Problem 23 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d y}{d x}=\frac{5 y}{x}, \quad y=3 \text { where } x=1$$

Step-by-Step Solution

Answer

The solution is \( y = 3x^5 \).

Step 1: Rearrange for Separation

Start by rewriting the given differential equation to separate the variables. We have \( \frac{dy}{dx} = \frac{5y}{x} \). To separate, write it as \( \frac{dy}{y} = \frac{5}{x} dx \). This separates the variables \( y \) and \( x \).

Step 2: Integrate Both Sides

Integrate both sides of the equation from Step 1. The left side becomes \( \int \frac{dy}{y} = \ln |y| + C_1 \). The right side is \( \int \frac{5}{x} \, dx = 5 \ln |x| + C_2 \).

Step 3: Combine Constants

Since both sides have integration constants, we can combine them into a single constant. After integration and simplification, the equation becomes \( \ln |y| = 5 \ln |x| + C \), where \( C = C_2 - C_1 \).

Step 4: Solve for y

Exponentiate both sides to solve for \( y \): \( |y| = e^{5 \ln |x| + C} = e^C \cdot (|x|)^5 = C' x^5 \). Thus, \( y = \pm C' x^5 \).

Step 5: Apply Initial Conditions

Use the initial condition \( y = 3 \) when \( x = 1 \) to find the specific solution. Substituting, \( 3 = C' \) since \( 1^5 = 1 \). So, \( y = 3x^5 \).

Step 6: Final Solution

The solution to the differential equation, given the initial condition, is \( y = 3x^5 \). Confirm this satisfies the conditions: when \( x = 1 \), \( y = 3 \).

Step 7: Verification

Substitute \( y = 3x^5 \) back into the original differential equation \( \frac{dy}{dx} = \frac{5y}{x} \): \( \frac{d}{dx}(3x^5) = 15x^4 \). This simplifies to \( \frac{5(3x^5)}{x} = 15x^4 \), which matches. Thus, the solution is verified.

Key Concepts

Differential Equations

A differential equation relates a function with its derivatives. In this exercise, we have the differential equation \( \frac{d y}{d x} = \frac{5 y}{x} \). This means we are working with a rate of change of \( y \) with respect to \( x \), which is represented by \( \frac{d y}{d x} \). Such equations play crucial roles in modeling the behavior of dynamic systems, such as growth or decay processes.

Differential equations can be complex, but they are often solvable by simplifying methods like separation of variables. This method allows us to solve equations by separating variables on either side of the equation, simplifying the integration process. The goal is to express each side in terms of a single variable, making the equation easier to manage.

Initial Conditions

Initial conditions provide specific values that a solution to a differential equation must satisfy at a given point. In this problem, the initial condition is \( y = 3 \) when \( x = 1 \).

Why are initial conditions vital? They help determine one specific solution out of potentially many possibilities for a differential equation. After finding a general solution, initial conditions allow us to calculate exact constants needed to meet these predefined conditions. Without such conditions, we would only have a family of solutions rather than a unique one.

Integration

Integration is the process of finding a function given its rate of change. In the context of differential equations, integration is used to solve the separated variables by finding an antiderivative.

In our problem, after separating the variables to \( \frac{dy}{y} = \frac{5}{x} dx \), we integrate both sides:

• The left-hand side integrates to \( \ln |y| \)
• The right-hand side integrates to \( 5 \ln |x| \)

By integrating, we find that the solution is the combination of these expressions, adjusted for the constant of integration \( C \). Once integrated, exponentiation helps eliminate the logarithms to solve for \( y \), making use of the properties of logarithms and exponents.

Exponential Functions

Exponential functions arise when solving linear differential equations like ours, particularly after integration. In the equation \( \ln |y| = 5 \ln |x| + C \), exponentiating both sides helps remove the logarithms, simplifying it to \( |y| = e^{C} \cdot (|x|)^5 \).

This showcases how exponential functions are deeply connected with the natural logarithm. They allow for solutions that involve growth or decay in complex systems.

Exponential functions are powerful tools in mathematics due to their unique properties, such as continuous growth and decay behaviors, which are evident in numerous real-world phenomena, like population growth and radioactive decay.