Problem 60 - Calculus for Biology and Medicine

Question

Solve the initial-value problem. $$ \frac{d y}{d x}=\frac{x^{2}}{3}, \text { for } x \geq 0 \text { with } y(0)=2 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{x^3}{9} + 2 \).

Step 1: Identify the Differential Equation

The given problem is to solve the differential equation \( \frac{dy}{dx} = \frac{x^2}{3} \). We also have the initial condition \( y(0) = 2 \). Our goal is to find a function \( y(x) \) that satisfies both the differential equation and the initial condition.

Step 2: Integrate the Differential Equation

To find \( y(x) \), we need to integrate the function on the right-hand side of the differential equation. Integrate with respect to \( x \):\[y(x) = \int \frac{x^2}{3} \, dx = \frac{1}{3} \int x^2 \, dx.\]

Step 3: Solve the Integral

Calculate the integral from Step 2:\[\frac{1}{3} \int x^2 \, dx = \frac{1}{3} \left( \frac{x^3}{3} \right) = \frac{x^3}{9}.\]Thus, the general solution is \( y(x) = \frac{x^3}{9} + C \), where \( C \) is the constant of integration.

Step 4: Apply the Initial Condition

Use the initial condition \( y(0) = 2 \) to find the constant \( C \). Substitute \( x = 0 \) and \( y = 2 \) in the equation:\[2 = \frac{0^3}{9} + C, \quad \text{which simplifies to} \quad 2 = C.\]Thus, the constant \( C = 2 \).

Step 5: Write the Particular Solution

Substitute the value of \( C \) back into the general solution:\[y(x) = \frac{x^3}{9} + 2.\]This is the particular solution to the initial-value problem.

Key Concepts

Understanding Differential Equations

A differential equation is essentially a mathematical equation that involves functions and their derivatives. In simpler terms, it implies a relationship between a function, which we may symbolically represent as \( y \), and its rate of change with respect to another variable, often \( x \). In our initial-value problem, the differential equation is given as \( \frac{dy}{dx} = \frac{x^2}{3} \). This indicates that the rate at which \( y \) changes in response to changes in \( x \) is proportional to \( \frac{x^2}{3} \).
Differential equations come in various forms, but they are generally used to model real-world phenomena where changes occur, such as physics, engineering, and biology. Solving a differential equation means finding a function that relates \( y \) and \( x \) that satisfies the given rate of change or derivative.
In our problem, to find the desired function \( y(x) \), we'll need to perform an integration of the differential expression.

The Art of Integration

Integration is often seen as the reverse process of differentiation. While differentiation concerns itself with rates of change, integration concerns accumulation or finding the original quantity from the derivative. In the case of our differential equation, \( \frac{dy}{dx} = \frac{x^2}{3} \), we integrate the derivative to find the original function \( y(x) \).
**Steps to Integrate:**

• Identify the expression to integrate: In this case, \( \frac{x^2}{3} \).
• Apply integration techniques: Since \( x^2 \) is a basic polynomial, we can use the power rule for integration.
• Integrate to find the antiderivative: \( \int \frac{x^2}{3} \, dx = \frac{1}{3} \int x^2 \, dx = \frac{x^3}{9} + C \), where \( C \) is the constant of integration.

The process of integration helps us reconstruct the original function \( y(x) \) from its rate of change. The result includes a constant of integration, which reflects the infinite set of functions that have the same derivative, differing only by a constant.

Decoding the Constant of Integration

Whenever we integrate a function, we encounter the constant of integration, often represented as \( C \). This constant is crucial because integration essentially undoes differentiation, and many different original functions could have the same derivative function. For example, both \( x^3 \) and \( x^3 + 5 \) differentiate to \( 3x^2 \). Adding a constant of integration accounts for this family of solutions.
In initial-value problems, like ours, this is where the initial condition \( y(0) = 2 \) comes into play. By substituting \( x = 0 \) and \( y = 2 \) into the integrated function \( y(x) = \frac{x^3}{9} + C \), we solve for \( C \). Here, we found \( 2 = \frac{0^3}{9} + C \), meaning \( C = 2 \). Adding this constant helps us to arrive at the particular solution that fits the specific condition, ensuring the function describes the scenario accurately.