Problem 60 - Calculus for Biology and Medicine

Question

In Problems 59-72, solve the initial-value problem. $$ \frac{d y}{d x}=\frac{x^{2}}{3}, \text { for } x \geq 0 \text { with } y=2 \text { when } x=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

The given differential equation is \( \frac{dy}{dx} = \frac{x^2}{3} \). We need to determine the function \( y(x) \) that satisfies this equation.

Step 2: Integrate the Differential Equation

To solve for \( y(x) \), integrate the right-hand side of the differential equation with respect to \( x \): \[ y(x) = \int \frac{x^2}{3} \, dx\] This integral simplifies to: \[ y(x) = \frac{1}{3} \int x^2 \, dx\] This further simplifies to:\[ y(x) = \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9} + C\] where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

We know from the problem that \( y = 2 \) when \( x = 0 \). Substitute these values into the equation obtained from integration to solve for the constant \( C \):\[ 2 = \frac{0^3}{9} + C \]Simplifying gives:\[ C = 2\] Thus, the particular solution to the initial-value problem is:\[ y(x) = \frac{x^3}{9} + 2\]

Step 4: Verify the Solution

Substitute \( x = 0 \) into the solution to verify it satisfies the initial condition. Substitute into:\[ y(x) = \frac{x^3}{9} + 2\] which gives \( y(0) = \frac{0^3}{9} + 2 = 2 \). The initial condition is verified. Hence, the solution is correct.

Key Concepts

Differential Equation

A differential equation is an equation that relates a function to its derivatives. In simple terms, it tells us how a particular function is changing at any given point. In this exercise, the **differential equation** provided is \( \frac{dy}{dx} = \frac{x^2}{3} \). This equation tells us the rate of change of the function \( y \) concerning the variable \( x \).
Knowing how to solve differential equations is crucial as they appear often in subjects like physics, engineering, and economics to model real-world situations like population growth or decay, heat distribution, and more.
The goal in our problem is to find a function \( y(x) \) that satisfies this equation. To achieve this, we'll follow a structured process of solving differential equations, which involves integration and applying known initial conditions.

Integration

The process of integration plays a vital role in solving differential equations. Integration is essentially the reverse operation of differentiation. When we integrate a function, we are trying to find the original function given its derivative.
In our specific problem, we tackled the integration of \( \frac{x^2}{3} \) to find \( y(x) \). This was done as follows:

• First, simplify the integral: \( y(x) = \int \frac{x^2}{3} \, dx \).
• This simplifies to: \( y(x) = \frac{1}{3} \int x^2 \, dx \).
• Further breaking it down, we get \( y(x) = \frac{1}{3} \cdot \frac{x^3}{3} = \frac{x^3}{9} + C \), where \( C \) is a critical piece called the constant of integration.

Thus, through integration, we derive a general solution, which represents infinitely many potential solutions before applying any constants.

Constant of Integration

When we integrate a function, we often introduce a **constant of integration,** denoted as \( C \). This constant accounts for the fact that when differentiating, any constant term would disappear, hence, integration has to consider this arbitrary constant.
In our problem-solving, this constant \( C \) appears in the expression \( y(x) = \frac{x^3}{9} + C \).
Without further information, \( C \) could take any value, representing different solutions. However, by incorporating an initial condition, we can solve for this constant, thereby fixing the particular solution that fits the given problem constraints.
It's crucial when solving initial-value problems, as it ensures the uniqueness of the specific solution required for the context of the problem.

Initial Condition

An **initial condition** in a differential equation problem provides specific values for the variables to aid in determining the constant of integration and finding the particular solution. In essence, it helps narrow down from the general solution to the one unique solution that meets the problem's criteria.
In our example, the initial condition given was \( y = 2 \) when \( x = 0 \). By substituting \( x = 0 \) into our expression \( y(x) = \frac{x^3}{9} + C \) and setting \( y = 2 \), we solve for \( C \):

• Plugging in these values yields \( 2 = \frac{0^3}{9} + C \).
• This equation simplifies to \( C = 2 \).

Thus, the initial condition tells us precisely which solution from the infinite possibilities applies to our problem, leading us to the specific, correct solution: \( y(x) = \frac{x^3}{9} + 2 \). Verifying this ensures it's consistent with our initial condition, making it a robust and reliable solution.