Integral Equations
In Exercises 7–12, write an equivalent first-order differential equation
and initial condition for y.

y = ∫₁ ͯ 1/t dt

Guida verificata passo dopo passo

Identify the given integral equation: \(y = \int_{1}^{x} \frac{1}{t} \, dt\).

Recall the Fundamental Theorem of Calculus, which states that if \(y = \int_{a}^{x} f(t) \, dt\), then \(\frac{dy}{dx} = f(x)\).

Apply the theorem to differentiate both sides with respect to \(x\): \(\frac{dy}{dx} = \frac{1}{x}\).

Write the equivalent first-order differential equation: \(\frac{dy}{dx} = \frac{1}{x}\).

Determine the initial condition by evaluating \(y\) at the lower limit of integration: since \(y = \int_{1}^{x} \frac{1}{t} \, dt\), then \(y(1) = 0\).

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Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if a function is defined as an integral with a variable upper limit, its derivative is the integrand evaluated at that limit. It allows converting integral expressions into differential equations by differentiating both sides.

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Fundamental Theorem of Calculus Part 1

Differentiation of Integral Functions

When a function is defined as an integral with a variable limit, differentiating it requires applying the Leibniz rule or the Fundamental Theorem of Calculus. This process transforms the integral equation into a differential equation involving the original function and its derivative.

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Integrals Resulting in Basic Trig Functions Example 1

Initial Conditions in Differential Equations

Initial conditions specify the value of the unknown function at a particular point, ensuring a unique solution to a differential equation. For integral-defined functions, the initial condition often comes from evaluating the integral at the lower limit.

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