Math Problem Statement

The query submitted by the user involves solving the differential equation \(\frac{dx}{\sqrt{x^2 - 1}} = dt\).

Solution

It seems like you've uploaded an image containing mathematical content. I will analyze it to provide an accurate interpretation and solution.

Give me a moment to extract the necessary information.The image contains the following mathematical expression:

dxx21=dt\frac{dx}{\sqrt{x^2 - 1}} = dt

To solve this differential equation, let's proceed with these steps:

1. Integration of both sides:

Integrating both sides with respect to xx and tt, respectively:

dxx21=dt\int \frac{dx}{\sqrt{x^2 - 1}} = \int dt

The right side integrates simply to t+Ct + C (where CC is the constant of integration). The left-hand side involves recognizing that the integral of 1x21\frac{1}{\sqrt{x^2 - 1}} is the inverse hyperbolic cosine function, cosh1(x)\cosh^{-1}(x).

2. Result:

Thus, the solution is:

cosh1(x)=t+C\cosh^{-1}(x) = t + C

Alternatively, you could express the solution as:

x=cosh(t+C)x = \cosh(t + C)

This is the general solution for the given equation.

Would you like further details or have any specific questions?

Here are 5 related questions to deepen your understanding:

  1. How is the inverse hyperbolic cosine function defined?
  2. What are the steps to solve integrals involving hyperbolic functions?
  3. Can the same solution method be applied for different forms like (\frac{1}{\sqrt{a

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differential Equations
Inverse Hyperbolic Functions

Formulas

\(\int \frac{dx}{\sqrt{x^2 - 1}} = \cosh^{-1}(x)\)
\(\cosh^{-1}(x) = t + C\)

Theorems

Integral of hyperbolic functions
Inverse hyperbolic functions

Suitable Grade Level

Grades 11-12 or University-level Calculus