Solve the initial value problems in Exercises 53–56 for y as a function of x.
(x² + 1)² (dy/dx) = √(x² + 1), where y(0) = 1

검증된 단계별 안내

Rewrite the given differential equation to isolate \( \frac{dy}{dx} \): \( (x^{2} + 1)^{2} \frac{dy}{dx} = \sqrt{x^{2} + 1} \) becomes \( \frac{dy}{dx} = \frac{\sqrt{x^{2} + 1}}{(x^{2} + 1)^{2}} \).

Simplify the right-hand side by expressing the square root as a power: \( \sqrt{x^{2} + 1} = (x^{2} + 1)^{\frac{1}{2}} \), so \( \frac{dy}{dx} = (x^{2} + 1)^{\frac{1}{2} - 2} = (x^{2} + 1)^{-\frac{3}{2}} \).

Set up the integral to find \( y \) by separating variables: \( dy = (x^{2} + 1)^{-\frac{3}{2}} dx \), then integrate both sides: \( y = \int (x^{2} + 1)^{-\frac{3}{2}} dx + C \).

Use an appropriate substitution or recognize the integral form to evaluate \( \int (x^{2} + 1)^{-\frac{3}{2}} dx \). For example, consider a trigonometric substitution such as \( x = \tan \theta \) or use a standard integral formula.

Apply the initial condition \( y(0) = 1 \) to solve for the constant of integration \( C \) after finding the antiderivative expression for \( y \).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

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10m

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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 the variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the solution.

Initial Value Problem (IVP)

An initial value problem specifies the value of the unknown function at a particular point, providing a condition to determine the unique solution of a differential equation. This ensures the solution curve passes through the given initial point.

Integration Techniques for Functions Involving Radicals

Solving the given differential equation requires integrating expressions with square roots and polynomial terms. Familiarity with substitution methods and recognizing integrable forms involving radicals is essential to find the antiderivative.

Solve the initial value problems in Exercises 53–56 for y as a function of x.(x² + 1)² (dy/dx) = √(x² + 1), where y(0) = 1