Problem 15 - Calculus Single Variable

Question

Solve the given differential equation. $$ \exp (2 x+3 y) \frac{d y}{d x}=1 $$

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{3} \ln \left(\frac{3}{2} \exp(-2x) - 3C \right) \).

Step 1: Separate the Variables

To solve the differential equation \( \exp(2x + 3y) \frac{dy}{dx} = 1 \), let's start by separating the variables. This means we need to have all terms involving \( y \) and \( dy \) on one side and all terms involving \( x \) and \( dx \) on the other side. Divide both sides by \( \exp(2x + 3y) \) and multiply both sides by \( dx \): \[ \frac{dy}{\exp(3y)} = \frac{dx}{\exp(2x)} \]

Step 2: Integrate Both Sides

Now integrate both sides of the equation. For the left side: \[ \int \frac{dy}{\exp(3y)} = \int e^{-3y} \, dy = -\frac{1}{3} \exp(-3y) + C_1 \]For the right side: \[ \int \frac{dx}{\exp(2x)} = \int e^{-2x} \, dx = -\frac{1}{2} \exp(-2x) + C_2 \]

Step 3: Introduce the Constant of Integration

Combine the constants of integration \( C_1 \) and \( C_2 \) into a single constant \( C \), giving \[ -\frac{1}{3} \exp(-3y) = -\frac{1}{2} \exp(-2x) + C \]

Step 4: Simplify and Solve for y

Multiply the entire equation by -3 to isolate \( \exp(-3y) \): \[ \exp(-3y) = \frac{3}{2} \exp(-2x) - 3C \]Take the reciprocal on both sides to isolate \( \exp(3y) \): \[ \exp(3y) = \frac{1}{\frac{3}{2} \exp(-2x) - 3C} \]Take the natural logarithm of both sides to solve for \( y \):\[ 3y = -\ln \left(\frac{3}{2} \frac{1}{\exp(2x)} - 3C \right) \]Thus, \[ y = -\frac{1}{3} \ln \left(\frac{3}{2} \exp(-2x) - 3C \right) \].

Key Concepts

Separation of Variables

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

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 \( y \).

Natural Logarithm

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.