In Exercises 129–132 solve the initial value problem.
129. dy/dx = e^(-x-y-2), y(0) = -2

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Identify the type of differential equation given: \( \frac{dy}{dx} = e^{-x - y - 2} \). Notice that the right side can be rewritten to separate variables.

Rewrite the equation by expressing the right side as \( e^{-x - y - 2} = e^{-2} \cdot e^{-x} \cdot e^{-y} \). This helps in separating variables \( x \) and \( y \).

Separate variables by rewriting the equation as \( e^{y} dy = e^{-2} e^{-x} dx \). This places all \( y \)-terms on one side and all \( x \)-terms on the other.

Integrate both sides: \( \int e^{y} dy = e^{-2} \int e^{-x} dx \). Remember to include the constant of integration after integrating.

Use the initial condition \( y(0) = -2 \) to solve for the constant of integration, then express \( y \) explicitly as a function of \( x \) if possible.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Initial Value Problem (IVP)

An initial value problem involves solving a differential equation with a given initial condition, which specifies the value of the unknown function at a particular point. This condition allows for finding a unique solution that fits both the differential equation and the initial value.

Separable Differential Equations

A separable differential equation can be written so that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side. This allows integration of both sides separately to find the general solution.

Integration of Exponential Functions

Integrating exponential functions often involves recognizing the form e^(ax+b) and applying substitution if necessary. Understanding how to integrate these functions is essential for solving differential equations involving exponentials.

In Exercises 129–132 solve the initial value problem.129. dy/dx = e^(-x-y-2), y(0) = -2