Problem 76 - Calculus for Biology and Medicine

Question

Solve by rewriting the differential equation as an equation for \(\frac{d x}{d y}\) : $$ \frac{d y}{d x}=\frac{1}{1-y}, \text { for } x \geq 0 \text { with } y(0)=0 $$

Step-by-Step Solution

Answer

The differential equation \( \frac{d y}{d x} = \frac{1}{1-y} \) becomes \( \frac{d x}{d y} = 1 - y \).

Step 1: Interpret the Differential Equation

The original differential equation is \( \frac{d y}{d x} = \frac{1}{1-y} \). This is an expression for \( \frac{d y}{d x} \), the rate of change of \( y \) with respect to \( x \). We need to express the rate of change of \( x \) with respect to \( y \), which is \( \frac{d x}{d y} \).

Step 2: Invert the Differential Equation

To find \( \frac{d x}{d y} \) from \( \frac{d y}{d x} \), we need to take the reciprocal of the given equation. Thus, \( \frac{d x}{d y} = 1 - y \).

Step 3: Check Conditions

The condition \( y(0) = 0 \) simply indicates the initial state of the system where, at \( x = 0 \), the value of \( y \) is 0. This condition can be used later in solving the differential equation but is not needed for finding \( \frac{d x}{d y} \).

Key Concepts

Understanding Rate of Change

In the context of differential equations, the **rate of change** represents how one quantity changes in relation to another. It is essentially a derivative, and it acts as a mathematical representation of how fast a variable changes over time or in relation to another variable. In our exercise, the given differential equation is \( \frac{d y}{d x} = \frac{1}{1-y} \). Here, \( \frac{d y}{d x} \) is the derivative of \( y \) with respect to \( x \), indicating the rate at which \( y \) changes as \( x \) changes.

To convert this into the rate of change of \( x \) with respect to \( y \), you essentially need to find the reciprocal of the derivative, which results in \( \frac{d x}{d y} \). Understanding rate of change is crucial for interpreting differential equations because it tells you how variables in the system interact with and affect each other.

Solving Differential Equations

**Solving differential equations** involves finding a function or a set of functions that satisfy the equation. In our exercise, the differential equation was initially \( \frac{d y}{d x} = \frac{1}{1-y} \). We then aimed to express it as \( \frac{d x}{d y} \). This process often involves manipulating the equation through algebraic methods or integration to find a solution that satisfies the conditions.

The step given here inverts the differential by taking the reciprocal of \( \frac{d y}{d x} \) to get \( \frac{d x}{d y} = 1 - y \). This kind of manipulation is a common technique when working with differential equations, particularly when attempting to solve or simplify them. The ultimate goal when solving differential equations is to find a function that makes the equation true, which often involves applying known mathematical techniques and recognizing the type of solution path that will work best for the given equation.

Grasping Initial Conditions

**Initial conditions** are requirements that a solution to a differential equation must satisfy at a particular point. They are used to specify the particular solution out of the family of possible solutions to a differential equation. In our problem, the initial condition is provided as \( y(0) = 0 \). This specifies that when \( x = 0 \), \( y \) must also equal 0.

Initial conditions are crucial in determining a unique solution to a differential equation. Without initial conditions, a differential equation typically has a general solution with an arbitrary constant. Once you have an initial condition, you can substitute it into the general solution to solve for the constant, thus narrowing down to the particular solution. In real-world applications, initial conditions often come from the physical or practical circumstances surrounding the problem being modeled.