91Ó°ÊÓ

Separation of Variables is a powerful technique used to solve differential equations. In these equations, you often have terms with both the dependent variable, say \( y \), and the independent variable, such as \( x \), together in the equation. The goal is to separate these variables so that each is on one side of the equation.Let's look at the example \( \exp(2x + 3y) \frac{dy}{dx} = 1 \). Here, we want to move all terms involving \( y \) and \( dy \) to one side and all terms involving \( x \) and \( dx \) to the other side:

• Divide by \( \exp(2x + 3y) \)
• Multiply by \( dx \)

This manipulation gives us the equation \( \frac{dy}{\exp(3y)} = \frac{dx}{\exp(2x)} \). Now each side of the equation involves only one variable making it ready for integration.

Integration is the process of finding a function that describes the area under a curve. In the context of differential equations, integration is used to solve for the original function when given its derivative.For our separated equation \( \frac{dy}{\exp(3y)} = \frac{dx}{\exp(2x)} \):

• Integrate the left side with respect to \( y \)
• Integrate the right side with respect to \( x \)

The left side simplifies to \( \int e^{-3y} \, dy = -\frac{1}{3} \exp(-3y) + C_1 \). The right side simplifies to \( \int e^{-2x} \, dx = -\frac{1}{2} \exp(-2x) + C_2 \). Combining the constants of integration \( C_1 \) and \( C_2 \) into a single constant, you form a connection between the two integrations. This combined result will lead us to find the explicit form of \( y \).

The natural logarithm is a fundamental concept in calculus, denoted as \( \ln \), and it is the inverse operation of exponentiation with the base \( e \). When solving differential equations, natural logarithms are often used to isolate terms and simplify results.In our solution, after integrating, we arrive at a step where we need to solve \( \exp(3y) = \frac{1}{\frac{3}{2} \exp(-2x) - 3C} \). To solve for \( y \), take the natural logarithm of both sides, leveraging its property as the inverse function of the exponential. Hence,

• Taking \( \ln \) ensures you can bring down exponents to the level of multiplication, making solving for \( y \) easier.

Thus, \( 3y = -\ln \left(\frac{3}{2} \frac{1}{\exp(2x)} - 3C \right) \). Eventually, dividing by 3 gives the final expression for \( y \), and showcases the utility of natural logarithms in finding solutions to differential equations.