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Chapter 8: Problem 9

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\). $$\frac{d y}{d x}-\frac{y^{2}-y}{\sin x}=0$$

Short Answer

Expert verified
The solution is \( y = \frac{k}{|\csc x + \cot x| + k} \), where \( k \) is a constant.

Step by step solution

01

Rearrange Terms

Start by rewriting the given differential equation \( \frac{dy}{dx} - \frac{y^2 - y}{\sin x} = 0 \). Move the term \( \frac{y^2 - y}{\sin x} \) to the right side to get: \( \frac{dy}{dx} = \frac{y^2 - y}{\sin x} \).
02

Separate Variables

Rewrite the equation in a form that separates variables: \( \frac{1}{y^2 - y} dy = \frac{1}{\sin x} dx \). This separates the \( y \) terms on one side and the \( x \) terms on the other.
03

Integrate Both Sides

Integrate both sides of the equation. For the left side, integrate \( \int \frac{1}{y^2 - y} dy \) and for the right side \( \int \frac{1}{\sin x} dx \).
04

Integrate the Left Side

Factor the denominator on the left: \( y^2 - y = y(y - 1) \). Use partial fraction decomposition: \( \frac{1}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1} \). Solve for \( A \) and \( B \) to get \( \frac{1}{y} - \frac{1}{y-1} \). Integrate both terms to get \( \ln|y| - \ln|y-1| = \ln\left|\frac{y}{y-1}\right| \).
05

Integrate the Right Side

The integral of \( \frac{1}{\sin x} \) is \( \int \csc x \, dx \), which equals \( -\ln|\csc x + \cot x| \) or equivalently \( \ln|\tan(x/2)| \) by using trigonometric identities.
06

Solve for Constant of Integration

Combine results and introduce a constant \( C \). The equation becomes \( \ln\left|\frac{y}{y-1}\right| = -\ln|\csc x + \cot x| + C \). Simplify: \( \left|\frac{y}{y-1}\right| = \frac{k}{|\csc x + \cot x|} \) where \( k = e^C \).
07

Express Solution Explicitly

Solve for \( y \): Multiply through by \( y - 1 \) and simplify: \( y = \frac{k}{|\csc x + \cot x| + k} \). This is the general solution expressed as an explicit function of \( x \).

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Key Concepts

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

Separation of Variables
Separation of variables is a powerful method used to solve certain types of differential equations. The main idea is to rewrite the equation so each side contains one variable. This allows us to integrate both sides separately. The key steps for separation of variables include:
  • Rearranging the equation to isolate terms involving the same variable on one side.
  • Setting up the expression such that each side of the equation only contains one variable, either plain or in a differential form.
  • Integrating both sides to find a general solution.
In our problem, we started with a differential equation: \( \frac{dy}{dx} = \frac{y^2 - y}{\sin x} \). We divided through to separate the variables, yielding \( \frac{1}{y^2 - y} dy = \frac{1}{\sin x} dx \). This separation is crucial as it allows straightforward integration later. Once separated, solving becomes a matter of integrating both sides individually.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions, which are easier to integrate or manipulate. Here's how it works:
  • Factor the denominator of the expression into simpler terms if possible.
  • Express the fraction as a sum of simpler fractions with unknown coefficients.
  • Solve for these coefficients by clearing the denominators and equating coefficients on both sides.
In the given solution, we decomposed \( \frac{1}{y^2 - y} \) by recognizing that \( y^2 - y = y(y - 1) \). This led to the expression \( \frac{1}{y} - \frac{1}{y-1} \). Decomposing complicated equations into such expressions helps in simplifying the integration process, making the mathematical work more manageable.
Trigonometric Integration
Trigonometric integration involves integrating functions that contain trigonometric expressions. Many integrals involving trigonometric functions require the use of identities or specific integral formulas.For our problem, we needed to integrate \( \frac{1}{\sin x} \) or \( \csc x \). Knowing that:
  • The integral of \( \csc x \) is \( \ln|\tan(x/2)| \), following from trigonometric identities.
  • Using properties of logarithms, these integrals can then be manipulated into more suitable forms, depending on the context.
These types of integrals are common in problems involving differential equations because of the periodic nature of trigonometric functions. When computing these integrals, keeping a list of identities and standard integrals can be especially helpful.
Integrating Factors
Integrating factors are typically used to solve linear first-order differential equations, but they are not specifically used in our problem. However, understanding them can be vital for broader studies. Key points to remember include:
  • An integrating factor is a function, usually in terms of the independent variable, which when multiplied by the entire differential equation, makes it easier to solve.
  • This technique enables turning a non-exact equation into an exact one, facilitating integration.
Although not directly applied in our current exercise, integrating factors provide a powerful tool for tackling linear differential equations where separation of variables may not be applicable. Recognizing when to use integrating factors can greatly expand your capability in solving a vast array of differential equations.

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Most popular questions from this chapter

Solve the initial-value problem. $$\frac{d y}{d x}-2 x y=2 x, \quad y(0)=3$$

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\). $$e^{-y} \sin x-y^{\prime} \cos ^{2} x=0$$

Use Euler's Method with the given step size \(\Delta x\) or \(\Delta t\) to approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph. Consider the initial-value problem $$ y^{\prime}=\sin \pi t, \quad y(0)=0 $$ Use Euler's Method with five steps to approximate \(y(1)\)

A glass of lemonade with a temperature of \(40^{\circ} \mathrm{F}\) is placed in a room with a constant temperature of \(70^{\circ} \mathrm{F}\), and 1 hour later its temperature is \(52^{\circ} \mathrm{F}\). Show that \(t\) hours after the lemonade is placed in the room its temperature is approximated by \(T=70-30 e^{-0.5 t}.\)

A rocket, fired upward from rest at time \(t=0,\) has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k\), the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t\) It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation $$m \frac{d v}{d t}=c k-m g$$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t.\) (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. \([\) Note: Take \(\left.g=9.8 \mathrm{m} / \mathrm{s}^{2} \text { and } c=2500 \mathrm{m} / \mathrm{s} .\right]\)
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