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Chapter 8: Problem 22

Solution of the differential equation \(\frac{d y}{d x}+\frac{1+y^{2}}{\sqrt{1-x^{2}}}=0\) is (1) \(\tan ^{-1} y+\sin ^{-1} x=c\) (2) \(\tan ^{-1} x+\sin ^{-1} y=c\) (3) \(\tan ^{-1} y \cdot \sin ^{-1} x=c\) (4) \(\tan ^{-1} \mathrm{y}-\sin ^{-1} x=c\)

Short Answer

Expert verified
The solution is (1) \( \tan^{-1}(y) + \sin^{-1}(x) = C \).

Step by step solution

01

Identify the differential equation

The given differential equation is: \( \frac{d y}{d x} + \frac{1 + y^{2}}{\sqrt{1 - x^{2}}} = 0 \).
02

Rewrite the equation

Rewrite the differential equation in the form where \( \frac{d y}{d x} \) is isolated: \( \frac{d y}{d x} = -\frac{1 + y^{2}}{\sqrt{1 - x^{2}}} \).
03

Recognize the standard form

This is a separable differential equation as it can be written in the form \( g(y) dy = f(x) dx \).
04

Separate the variables

Separate the variables by rewriting the equation: \( \frac{1}{1 + y^{2}} \, dy = -\frac{1}{\sqrt{1 - x^{2}}} \, dx \).
05

Integrate both sides

Integrate both sides: \( \int \frac{1}{1 + y^{2}} \, dy = \int -\frac{1}{\sqrt{1 - x^{2}}} \, dx \). The left side integrates to \( \tan^{-1}(y) \) and the right side integrates to \( -\sin^{-1}(x) \).
06

Solve the integrals

Solving these integrals, we get: \( \tan^{-1}(y) + \sin^{-1}(x) = C \), where \( C \) is the constant of integration.
07

Write the general solution

The general solution of the differential equation is: \( \tan^{-1}(y) + \sin^{-1}(x) = C \).
08

Match with given options

Compare the obtained solution with the given options. The correct option is (1) \( \tan^{-1}(y) + \sin^{-1}(x) = C \).

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Key Concepts

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

Integration Techniques
When solving separable differential equations, integration is a vital step. This involves finding the antiderivatives of both sides of the equation. In the given exercise, we separate the differential equation into \(\frac{1}{1 + y^{2}} \, dy = -\frac{1}{\text{鈭殅(1 - x^{2})} \, dx\).

Next, we integrate both sides separately. The integral of \(\frac{1}{1 + y^{2}} \, dy\) is \(\tan^{-1}(y)\), and the integral of \(-\frac{1}{\text{鈭殅(1 - x^{2})} \, dx\) is \(-\text{sin}^{-1}(x)\).

Integrals are powerful tools. Here are some tips for handling them:
  • Recognize common integrals like \(\tan^{-1}(y)\) and \(\text{sin}^{-1}(x)\).
  • When faced with more complex integrals, consider substitution methods.
  • Always include the constant of integration, represented by \(C\).
Using these techniques correctly transforms differential equations into solvable forms.
Initial Conditions in Differential Equations
Initial conditions specify the value of a solution at a particular point. These can be used to determine the constant of integration \(C\). In this exercise, we derived the general solution as \(\tan^{-1}(y) + \text{sin}^{-1}(x) = C\).

Without initial conditions, we leave \(C\) as an arbitrary constant. But if we know that, for example, \(\tan^{-1}(y_0) + \text{sin}^{-1}(x_0) = 0\) at some initial point \(x_0, y_0\):
  • Plug \(x_0, y_0\) into the general solution.
  • Solve for \(C\).
Managing initial conditions effectively allows us to find precise solutions to differential equations, fitting them to real-world data.
Trigonometric Identities
Understanding trigonometric identities is crucial to solve and simplify differential equations involving trigonometric functions. In the exercise, we identified integrals that led to \( \tan^{-1}(y) + \text{sin}^{-1}(x) = C\).

Several useful trigonometric identities include:
  • The Pythagorean identities: \( \text{sin}^2(x) + \text{cos}^2(x) = 1 \).
  • Inverse trigonometric functions: \( \tan^{-1}(y) , \text{sin}^{-1}(x) \).
  • Angle sum and difference identities: \( \text{sin}(a \text{卤} b) \).
Through using trigonometric identities, we can simplify complex trigonometric expressions, making them easier to integrate or differentiate. These identities allow a deeper understanding and flexibility in manipulating equations involving trigonometric functions.

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Most popular questions from this chapter

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