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Chapter 4: Problem 51

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

Short Answer

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

Step by step solution

01

Identify the Equation Type

The given differential equation is \( \frac{d y}{d x} + \frac{1+y^2}{\sqrt{1-x^2}} = 0 \). It is a first-order differential equation because it involves \( \frac{d y}{d x} \).
02

Rearrange the Equation

Rearrange the equation by isolating \( \frac{d y}{d x} \): \( \frac{d y}{d x} = -\frac{1+y^2}{\sqrt{1-x^2}} \).
03

Integrate Both Sides

Rewrite the equation as \( d y = -\frac{1+y^2}{\sqrt{1-x^2}} d x \). Integrate both sides: \( \int 1 d y = \int -\frac{1+y^2}{\sqrt{1-x^2}} d x \).
04

Integrate the Left Side

The integral of \( 1 d y \) is \( y \).
05

Change Variable for the Right Side

Set \( y = \tan z \), then \( dy = \sec^2 z dz \). Substitute and integrate to get \( \int \sec^2 z dz = \int -\frac{1+ an^2 z}{\sqrt{1-x^2}} d x \).
06

Simplify Using Trigonometric Identities

Since \( 1 + \tan^2 z = \sec^2 z \), it simplifies to \( \int \sec^2 z dz = \int -\sec^2 z d x/\sqrt{1-x^2} \).
07

Solve the Integral

Realize \( \int -\sec^2 z dz = -\tan^{-1} y \) and solve \( \int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x \).
08

Combine the Results

Combining the results of the integration \( \tan^{-1} y = -\sin^{-1} x + C \) or equivalently \( \tan^{-1} y + \sin^{-1} x = C \).
09

Choose the Correct Option

Comparing with the given options, the solution matches with option (a): \( \tan^{-1} y + \sin^{-1} x = C \).

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Key Concepts

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

Integration Techniques
Integration is a fundamental operation in calculus, and it involves finding the antiderivative of a function. In this context, solving first-order differential equations often requires integrating both sides of an equation. Here are some key techniques that are useful:
  • Direct Integration: This involves integrating both sides of an equation without changing variables. It's suitable for simpler forms where the integrand is easy to work with.

  • Substitution: When direct integration is challenging, substitution can help by simplifying the equation. By introducing a new variable, like letting y = tan z, the equation becomes more manageable.

  • Partial Fractions: Breaking down a complex fraction into simpler parts can make integration possible when dealing with rational expressions.
These techniques often go hand-in-hand, and choosing the right one depends on the structure of the equation at hand. Mastering them is crucial for solving differential equations effectively.
First-Order Differential Equations
First-order differential equations are equations involving the first derivative of a function. Recognizing the type of differential equation is key to deciding the most appropriate method for solving it. The equation given in the exercise is an example of a first-order differential equation: \( \frac{d y}{d x} + \frac{1+y^2}{\sqrt{1-x^2}} = 0 \).
Here are some common characteristics:
  • They involve only the first derivative \( \frac{d y}{d x} \).

  • Such equations often require rearranging before integration, which is a typical step in solving them.
Once we identify it as a first-order differential equation, we can integrate both sides to find the solution. First-order differential equations are foundational in calculus and applied mathematics, providing models for various real-world problems such as growth rates, decay processes, and more.
Trigonometric Substitution
Trigonometric substitution is a powerful method to simplify the integration of functions involving radicals. In calculus, this technique is useful when solving integrals that involve square roots, like \( \sqrt{1 - x^2} \). Here is how it's typically applied:
  • Purpose: Transform a variable in an integral using a trigonometric identity, so the resulting integral is easier to solve.

  • Example: If you have \( \sqrt{1 - x^2} \), you might substitute \( x = \sin \theta \), thus simplifying the expression using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \).

  • Application: After substituting, integrate with respect to the new variable. Finally, reverse the substitution to express the result in terms of the original variable.
By understanding and leveraging trigonometric identities, substitution facilitates the solution of integrals that are otherwise difficult to solve directly. It's especially useful in this exercise where restricting the scope of the integration to known identities yields a straightforward solution.

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Most popular questions from this chapter

Let \(f:(0, \infty) \rightarrow R\) be a differentiable function such that \(f^{\prime}(x)=2-\frac{f(x)}{x}\) for all \(x \in(0, \infty)\) and \(f(1) \neq 1\). Then (a) \(\lim _{x \rightarrow 0+} f\left(\frac{1}{x}\right)=1\) (b) \(\lim _{x \rightarrow 0+} x f\left(\frac{1}{x}\right)=2\) (c) \(\lim _{x \rightarrow 0+} x^{2} f^{\prime}(x)=0\) (d) \(|f(x)| \leq 2\) for all \(x \in(0,2)\)

The hemispherical tank of radius \(2 \mathrm{~m}\) is initially full of water and has an outlet of \(12 \mathrm{~cm}^{2}\) cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law \(v(t)=0.6 \sqrt{2 g h(t)}\), where \(v(t)\) and \(h(t)\) are respectively velocity of the flow through the outlet and the height of water level above the outlet at time \(t\) and \(g\) is the acceleration due to gravity. Find the time it takes to empty the tank.

If \(\frac{d y}{d x}=y+3>0\) and \(y(0)=2\), then \(y(\log 2)\) is equal to (a) 5 (b) 13 (c) \(-2\) (d) 7

Let \(y(x)\) be a solution of the differential equation \(\left(1+e^{x}\right) y^{\prime}+y e^{x}=1 .\) If \(y(0)=2\), then which of the following statement(s) is/are true? (a) \(y(-4)=0\) (b) \(y(-2)=0\) (c) \(y(x)\) has a critical point in the interval \((-1,0)\) (d) \(y(x)\) has no critical point in the interval \((-1,0)\)

Which of the following pair(s) is/are orthogonal? (a) \(16 x^{2}+y^{2}=C\) and \(y^{16}=k x\) (b) \(y=x+C e^{-x}\) and \(x+2=y+k e^{-y}\) (c) \(y=C x^{2}\) and \(x^{2}+2 y^{2}=k\) (d) \(x^{2}-y^{2}=C\) and \(x y=k\)
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