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

Verified step by step guidance

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 \(y\).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.

Recommended video:

05:03

Initial Value Problems

Related Practice

Textbook Question

Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.

f(x) = x² − 2x, x ≤ 1

Textbook Question

Evaluate the integrals in Exercises 77–90.

79. ∫(from -1 to 0)6dt/√(3-2t-t²)

Textbook Question

Use l’Hôpital’s rule to find the limits in Exercises 7–52.

24. lim (x → π/2) (ln(csc x)) / (x - (π/2))²

views

Textbook Question

In Exercises 5–8, show that each function is a solution of the given initial value problem.

5. Differential Equation: 2y + y' = 4x + 2

Initial condition: y(-1) = e² - 2