What integral equation is equivalent to the initial value problem y' = f(x), y(x₀) = y₀?

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Start with the given initial value problem (IVP): \(y' = f(x)\) with the initial condition \(y(x_0) = y_0\).

Recall that \(y' = \frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\).

To find an equivalent integral equation, integrate both sides of the differential equation from \(x_0\) to \(x\): \(\int_{x_0}^{x} y'(t) \, dt = \int_{x_0}^{x} f(t) \, dt\).

By the Fundamental Theorem of Calculus, the left integral simplifies to \(y(x) - y(x_0)\), so we have \(y(x) - y_0 = \int_{x_0}^{x} f(t) \, dt\).

Rearranging, the integral equation equivalent to the IVP is \(y(x) = y_0 + \int_{x_0}^{x} f(t) \, dt\).

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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 specifies a differential equation along with a value of the unknown function at a particular point. For y' = f(x), y(x₀) = y₀, the solution must satisfy both the differential equation and the initial condition y(x₀) = y₀.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if a function is continuous, its integral can be used to recover the original function up to a constant. It allows rewriting the differential equation y' = f(x) as an integral equation involving the integral of f.

Integral Equation Equivalent to IVP

The initial value problem y' = f(x), y(x₀) = y₀ can be expressed as y(x) = y₀ + ∫ₓ₀ˣ f(t) dt. This integral equation states that the solution y(x) equals the initial value plus the accumulated area under f from x₀ to x.