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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.4.13

Solve the differential equation in Exercises 9–22.
13. (dy/dx) = √y cos²√y

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1
Recognize that the given differential equation is \( \frac{dy}{dx} = \sqrt{y} \cos^{2}(\sqrt{y}) \). Our goal is to separate variables to integrate both sides.
Make the substitution \( t = \sqrt{y} \), which implies \( y = t^{2} \) and \( \frac{dy}{dx} = 2t \frac{dt}{dx} \). Rewrite the differential equation in terms of \( t \): \( 2t \frac{dt}{dx} = t \cos^{2}(t) \).
Simplify the equation by dividing both sides by \( t \) (assuming \( t \neq 0 \)): \( 2 \frac{dt}{dx} = \cos^{2}(t) \). Then, solve for \( \frac{dt}{dx} \): \( \frac{dt}{dx} = \frac{1}{2} \cos^{2}(t) \).
Separate variables to isolate \( t \) and \( x \): \( \frac{dt}{\cos^{2}(t)} = \frac{1}{2} dx \). Recall that \( \frac{1}{\cos^{2}(t)} = \sec^{2}(t) \), so the equation becomes \( \sec^{2}(t) dt = \frac{1}{2} dx \).
Integrate both sides: \( \int \sec^{2}(t) dt = \int \frac{1}{2} dx \). Use the integral formula \( \int \sec^{2}(t) dt = \tan(t) + C \) to find the implicit solution, then substitute back \( t = \sqrt{y} \) to express the solution in terms of \( y \) and \( x \).

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주요 개념

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

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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.
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Solving the equation requires integrating functions involving square roots and trigonometric expressions. Familiarity with substitution methods and standard integral formulas is essential to handle integrals like ∫(1/√y cos²√y) dy or ∫ cos²(u) du.
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