11–16. Initial value problems Solve the following initial value problems.

y'(x) = −y + 2, y(0) = −2

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Identify the type of differential equation given. Here, the equation is a first-order linear ordinary differential equation of the form \(y'(x) + y = 2\).

Rewrite the differential equation in standard linear form: \(y' + p(x)y = q(x)\), where \(p(x) = 1\) and \(q(x) = 2\).

Find the integrating factor \(\mu(x)\) using the formula \(\mu(x) = e^{\int p(x) \, dx}\). Since \(p(x) = 1\), calculate \(\mu(x) = e^{\int 1 \, dx} = e^{x}\).

Multiply both sides of the differential equation by the integrating factor \(e^{x}\) to get \(e^{x} y' + e^{x} y = 2 e^{x}\). Recognize that the left side is the derivative of \(e^{x} y\) with respect to \(x\).

Integrate both sides with respect to \(x\): \(\int \frac{d}{dx} (e^{x} y) \, dx = \int 2 e^{x} \, dx\). Then solve for \(y(x)\) and apply the initial condition \(y(0) = -2\) to find the constant of integration.

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주요 개념

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First-Order Differential Equations

A first-order differential equation involves the first derivative of a function and the function itself. Solving such equations means finding a function y(x) that satisfies the given relationship between y' and y. In this problem, y'(x) = -y + 2 is a linear first-order differential equation.

Initial Value Problems (IVP)

An initial value problem specifies the value of the unknown function at a particular point, here y(0) = -2. This condition allows us to find a unique solution to the differential equation by determining the constant of integration after solving the general solution.

Solving Linear Differential Equations Using Integrating Factors

Linear first-order differential equations can be solved using an integrating factor, which is a function that simplifies the equation into an exact derivative. Multiplying both sides by this factor allows integration and leads to the general solution, which can then be adjusted using the initial condition.

11–16. Initial value problems Solve the following initial value problems.y'(x) = −y + 2, y(0) = −2