Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 134

How do differential equations with constant coefficients differ from those with variable coefficients? Give an example for each type.

Short Answer

Expert verified
Provide an example for each type. The main difference between differential equations with constant coefficients and those with variable coefficients is the nature of the coefficient term. In differential equations with constant coefficients, the coefficient term remains the same throughout the equation, while in those with variable coefficients, it changes with respect to the independent variable. Example with constant coefficients: Consider the differential equation: \(\frac{dy}{dx} + 2y = 4\) The general solution to this first-order linear differential equation with constant coefficients is: \(y(x) = 2 + \frac{C}{e^{2x}}\) Example with variable coefficients: Consider the differential equation: \(\frac{dy}{dx} + \frac{1}{x}y = x^2\) The general solution to this first-order linear differential equation with variable coefficients is: \(y(x) = \frac{1}{4}x^2 + \frac{C}{x^2}\) In summary, the main differences between these types of differential equations are both the form of the coefficient term and the methods used to solve them.

Step by step solution

01

Introduction

A differential equation of the form \(\frac{dy}{dx} + a(x)y = f(x)\) is called a first-order linear differential equation. When the term \(a(x)\) is a constant, then the equation has constant coefficients. If the term \(a(x)\) depends on \(x\), then the equation has variable coefficients.
02

Example with constant coefficients

Let's consider a first-order linear differential equation with constant coefficients: $$\frac{dy}{dx} + 2y = 4$$ To solve this equation, we can use an integrating factor, which is given by \(e^{\int a(x)dx}\). In this case, \(a(x) = 2\), so the integrating factor is: $$e^{\int 2 dx} = e^{2x}$$ Now, we can multiply the equation by the integrating factor: $$e^{2x} \frac{dy}{dx} + 2e^{2x}y = 4e^{2x}$$ Observe that the left-hand side of the equation is the derivative of the product \(e^{2x}y\). Thus, we can write: $$\frac{d}{dx}(e^{2x}y) = 4e^{2x}$$ Now, integrate both sides with respect to \(x\): $$\int \frac{d}{dx}(e^{2x}y) dx = \int 4e^{2x} dx$$ $$e^{2x}y = 2e^{2x} + C$$ Lastly, divide both sides by \(e^{2x}\): $$y(x) = 2 + \frac{C}{e^{2x}}$$ This is the general solution to our first-order linear differential equation with constant coefficients.
03

Example with variable coefficients

Now let's consider a first-order linear differential equation with variable coefficients: $$\frac{dy}{dx} + \frac{1}{x}y = x^2$$ The integrating factor in this case is: $$e^{\int \frac{1}{x} dx} = e^{\ln x} = x$$ Multiplying the equation by the integrating factor, we get: $$x \frac{dy}{dx} + xy = x^3$$ The left-hand side is the derivative of the product \(x^2y\). Thus, we can write: $$\frac{d}{dx}(x^2y) = x^3$$ Now, integrate both sides with respect to \(x\): $$\int \frac{d}{dx}(x^2y) dx = \int x^3 dx$$ $$x^2y = \frac{1}{4}x^4 + C$$ Lastly, divide both sides by \(x^2\): $$y(x) = \frac{1}{4}x^2 + \frac{C}{x^2}$$ This is the general solution to our first-order linear differential equation with variable coefficients.
04

Conclusion

The main difference between differential equations with constant coefficients and those with variable coefficients is the nature of the coefficient term. The constant coefficient remains the same throughout the equation, while the variable coefficient changes with respect to the independent variable. The examples provided demonstrate this difference in form and solution methods.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A stainless steel spherical container, with \(k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used for storing chemicals undergoing exothermic reaction. The reaction provides a uniform heat flux of \(60 \mathrm{~kW} / \mathrm{m}^{2}\) to the container's inner surface. The container has an inner radius of \(50 \mathrm{~cm}\) and a wall thickness of \(5 \mathrm{~cm}\) and is situated in a surrounding with an ambient temperature of \(23^{\circ} \mathrm{C}\). The container's outer surface is subjected to convection heat transfer with a coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For safety reasons to prevent thermal burn to individuals working around the container, it is necessary to keep the container's outer surface temperature below \(50^{\circ} \mathrm{C}\). Determine the variation of temperature in the container wall and the temperatures of the inner and outer surfaces of the container. Is the outer surface temperature of the container safe to prevent thermal burn?

A spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and dropped into a large body of water at \(T_{\infty}\) where it is cooled by convection with an average convection heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of \(70^{\circ} \mathrm{C}\) and flow in subsea surrounding of \(5^{\circ} \mathrm{C}\). As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of \(0.5 \mathrm{~m}\) and wall thickness of \(8 \mathrm{~mm}\) is used for transporting liquid hydrocarbon at an average temperature of \(70^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The subsea surrounding has a temperature of \(5^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipeline is made of material with thermal conductivity of \(60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, \((b)\) determine the inner surface temperature of the pipeline, \((c)\) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and \((d)\) determine the heat flux through the outer pipeline surface.

A circular metal pipe has a wall thickness of \(10 \mathrm{~mm}\) and an inner diameter of \(10 \mathrm{~cm}\). The pipe's outer surface is subjected to a uniform heat flux of \(5 \mathrm{~kW} / \mathrm{m}^{2}\) and has a temperature of \(500^{\circ} \mathrm{C}\). The metal pipe has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=7.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0012 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). Determine the inner surface temperature of the pipe.

In a manufacturing plant, a quench hardening process is used to treat steel ball bearings \((c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\) \(60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\) ) of \(25 \mathrm{~mm}\) in diameter. After being heated to a prescribed temperature, the steel ball bearings are quenched. Determine the rate of heat loss if the rate of temperature decrease in the ball bearing at a given instant during the quenching process is \(50 \mathrm{~K} / \mathrm{s}\).
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Torque and Rotational Motion

Read Explanation

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Radiation

Read Explanation

Rotational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free