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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
9장, 문제 9.PE.17

In Exercises 1–22, solve the differential equation.


y' = sin³ x cos² y

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1
Recognize that the given differential equation is separable since it can be written as a product of a function of x and a function of y: \(y' = \sin^{3} x \cos^{2} y\).
Rewrite the differential equation in differential form: \(\frac{dy}{dx} = \sin^{3} x \cos^{2} y\).
Separate the variables by dividing both sides by \(\cos^{2} y\) and multiplying both sides by \(dx\): \(\frac{dy}{\cos^{2} y} = \sin^{3} x \, dx\).
Express the left side in terms of \(\sec^{2} y\) to facilitate integration: \(\sec^{2} y \, dy = \sin^{3} x \, dx\).
Integrate both sides separately: \(\int \sec^{2} y \, dy = \int \sin^{3} x \, dx\), then solve each integral using appropriate techniques (such as substitution or trigonometric identities).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Separable Differential Equations

A separable differential equation can be written as a product of a function of x and a function of y, allowing variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the solution.
추천 영상:
06:06
Solving Separable Differential Equations

Integration of Trigonometric Functions

Solving the equation involves integrating powers of sine and cosine functions. Familiarity with trigonometric identities and integration techniques, such as substitution or power-reduction formulas, is essential to evaluate these integrals correctly.
추천 영상:
6:04
Introduction to Trigonometric Functions

Implicit Solutions and Initial Conditions

After integration, solutions may be implicit, involving both x and y in an equation. Understanding how to interpret implicit solutions and apply initial conditions, if given, helps in finding explicit solutions or particular solutions to the differential equation.
추천 영상:
05:03
Initial Value Problems