Solve the initial value problems in Exercises 115–120.
117. dy/dx = 1/(x√(x² - 1)), x > 1; y(2) = π

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Identify the given differential equation: \(\frac{dy}{dx} = \frac{1}{x \sqrt{x^{2} - 1}}\) with the initial condition \(y(2) = \pi\) and domain \(x > 1\).

Rewrite the differential equation in differential form: \(dy = \frac{1}{x \sqrt{x^{2} - 1}} \, dx\).

Integrate both sides with respect to \(x\): \(y = \int \frac{1}{x \sqrt{x^{2} - 1}} \, dx + C\), where \(C\) is the constant of integration.

To solve the integral \(\int \frac{1}{x \sqrt{x^{2} - 1}} \, dx\), consider using a trigonometric substitution such as \(x = \sec \theta\), which simplifies the square root expression.

After finding the antiderivative, apply the initial condition \(y(2) = \pi\) to solve for the constant \(C\), then write the explicit solution for \(y\).

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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 dy/dx = g(x)h(y), allowing the variables y and x to be separated on opposite sides of the equation. This enables integration with respect to each variable independently, facilitating the solution of the differential equation.

Integration Techniques Involving Inverse Trigonometric Functions

Integrals involving expressions like 1/(x√(x² - 1)) often lead to inverse trigonometric functions such as arcsec or arccos. Recognizing these forms and applying appropriate substitution or standard integral formulas is essential to solve the integral correctly.

Initial Value Problems (IVP)

An initial value problem specifies a differential equation along with a condition y(x₀) = y₀. Solving an IVP involves finding the general solution and then using the initial condition to determine the particular constant, yielding a unique solution curve.

Solve the initial value problems in Exercises 115–120.117. dy/dx = 1/(x√(x² - 1)), x > 1; y(2) = π