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Chapter 19: Problem 32

Find the particular solution of the differential equation that satisfies the given condition. \(x d y-\sqrt{1-y^{2}} d x=0 ; \quad y=\frac{1}{2}\) when \(x=1\)

Short Answer

Expert verified
The particular solution is \( \sin^{-1}(y) = \ln|x| + \frac{\pi}{6} \).

Step by step solution

01

Identify the Differential Equation

The given differential equation is expressed as \( x \, dy - \sqrt{1-y^2} \, dx = 0 \), which we need to manipulate and solve.
02

Separate Variables

Rewrite the equation to separate variables: \( x \, dy = \sqrt{1-y^2} \, dx \). Now, divide both sides by \(\sqrt{1-y^2} \cdot x\) to get \( \frac{dy}{\sqrt{1-y^2}} = \frac{dx}{x} \).
03

Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{dy}{\sqrt{1-y^2}} = \int \frac{dx}{x} \). The left side integrates to \( \sin^{-1}(y) \) and the right side to \( \ln|x| + C \).
04

Solve for the Integration Constant

The general solution is \( \sin^{-1}(y) = \ln|x| + C \). Use the initial condition \( y=\frac{1}{2} \) when \( x=1 \) to find \( C \): \( \sin^{-1}\left(\frac{1}{2}\right)=\ln(1)+C \). Since \( \sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \), it follows that \( C=\frac{\pi}{6} \).
05

Write the Particular Solution

Substitute \( C=\frac{\pi}{6} \) back into the general solution: \( \sin^{-1}(y) = \ln|x| + \frac{\pi}{6} \). This is the particular solution satisfying the given condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation
A differential equation is an equation that relates a function with its derivatives. It plays a crucial role in describing various phenomena such as motion, growth, and decay processes. In the context of our problem, the given equation is \( x \, dy - \sqrt{1-y^2} \, dx = 0 \). This is an example of a first-order differential equation because it involves the first derivatives of the unknown function \( y \). Understanding differential equations is fundamental as they often model the behavior of real-world systems.
Separation of Variables
Separation of variables is a technique used to solve differential equations by isolating variables on different sides of the equation. In our problem, we start with the equation \( x \, dy = \sqrt{1-y^2} \, dx \). The goal is to rearrange the terms so that each variable and its differential are on opposite sides. By dividing each side by \( \sqrt{1-y^2} \cdot x \), we get \( \frac{dy}{\sqrt{1-y^2}} = \frac{dx}{x} \). This separation allows us to integrate each side with respect to its variable, paving the way to find the solution.
Initial Condition
An initial condition gives a specific value of the function at a particular point, which helps to find a particular solution from a general solution of a differential equation. In our exercise, the initial condition provided is \( y=\frac{1}{2} \) when \( x=1 \). By applying this condition, we are able to solve for the integration constant, ensuring that our solution satisfies the original problem itself. Initial conditions are essential to find the exact equation that models the specific scenario under consideration.
Integration Constant
The integration constant \( C \) arises when you integrate functions to solve a differential equation. Integration generally leads to an infinite number of possible functions, hence we call them the general solutions. To find a particular solution, specific conditions such as initial conditions are used to determine the value of \( C \). In the example provided, after integrating, our equation was \( \sin^{-1}(y) = \ln|x| + C \). By using the initial condition, \( y = \frac{1}{2} \) when \( x = 1 \), we calculated that \( C = \frac{\pi}{6} \), thus giving us a unique solution that fits the particular case.

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Most popular questions from this chapter

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