Problem 46 - Calculus for Biology and Medicine

Question

In Problems 46-54, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=2 \frac{y}{x}, \text { with } y(1)=1 $$

Step-by-Step Solution

Answer

The solution is \( y = x^2 \).

Step 1 - Rearrange the differential equation

The given differential equation is \( \frac{d y}{d x} = 2 \frac{y}{x} \). Notice that this is a separable differential equation. To solve it, we first rewrite it as \( \frac{d y}{y} = 2 \frac{d x}{x} \), which separates the variables.

Step 2 - Integrate both sides

Integrate both sides of the equation: \( \int \frac{1}{y} \, d y = \int 2 \frac{1}{x} \, d x \). The left side integrates to \( \ln |y| \) and the right side integrates to \( 2 \ln |x| + C \), where \( C \) is the integration constant. Thus, we have \( \ln |y| = 2 \ln |x| + C \).

Step 3 - Exponentiate to solve for \( y \)

Exponentiate both sides of \( \ln |y| = 2 \ln |x| + C \) to solve for \( y \). This gives \( |y| = e^C \cdot x^2 \). We can write \( y = A x^2 \) where \( A = \pm e^C \), representing a constant.

Step 4 - Use the initial condition

Use the initial condition \( y(1) = 1 \) to find \( A \). Substituting \( x = 1 \) and \( y = 1 \) into \( y = A x^2 \) gives \( 1 = A \cdot 1^2 \), so \( A = 1 \). Thus, \( y = x^2 \).

Key Concepts

Separable Differential Equations

Separable differential equations are a special type of differential equations where you can separate the variables. This makes them relatively simple to solve compared to other types. The key idea is to manipulate the equation such that all terms involving one variable (let's say it's \( y \)) are on one side of the equation, and all terms involving another variable (such as \( x \)) are on the other side. This allows each side to be integrated separately.

For example, consider the differential equation from the exercise: \( \frac{d y}{d x} = 2 \frac{y}{x} \). You can re-arrange it to be \( \frac{d y}{y} = 2 \frac{d x}{x} \). Now, terms involving \( y \) and \( x \) are separated, which is the essence of a separable differential equation.

To solve these equations:

• Rearrange the equation to separate the variables.
• Integrate both sides separately with respect to their respective variables.
• Solve for the variable in question, often using additional information like initial conditions.

This method leverages integration as a stepping stone to the solution, making it a straightforward process.

Initial Conditions

Initial conditions are crucial in differential equations because they allow us to find particular solutions. Without them, we would only have a general solution which includes an arbitrary constant. This constant can represent an infinite number of solutions. The initial condition helps to 'pin down' which one of these infinite solutions is relevant to the specific scenario we are looking at.

In this exercise, the initial condition given is \( y(1) = 1 \). This tells us that when \( x = 1 \), \( y \) must be equal to 1. By substituting \( x = 1 \) and \( y = 1 \) into the equation \( y = A x^2 \), we can solve for \( A \). It turns out \( A = 1 \), so our particular solution is \( y = x^2 \).

Initial conditions are essential for:

• Specifying a unique solution to a differential equation.
• Making the abstract solution concrete and applicable to real-world situations.
• Testing the solution manually by plugging in values to ensure accuracy.

They transform general solutions into specific answers tailored to the problem at hand.

Integration

Integration is a fundamental technique in calculus used for finding a function based on its derivative. In the context of differential equations, integration is employed to solve equations where derivatives are given and you need to find the original function.

For separable differential equations, once the equation is rearranged to separate the variables, both sides can be integrated with respect to their respective variables. Using our example, after separating the variables: \( \int \frac{1}{y} \, dy = \int 2 \frac{1}{x} \, dx \), the integration gives us the general solution: \( \ln |y| = 2 \ln |x| + C \). Here, \( C \) represents the integration constant.

Integration is used:

• To reverse the process of differentiation, giving us the original function from its derivative.
• To uncover the solutions of differential equations, critical in fields like physics and engineering.
• To incorporate initial conditions, which allow the conversion from general to particular solutions.
  
  Through integration, the seemingly complex process of solving differential equations becomes manageable, enabling us to uncover valuable insights into natural and mathematical phenomena.