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 \(\frac{1}{1 + y^{2}} \, dy = -\frac{1}{\text{√}(1 - x^{2})} \, dx\).
Next, we integrate both sides separately. The integral of \(\frac{1}{1 + y^{2}} \, dy\) is \(\tan^{-1}(y)\), and the integral of \(-\frac{1}{\text{√}(1 - x^{2})} \, dx\) is \(-\text{sin}^{-1}(x)\).
Integrals are powerful tools. Here are some tips for handling them:
Next, we integrate both sides separately. The integral of \(\frac{1}{1 + y^{2}} \, dy\) is \(\tan^{-1}(y)\), and the integral of \(-\frac{1}{\text{√}(1 - x^{2})} \, dx\) 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 \(C\).
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 \(x_0, y_0\):
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 \(C\).
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:
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: \( \text{sin}(a \text{±} b) \).