/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Solve the differential equation ... [FREE SOLUTION] | 91Ó°ÊÓ

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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}{x}

Short Answer

Expert verified
The solution is the family of functions \( y = Cx \).

Step by step solution

01

Rewrite the Differential Equation

Start with the given differential equation: \( \frac{dy}{dx} = \frac{y}{x} \). The aim is to separate the variables, \( y \) and \( x \), to opposite sides. We can do this by multiplying both sides by \( dx \) and then dividing both sides by \( y \), resulting in \( \frac{dy}{y} = \frac{dx}{x} \).
02

Integrate Both Sides

Now that the variables are separated, we can integrate both sides. Integrating the left side with respect to \( y \) gives \( \int \frac{1}{y} \, dy \), and integrating the right side with respect to \( x \) gives \( \int \frac{1}{x} \, dx \). This results in \( \ln |y| = \ln |x| + C \), where \( C \) is the constant of integration.
03

Solve for y

To express \( y \) as an explicit function of \( x \), exponentiate both sides of the equation to eliminate the natural logarithm. This yields \( |y| = e^{\ln |x| + C} = e^{\ln |x|} \cdot e^C = |x| \cdot e^C \). Since \( e^C \) is a constant, it can be denoted by \( C_1 = e^C \) (arbitrary positive constant). So, the solution can be written as \( y = C_1 \cdot x \) or \( y = \pm C_1 \cdot x \).
04

Express the General Solution

Finally, to express the family of solutions explicitly, we conclude that \( y = Cx \), where \( C \) is any constant. This general form shows the relationship between \( y \) and \( x \) for any constant \( C \).

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Key Concepts

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

Understanding Separation of Variables
The method of separation of variables is a fundamental approach used to solve first-order differential equations. It involves rearranging the given differential equation so that each type of variable appears on a different side of the equation.

In the example provided, the differential equation is \( \frac{dy}{dx} = \frac{y}{x} \). We begin by multiplying both sides by \( dx \) to transition from a differential equation to a form where integration is possible. This gives us \( dy = \frac{y}{x} \, dx \).

To separate variables, we divide both sides by \( y \), leading to \( \frac{dy}{y} = \frac{dx}{x} \). Now, the equation is in a form where the left side contains only \( y \) and the right side contains only \( x \).

By separating the variables, we simplify the process of solving the differential equation by making each side independent, facilitating easier integration.
The Role of Integration in Solving Equations
Integration is a critical step when solving a differential equation after separating variables. Once the equation \( \frac{dy}{y} = \frac{dx}{x} \) is achieved, the next step is to integrate both sides to find the solutions.

The left side, \( \int \frac{1}{y} \, dy \), leads to \( \ln |y| \), while the right side, \( \int \frac{1}{x} \, dx \), results in \( \ln |x| \), plus a constant of integration \( C \).

Integration plays the role of reversing differentiation, allowing us to move from a rate of change to a function. It effectively unwinds the differential equation to reveal relationships between the variables.

Once the integration is complete, the constants like \( C \) account for the indefinite nature of integration, signifying that there are infinite possible solutions differing by a constant.
Demystifying the Natural Logarithm
The natural logarithm, denoted by \( \ln \), is essential in solving differential equations involving separation of variables.

In this context, after integrating \( \frac{1}{y} \) and \( \frac{1}{x} \), we obtain \( \ln |y| \) and \( \ln |x| \), respectively. The properties of logarithms allow these expressions to be manipulated algebraically.

To clear the logarithms, we exponentiate both sides, utilizing the exponential function as the inverse of the natural logarithm. This transformation changes \( \ln |y| = \ln |x| + C \) into \( |y| = e^{C} \cdot |x| \).

The beauty of using logarithms here is that they simplify integrations of reciprocal functions and align nicely with exponentials for solving differential equations. Ultimately, this helps express the relationship between \( y \) and \( x \) in a clear, multiplicative manner.
Expressing Explicit Functions from Solutions
Transforming an implicit solution into an explicit expression is often the final step in solving differential equations using separation of variables.

After obtaining \( |y| = C_1 \cdot |x| \) from exponentiating the integrated equation, we express \( y \) as an explicit function of \( x \): \( y = C \cdot x \), where \( C \) encompasses \( \pm C_1 \).

An explicit function is one where the dependent variable, \( y \), is isolated on one side of the equation, written directly in terms of the independent variable, \( x \). This form is advantageous as it makes the interpretation of the relation between \( y \) and \( x \) more direct and accessible.

Such clarity in functions allows for easier graphing, solution interpretation, and analysis, enhancing the understanding of how changes in \( x \) affect \( y \). Thus, expressing solutions explicitly bridges the gap from theoretical solutions to practical understanding in mathematics.

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Most popular questions from this chapter

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. \([\)Take \(g=9.8\) \(\left.\mathrm{m} / \mathrm{s}^{2} \text { and } c=2500 \mathrm{m} / \mathrm{s} .\right]\)

Consider the initial-value problem $$ y^{\prime}=\cos 2 \pi t, \quad y(0)=1 $$ Use Euler's Method with five steps to approximate \(y(1)\)

Solve the differential equation by the method of integrating factors. \(y^{\prime}+y=\cos \left(e^{x}\right)\)

Suppose that the growth of a population \(y=y(t)\) is given by the logistic equation $$ y=\frac{60}{5+7 e^{-t}} $$ (a) What is the population at time \(t=0 ?\) (b) What is the carrying capacity \(L ?\) (c) What is the constant \(k ?\) (d) When does the population reach half of the carrying capacity? (c) Find an initial-value problem whose solution is \(y(t)\)

Solve the initial-value problem by any method. y^{\prime}-x e^{y}=2 e^{y}, \quad y(0)=0

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