Show that the solution of the initial value problem
y' = x + y, y(x₀) = y₀
is
y = -1 -x + (1 + x₀ + y₀) exp(x-x₀).

검증된 단계별 안내

Recognize that the given differential equation is a first-order linear ordinary differential equation of the form \(y' - y = x\).

Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int -1 \, dx} = e^{-x}\).

Multiply both sides of the differential equation by the integrating factor to get \(e^{-x} y' - e^{-x} y = x e^{-x}\), which simplifies to \(\frac{d}{dx}(e^{-x} y) = x e^{-x}\).

Integrate both sides with respect to \(x\): \(\int \frac{d}{dx}(e^{-x} y) \, dx = \int x e^{-x} \, dx\), leading to \(e^{-x} y = \int x e^{-x} \, dx + C\).

Solve the integral on the right side using integration by parts, then multiply both sides by \(e^{x}\) to isolate \(y\). Finally, apply the initial condition \(y(x_0) = y_0\) to solve for the constant \(C\) and write the solution in the form \(y = -1 - x + (1 + x_0 + y_0) e^{x - x_0}\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

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

First-Order Linear Differential Equations

A first-order linear differential equation has the form y' + p(x)y = q(x). Such equations can be solved using an integrating factor, which simplifies the equation into an exact derivative, allowing integration to find the general solution.

Integrating Factor Method

The integrating factor is a function, usually denoted μ(x) = exp(∫p(x) dx), used to multiply both sides of a linear differential equation. This transforms the left side into the derivative of (μ(x)y), enabling straightforward integration to solve for y.

Initial Value Problem (IVP)

An initial value problem specifies the value of the solution at a particular point, y(x₀) = y₀. After finding the general solution, the initial condition is used to determine the constant of integration, yielding the unique solution that satisfies the problem.

Show that the solution of the initial value problemy' = x + y, y(x₀) = y₀isy = -1 -x + (1 + x₀ + y₀) exp(x-x₀).