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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.15

In Exercises 1–22, solve the differential equation.


(x + 3y²) dy + y dx = 0 (Hint: d(xy) = y dx + x dy)

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Rewrite the given differential equation \((x + 3y^{2}) \, dy + y \, dx = 0\) to isolate terms involving \(dx\) and \(dy\). This can be expressed as \(y \, dx + (x + 3y^{2}) \, dy = 0\).
Recognize the hint \(d(xy) = y \, dx + x \, dy\) and try to express part of the equation in terms of \(d(xy)\) to simplify the problem.
Rewrite the equation as \(y \, dx + x \, dy + 3y^{2} \, dy = 0\), which becomes \(d(xy) + 3y^{2} \, dy = 0\).
Integrate both sides with respect to the appropriate variables: integrate \(d(xy)\) directly and integrate \(3y^{2} \, dy\) separately.
After integration, combine the results to form an implicit solution involving \(xy\) and \(y^{3}\), then solve for the general solution of the differential equation.

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Exact Differential Equations

An exact differential equation can be written in the form M(x,y) dx + N(x,y) dy = 0, where there exists a function F(x,y) such that dF = M dx + N dy. Solving involves finding F(x,y) = C, which implicitly defines the solution. Recognizing exactness often simplifies solving differential equations.
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Classifying Differential Equations

Using the Product Rule for Differentials

The product rule states d(uv) = u dv + v du for functions u and v. In this problem, the hint d(xy) = y dx + x dy helps rewrite terms to identify exact differentials or simplify the equation, facilitating integration and solution.
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The Product Rule

Separation and Rearrangement of Variables

Rearranging terms to isolate dy and dx or grouping expressions helps in identifying integrable forms or exact differentials. This step is crucial for transforming the given equation into a solvable form, either by direct integration or applying known solution methods.
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Separation of Variables
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