Problem 32 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ \frac{d y}{d x}=\exp (x+y) \quad y(0)=0 $$

Step-by-Step Solution

Answer

The solution is \( y = -\ln(2 - e^{x}) \).

Step 1: Separate Variables

We start by separating the variables in the differential equation. The original differential equation is \( \frac{d y}{d x}=\exp(x+y) \). We rewrite it in a separable form: \( y'(x) = e^x \cdot e^y \). By separation of variables, we have \( \frac{dy}{e^y} = e^x \, dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation from Step 1: \( \int \frac{dy}{e^y} = \int e^x \, dx \). This gives us \( \int e^{-y} \, dy = \int e^x \, dx \).

Step 3: Compute Integrals

The integral of \( e^{-y} \, dy \) is \( -e^{-y} \) and the integral of \( e^{x} \, dx \) is \( e^{x} \). Thus, after integrating, we have: \( -e^{-y} = e^{x} + C \), where \( C \) is the integration constant.

Step 4: Apply Initial Condition

Use the initial condition \( y(0) = 0 \) to find the value of \( C \). Plugging this into the equation, \( -e^{-0} = e^{0} + C \) gives \( -1 = 1 + C \), so \( C = -2 \).

Step 5: Solve for \( y \)

Substitute \( C = -2 \) back into the integrated equation: \( -e^{-y} = e^{x} - 2 \). Solve for \( y \) by taking negative inverse and logarithm: \( e^{-y} = 2 - e^{x} \), implying \( -y = \ln(2 - e^{x}) \). Hence, \( y = -\ln(2 - e^{x}) \).

Key Concepts

Initial Value Problem

An initial value problem in differential equations represents a problem where one seeks a solution to a differential equation that not only adheres to the equation itself but also satisfies a given initial condition. This initial condition specifies the value of the solution at a particular point in the domain.
In our exercise, the differential equation provided is \( \frac{dy}{dx} = \exp(x + y) \), with the initial condition \( y(0) = 0 \). This condition allows us to determine the constant of integration, ensuring our solution is precise not just in form but also in context.

• The given initial condition \( y(0) = 0 \) helps in finding the unique solution to the differential equation among the infinitely many possible solutions.
• Without the initial condition, the solution would have an arbitrary constant \( C \), representing a family of functions.

These initial value problems are common in mathematical modeling of real-world scenarios where the state of the system is known at the beginning.

Separation of Variables

Separation of variables is a powerful method for solving differential equations, making it particularly useful for those expressed in separable form.
This technique involves rewriting the differential equation so that each different variable—\( x \) and \( y \)—appears on opposite sides of the equation. In our provided differential equation \( \frac{dy}{dx} = \exp(x + y) \), the challenge is to isolate the variables:

• We express the differential equation as \( y'(x) = e^x \cdot e^y \), which aids in separating the variables.
• Upon separation, we reach \( \frac{dy}{e^y} = e^x \, dx \). This configuration allows us to integrate each side with respect to its own variable.

By strategically moving terms, typical non-linear equations can be transformed into simpler integrable forms. This method is highly advantageous due to its direct approach and is effective when a differential equation allows for such manipulation.

Integration

Once the variables in the differential equation are separated, integration is used to solve for the unknowns. This step involves calculating the antiderivative of each side of the equation after separation.
For our equation, we begin by integrating both sides once they are properly separated:

• We solve \( \int \frac{dy}{e^y} = \int e^x \, dx \).
• The integral on the left gives \( -e^{-y} \), and the integral on the right results in \( e^x \).

Integration also involves determining the constant of integration \( C \). Applying the initial condition aids in solving for \( C \), making the solution to both sides definite rather than arbitrary.
Integration, being a fundamental concept in calculus, bridges the process of finding solutions to differential equations by reversing differentiation.

Exponential Function

The exponential function, \( e^{x} \), is significant when solving differential equations, particularly those involving growth and decay processes. In the given problem, the exponential function appears in the equation \( \frac{dy}{dx} = \exp(x + y) \). The role of the exponential function in the equation is multifaceted:

• It defines the growth rate of the function \( y \) with respect to \( x \), which influences how \( y \) changes as \( x \) varies.
• The exponentiation allows for transformation during the separation stage, where \( \exp(x+y) \) becomes \( e^x \cdot e^y \).

The exponential function is particularly useful because of its unique properties:

• It is its own derivative, which simplifies calculations during integration.
• Exponential growth is ubiquitous in many natural processes, such as population growth and radioactive decay, thus understanding it deeply can blend mathematical theory with real-world applications.

Mastering the properties of the exponential function facilitates a deeper understanding of both its standalone importance and how it integrates into solving differential equations.