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Chapter 2: Problem 2

Find the general solution of the given differential equation. Give the largest interval over which the general solution is defined. Determine whether there are any transient terms in the general solution. $$ \frac{d y}{d x}+2 y=0 $$

Short Answer

Expert verified
The general solution is \( y = Ce^{-2x} \), defined over \((-\infty, \infty)\). Transient term: \( Ce^{-2x} \).

Step by step solution

01

Identify the type of differential equation

The given differential equation \( \frac{d y}{d x} + 2y = 0 \) is a first-order linear homogeneous differential equation. It can be written in the standard form \( \frac{d y}{d x} + P(x) y = Q(x) \), where \( P(x) = 2 \) and \( Q(x) = 0 \).
02

Determine the integrating factor

For a first-order linear differential equation, the integrating factor \( \mu(x) \) is given by the formula \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = 2 \), so we calculate \( \mu(x) = e^{\int 2 \, dx} = e^{2x} \).
03

Multiply the entire equation by the integrating factor

We multiply the entire differential equation by the integrating factor \( e^{2x} \), yielding \( e^{2x} \frac{d y}{d x} + 2e^{2x} y = 0 \). This transforms the equation into the form \( \frac{d}{dx}(e^{2x}y) = 0 \).
04

Integrate both sides with respect to \( x \)

Since \( \frac{d}{dx}(e^{2x}y) = 0 \), integrating both sides gives \( e^{2x}y = C \), where \( C \) is the constant of integration.
05

Solve for \( y \)

To solve for \( y \), divide both sides by \( e^{2x} \): \( y = Ce^{-2x} \). This is the general solution of the differential equation.
06

Determine the interval of definition

The general solution \( y = Ce^{-2x} \) is valid for all \( x \) because the exponential function is defined for all real numbers. Therefore, the largest interval over which the solution is defined is \( (-\infty, \infty) \).
07

Check for transient terms

A transient term is a term in the solution that diminishes to zero as \( x \to \pm\infty \). Here, \( Ce^{-2x} \) tends to zero as \( x \to \infty \), implying that the term is transient in the positive infinity direction.

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Key Concepts

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

General Solution
When dealing with first-order linear differential equations, like our example \( \frac{d y}{d x}+2 y=0 \), finding the general solution is vital. This gives us a family of solutions that describe how a system can behave under a variety of circumstances.

The general solution is derived by rewriting the differential equation in a standard form and determining an appropriate integrating factor. By doing so, we can integrate and find a solution that includes a constant of integration \( C \).

This constant \( C \) allows for a range of potential solutions, which is why the solution is considered "general." It represents a collective of particular solutions that are influenced by initial conditions or additional constraints. In our specific case, the general solution is \( y = Ce^{-2x} \), applicable for all \( x \) values.
Integrating Factor
The integrating factor is a tool that simplifies solving first-order linear differential equations. To determine it, we use the formula \( \mu(x) = e^{\int P(x) \, dx} \). This multiplying factor transforms the equation into a new form that's easier to solve.

Let's look at our example: the function \( P(x) \) is equal to 2. Therefore, the integrating factor becomes \( \mu(x) = e^{2x} \). By multiplying the whole differential equation by this factor, it helps us rewrite it and eventually isolate a derivative form of \( y \), making integration straightforward.

In essence, the integrating factor works like a key that unlocks a complex equation, turning it into a simple one where traditional calculus techniques like integrating can be used effectively.
Transient Terms
Understanding transient terms in a solution helps us determine how the behavior of a system changes over time, often approaching a steady state. In mathematical terms, a transient term vanishes as time goes to infinity or negative infinity.

In our solution \( y = Ce^{-2x} \), \( Ce^{-2x} \) fades away as \( x \to \infty \). This property categorizes it as a transient term when considering the equation over a long period.

Transient terms are temporary by nature and primarily affect the short-term response of a system. They die out, leaving only any persistent terms which define the system's long-term behavior. In our example, since no persistent terms remain, the entire solution is transient. By identifying such terms, you predict the system's eventual stabilization as transient responses die away.

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Most popular questions from this chapter

When interest is compounded continuously, the amount of money increases at a rate proportional to the amount \(S\) present at time \(t\), that is, \(d S / d t=r S\), where \(r\) is the annual rate of interest. (a) Find the amount of money accrued at the end of 5 years when \(\$ 5000\) is deposited in a savings account drawing \(5 \frac{3}{4} \%\) annual interest compounded continuously. (b) In how many years will the initial sum deposited bave doubled? (c) Use a calculator to compare the amount obtained in part (a) with the amount \(S=5000\left(1+\frac{1}{4}(0.0575)\right)^{5(4)}\) that is arcrued when interest is compounded quarterly.

The differential equation $$ \frac{d y}{d x}=\frac{-x+\sqrt{x^{2}+y^{2}}}{y} $$ describes the shape of a plane curve \(C\) that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises \(1.3\). There are several ways of solving this \(\mathrm{DE}\). (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution \(y=u x\) yields $$ \frac{u d u}{\sqrt{1+u^{2}}\left(1-\sqrt{\left.1+u^{2}\right)}\right.}=\frac{d x}{x}. $$ Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve \(C\) must be a parabola with focus at the origin and is symmetric with respect to the \(x\) -axis. (b) Show that the first differential equation can also be solved by means of the substitution \(u=x^{2}+y^{2}\).

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ L \frac{d i}{d t}+R i=E ; \quad i(0)=i_{0}, L, R, E, \text { and } i_{0} \text { constants } $$

When forgetfulness is taken into account, the rate of memorization of a subject is given by $$ \frac{d A}{d t}=k_{1}(M-A)-k_{2} A $$ where \(k_{1}>0, k_{2}>0, A(t)\) is the amount to be memorized in time \(t, M\) is the total amount to be memorized, and \(M-A\) is the amount remaining to be memorized. See Problems 25 and 26 in Exercises \(1.3\). (a) Since the DE is autonomous, use the phase portrait concept of Section \(2.1\) to find the limiting value of \(A(t)\) as \(t \rightarrow \infty\). Interpret the result. (b) Solve for \(A(t)\) subject to \(A(0)=0\). Sketch the graph of \(A(t)\) and verify your prediction in part (a).

Consider the initial-value problem \(y^{\prime}+e^{x} y=f(x), y(0)=1\) Express the solution of the IVP for \(x>0\) as a nonelementary integral when \(f(x)=1\). What is the solution when \(f(x)=0 ?\) When \(f(x)=e^{x} ?\)
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