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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 9 - First-Order Differential EquationsProblema 9.PE.12
Capitolo 9, Problema 9.PE.12

In Exercises 1–22, solve the differential equation.
xy' - y = 2x ln x

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1
Rewrite the given differential equation in the standard linear form. Starting with \(xy' - y = 2x \ln x\), divide both sides by \(x\) (assuming \(x \neq 0\)) to get \(y' - \frac{1}{x} y = 2 \ln x\).
Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int P(x) \, dx}\) where \(P(x)\) is the coefficient of \(y\) in the standard form. Here, \(P(x) = -\frac{1}{x}\), so calculate \(\mu(x) = e^{\int -\frac{1}{x} \, dx}\).
Simplify the integrating factor \(\mu(x)\) by evaluating the integral \(\int -\frac{1}{x} \, dx\) and then exponentiating the result.
Multiply the entire differential equation by the integrating factor \(\mu(x)\) to transform the left side into the derivative of the product \(\mu(x) y\). This gives \(\frac{d}{dx} [\mu(x) y] = \mu(x) \cdot 2 \ln x\).
Integrate both sides with respect to \(x\) to find \(\mu(x) y = \int \mu(x) \cdot 2 \ln x \, dx + C\), where \(C\) is the constant of integration. Finally, solve for \(y\) by dividing both sides by \(\mu(x)\).

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A first-order linear differential equation has the form y' + P(x)y = Q(x). Solving it involves finding an integrating factor to simplify the equation into an exact derivative, allowing integration to find the general solution.
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The integrating factor is a function, usually denoted μ(x), used to multiply both sides of a linear differential equation to make the left side an exact derivative. It is typically μ(x) = e^(∫P(x) dx), facilitating straightforward integration.
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