In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.
3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x

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

Apply the product rule and the Fundamental Theorem of Calculus to find \(y'\). Since \(y\) is a quotient involving an integral, write \(y\) as \(y = \frac{1}{x} \cdot F(x)\) where \(F(x) = \int_{1}^{x} \frac{e^{t}}{t} \, dt\).

Differentiate \(y\) using the quotient rule or product rule: \(y' = \frac{d}{dx} \left( \frac{1}{x} F(x) \right) = -\frac{1}{x^{2}} F(x) + \frac{1}{x} F'(x)\).

Use the Fundamental Theorem of Calculus to find \(F'(x) = \frac{e^{x}}{x}\), then substitute back into the expression for \(y'\).

Substitute \(y\) and \(y'\) into the differential equation \(x^{2} y' + x y\) and simplify to verify that it equals \(e^{x}\).

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주요 개념

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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 us to differentiate functions defined by integrals, which is essential for finding y' in the given problem.

Differentiation of Functions Defined by Integrals

When a function is given as an integral with variable limits, its derivative involves applying the Fundamental Theorem of Calculus and the product or chain rule if multiplied by other functions. Understanding this helps compute y' for y = (1/x) ∫(1 to x) (e^t / t) dt.

Verification of Solutions to Differential Equations

To verify a function solves a differential equation, substitute the function and its derivative into the equation and simplify. If both sides match, the function is a solution. This process confirms that y and y' satisfy x²y' + xy = e^x.

In Exercises 1–4, show that each function y=f(x) is a solution of the accompanying differential equation.3. y = 1/x ∫(from 1 to x) e^t/t dt, x²y' + xy = e^x