomes \( \int \ dx \). The left-hand side requires partial fraction decomposition: \[ \frac{1}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y} \] Solve for \( A \) and \( B \): \[ 1 = A(1+y) + By \]Set \( y = 0 \): \( A = 1 \).Set \( y = -1 \): \( B = -1 \). This gives the partial fractions: \[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy = \int \, dx \]

Step 3: Evaluate Integrals

For the left side, integrate each part:\[ \int \frac{1}{y} \, dy = \ln |y| \]\[ \int \frac{1}{1+y} \, dy = \ln |1+y| \]Therefore, the integral is:\[ \ln |y| - \ln |1+y| = x + C \], where \( C \) is the constant of integration.

Step 4: Simplify the Result

Combine the logarithms using the property \( \ln a - \ln b = \ln \frac{a}{b} \):\[ \ln \frac{|y|}{|1+y|} = x + C \]To solve for \( y \), remove the natural logarithm by exponentiating both sides:\[ \frac{|y|}{|1+y|} = Ce^x \], where \( C = e^C \).

Step 5: Solve for \( y \) Explicitly

Assume \( y \) is positive, which simplifies the absolute values, and multiply both sides by \( |1+y| \):\[ |y| = Ce^x |1+y| \]Assuming \( y = Ce^x (1+y) \) simplifies to solve for \( y(x) \), considering the scenario \( y > 0 \), results in the explicit solution:\[ y = \frac{Ce^x}{1 - Ce^x} \] for \( Ce^x < 1 \).

Step 6: Final Step: Conclusion of Problem

The solution to the differential equation is given by \( y(x) = \frac{Ce^x}{1 - Ce^x} \), where \( C \) is an arbitrary constant dependent on initial conditions.

Key Concepts

Separation of Variables

Separation of variables is a technique commonly used to solve differential equations. The fundamental idea is to rearrange the equation so that each variable appears on a different side of the equation.
For the given differential equation \( \frac{dy}{dx} = y(1+y) \), this means rewriting it such that all terms involving \( y \) are on one side, and all terms involving \( x \) are on the other side:

• Start with the equation: \( \frac{dy}{dx} = y(1+y) \).
• Rearrange to isolate terms: \( \frac{1}{y(1+y)} dy = dx \).

This separates "y " and "x" making it easier to integrate, helping us progress towards finding a solution.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. Once variables are separated, the next step is to integrate both sides.
In the exercise, the left-hand side \( \int \frac{1}{y(1+y)} \, dy \) can be integrated using partial fraction decomposition, while the right-hand side involves a direct integration \( \int \, dx \), which is straightforward:

• The right side integrates to \( x + C \), where \( C \) is the constant of integration.
• Partial fractions helps simplify the left side for easier integration.

Integration transforms our separated equation into a solvable form, moving us closer to the solution.

Partial Fraction Decomposition

Partial fraction decomposition is a method for breaking down complex rational expressions into simpler fractions that are easier to integrate.
Consider the expression \( \frac{1}{y(1+y)} \).

• Express it as \( \frac{A}{y} + \frac{B}{1+y} \).
• Determine constants \( A \) and \( B \) by equating \( A(1+y) + By = 1 \).
• Solving yields: \( A = 1 \) and \( B = -1 \).

This decomposition simplifies our integral, allowing us to solve it as \( \int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy \), which is straightforward to evaluate.