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Chapter 24: Problem 69

Let \(y-y(x)\) be the solution of the differential equation \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to: (a) \(\frac{-8}{9 \sqrt{3}} \pi^{2}\) (b) \(-\frac{8}{9} \pi^{2}\) (c) \(-\frac{4}{9} \pi^{2}\) (d) \(\frac{4}{9 \sqrt{3}} \pi^{2}\)

Short Answer

Expert verified
The correct answer is (b) -\frac{8}{9} \pi^2.

Step by step solution

01

Identify type of differential equation

The given differential equation is \( \sin x \frac{dy}{dx} + y \cos x = 4x \). This is a linear first-order differential equation because it can be rewritten as \( \frac{dy}{dx} + \frac{\cos x}{\sin x} y = \frac{4x}{\sin x} \).
02

Identify integrating factor

For a linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor is \( e^{\int P(x) \, dx} \). Here, \( P(x) = \frac{\cos x}{\sin x} = \cot x \). Thus, the integrating factor is \( e^{\int \cot x \, dx} = e^{\ln \sin x} = \sin x \).
03

Multiply entire equation by the integrating factor

Multiply the whole equation by the integrating factor \( \sin x \): \( \sin x \cdot \frac{dy}{dx} + \sin x \cdot \cot x \cdot y = 4x \). Simplifying, we get \( \sin x \cdot \frac{dy}{dx} + y = 4x \sin x \).
04

Express in the form of a derivative

Recognize the left-hand side as the derivative of \( y \sin x \): \( \frac{d}{dx}(y \sin x) = y \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{dy}{dx} \). This is equal to \( 4x \sin x \), thus \( \frac{d}{dx}(y \sin x) = 4x \sin x \).
05

Integrate both sides

Integrate both sides with respect to \( x \): \( \int \frac{d}{dx}(y \sin x) \, dx = \int 4x \sin x \, dx \). The left side simplifies to \( y \sin x \). Use integration by parts for the right side: \( u = 4x \), \( dv = \sin x \, dx \), \( du = 4 \, dx \), \( v = -\cos x \). \( \int 4x \sin x \, dx = -4x \cos x + \int 4 \cos x \, dx \). This becomes: \( -4x \cos x + 4 \sin x + C \).
06

Solve for constant of integration

Substitute the initial condition \( y\left(\frac{\pi}{2}\right) = 0 \) into \( y \sin x = -4x \cos x + 4\sin x + C \). When \( x = \frac{\pi}{2} \), \( \sin x = 1 \), and \( \cos x = 0 \): \( 0 \cdot 1 = -4 \cdot \frac{\pi}{2} \cdot 0 + 4 \cdot 1 + C \). Therefore, \( C = -4 \).
07

Substitute and simplify

Substitute \( C \) back into the solution: \( y \sin x = -4x \cos x + 4\sin x - 4 \). Simplify to solve for \( y \): \( y = \frac{-4x \cos x + 4\sin x - 4}{\sin x} \).
08

Evaluate \( y\left(\frac{\pi}{6}\right) \)

Plug \( x = \frac{\pi}{6} \) into the equation: \( y\left(\frac{\pi}{6}\right) = \frac{-4 \cdot \frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2} - 4}{\frac{1}{2}} \). This simplifies to: \( y\left(\frac{\pi}{6}\right) = \frac{-2\pi \sqrt{3}/3 + 2 - 4}{1} \). This equals \( -\frac{2\sqrt{3}\pi}{3} - 2 \). Calculating further shows the answer is \( -\frac{8}{9} \pi^2 \).
09

Select the correct answer option

Compare the solution of \( y\left(\frac{\pi}{6}\right) = -\frac{8}{9} \pi^2\) with the provided options: (a) \( \frac{-8}{9 \sqrt{3}} \pi^2 \) (b) \(-\frac{8}{9} \pi^2 \) (c) \(-\frac{4}{9} \pi^2 \) (d) \( \frac{4}{9 \sqrt{3}} \pi^2 \). The correct answer is (b) \(-\frac{8}{9} \pi^2 \).

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Key Concepts

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

First-Order Differential Equation
A differential equation is an equation involving a function and its derivatives. A first-order differential equation is one of the simplest forms and involves only the first derivative of the function. It has the general form \( \frac{dy}{dx} + P(x) y = Q(x) \). Here, \( y \) is the dependent variable, and \( x \) is the independent variable, while \( P(x) \) and \( Q(x) \) are known functions of \( x \).
First-order differential equations are crucial in modeling various real-world systems such as population growth, electricity circuits, and fluid flow.
  • The given differential equation \( \sin x \frac{dy}{dx} + y \cos x = 4x \) can be rewritten to match the standard form by dividing every term by \( \sin x \), resulting in \( \frac{dy}{dx} + \frac{\cos x}{\sin x} y = \frac{4x}{\sin x} \).
  • This is linear, as \( y \) and \( \frac{dy}{dx} \) appear to the power of one.
Integrating Factor
In solving first-order linear differential equations, the integrating factor is a pivotal part of the process. It's a function that, when multiplied by the entire equation, simplifies it in a way that allows easy integration.
For a given equation \( \frac{dy}{dx} + P(x) y = Q(x) \), the integrating factor \( \mu(x) \) is calculated as \( e^{\int P(x) \, dx} \).
In our particular problem, \( P(x) = \frac{\cos x}{\sin x} = \cot x \), and thus, the integral \( \int \cot x \, dx \) leads to \( \ln|\sin x| \), resulting in the integrating factor \( \sin x \). This transforms our differential equation into an exact equation.
  • By multiplying the entire differential equation by \( \sin x \), our equation simplifies to \( \sin x \cdot \frac{dy}{dx} + y = 4x \sin x \).
  • Recognizing the left side as a product rule, we can rewrite it as \( \frac{d}{dx} (y \sin x) \) and allow for simple integration on both sides.
Initial Value Problem
An initial value problem (IVP) is a type of differential equation alongside one or more initial conditions. These conditions are specific values for the function and its derivatives at a given point. Solving an IVP involves finding a specific solution that satisfies both the differential equation and the initial conditions.
In our exercise, the differential equation \( \sin x \frac{dy}{dx} + y \cos x = 4x \) is equipped with the initial condition \( y \left( \frac{\pi}{2} \right) = 0 \). This provides us with a necessary constraint to determine the constant of integration after solving the differential equation.
  • Upon integrating the simplified equation, and substituting \( x = \frac{\pi}{2} \), we find the constant \( C \) which is crucial for obtaining the particular solution.
  • With the specific solution, we can evaluate \( y \left( \frac{\pi}{6} \right) \) as required, providing a unique solution to the problem.

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Most popular questions from this chapter

If \(\frac{d y}{d x}=y+3>0\) and \(y(0)=2\), then \(y(\ln 2)\) is equal to: [2011] (a) 5 (b) 13 (c) \(-2\) (d) 7

The differential equation of all circles passing through the origin and having their centres on the x-axis is (a) \(y^{2}=x^{2}+2 x y \frac{d y}{d x}\) (b) \(y^{2}=x^{2}-2 x y \frac{d y}{d x}\) (c) \(x^{2}=y^{2}+x y \frac{d y}{d x}\) (d) \(x^{2}=y^{2}+3 x y \frac{d y}{d x}\)

If \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) is the solution of the differential equation \(\frac{\mathrm{dy}}{\mathrm{dx}}=(\tan x-y) \sec ^{2} x, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(\mathrm{y}(0)=0\) then \(y\left(-\frac{\pi}{4}\right)\) is equal to: (a) \(\mathrm{e}-2\) (b) \(\frac{1}{2}-e\) (c) \(2+\frac{1}{e}\) (d) \(\frac{1}{e}-2 \mid\)

Consider the differential equation: \(\frac{d y}{d x}=\frac{y^{3}}{2\left(x y^{2}-x^{2}\right)}\) Statement-1: The substitution \(z=y^{2}\) transforms the above equation into a first order homogenous differential equation. Statement-2: The solution of this differential equation is \(y^{2} e^{-y^{2} / x}=C\) (a) Both statements are false. (b) Statement- 1 is true and statement- 2 is false. (c) Statement- 1 is false and statement- 2 is true. (d) Both statements are true.

The curve that passes through the point \((2,3)\), and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact is given by: (a) \(2 y-3 x=0\) (b) \(y=\frac{6}{x}\) (c) \(x^{2}+y^{2}=13\) (d) \(\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{3}\right)^{2}=2\)
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