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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
Capitolo 9, Problema 9.PE.20

In Exercises 1–22, solve the differential equation.
y' + 3x²y = 7x²

Guida verificata passo dopo passo
1
Identify the type of differential equation. The given equation is a first-order linear differential equation of the form \(y' + P(x)y = Q(x)\), where \(P(x) = 3x^{2}\) and \(Q(x) = 7x^{2}\).
Find the integrating factor (IF), which is given by \(\mu(x) = e^{\int P(x) \, dx}\). Calculate \(\int 3x^{2} \, dx\) to determine the integrating factor.
Multiply both sides of the differential equation by the integrating factor \(\mu(x)\) to make the left side an exact derivative of the product \(\mu(x)y\).
Rewrite the left side as \(\frac{d}{dx}[\mu(x)y]\) and set it equal to the right side multiplied by \(\mu(x)\). This gives \(\frac{d}{dx}[\mu(x)y] = \mu(x) Q(x)\).
Integrate both sides with respect to \(x\) to find \(\mu(x)y = \int \mu(x) Q(x) \, dx + C\). Finally, solve for \(y\) by dividing both sides by \(\mu(x)\).

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First-Order Linear Differential Equations

A first-order linear differential equation has the form y' + P(x)y = Q(x). It can be solved using an integrating factor, which simplifies the equation into an exact derivative, allowing integration to find the solution.
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The integrating factor is typically e^(∫P(x)dx). Multiplying the entire differential equation by this factor transforms the left side into the derivative of (integrating factor × y), enabling straightforward integration to solve for y.
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