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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.4.19

Solve the differential equation in Exercises 9–22.
19. y²(dy/dx) = 3x²y³ - 6x²

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Rewrite the given differential equation \(y^{2} \frac{dy}{dx} = 3x^{2} y^{3} - 6x^{2}\) to isolate \(\frac{dy}{dx}\). Divide both sides by \(y^{2}\) (assuming \(y \neq 0\)) to get \(\frac{dy}{dx} = 3x^{2} y - \frac{6x^{2}}{y^{2}}\).
Recognize that the equation is separable or can be manipulated into a separable form. Try to express all terms involving \(y\) on one side and all terms involving \(x\) on the other side.
Rewrite the equation as \(\frac{dy}{dx} = 3x^{2} y - 6x^{2} y^{-2}\). This suggests grouping terms to separate variables: \(\frac{dy}{dx} = x^{2} (3y - 6 y^{-2})\).
Express the differential equation in differential form: \(dy = x^{2} (3y - 6 y^{-2}) dx\). Then rearrange to isolate \(y\) terms with \(dy\) and \(x\) terms with \(dx\): \(\frac{dy}{3y - 6 y^{-2}} = x^{2} dx\).
Integrate both sides: integrate \(\int \frac{dy}{3y - 6 y^{-2}}\) with respect to \(y\) and \(\int x^{2} dx\) with respect to \(x\). After integration, include the constant of integration and solve for \(y\) if possible.

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주요 개념

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A separable differential equation can be written as a product of a function of y and a function of x, allowing the variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the solution.
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