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Chapter 2: Problem 2

For \(y^{\prime}+2 y=0\) an integrating factor is \(e^{\int 2 d x}=e^{2 x}\) so that \(\frac{d}{d x}\left[e^{2 x} y\right]=0\) and \(y=c e^{-2 x}\) for \(-\infty < x<\infty\). The transient term is \(c e^{-2 x}\).

Short Answer

Expert verified
The transient term is \(ce^{-2x}\) in the solution \(y=ce^{-2x}\).

Step by step solution

01

Recognize the Differential Equation Type

The given differential equation \(y' + 2y = 0\) is a first-order linear homogeneous differential equation. Our objective is to find the general solution using an integrating factor.
02

Determine the Integrating Factor

The standard form of a first-order linear differential equation is \(y' + P(x)y = Q(x)\). Here, \(P(x) = 2\) and \(Q(x) = 0\). The integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x}\).
03

Multiply by the Integrating Factor

Multiply both sides of the differential equation by the integrating factor \(e^{2x}\) to obtain: \(e^{2x} y' + 2e^{2x} y = 0\). This allows us to express the left-hand side as a derivative of a product.
04

Simplify to Form a Derivative

Rewrite the left side of the equation as the derivative of the product: \(\frac{d}{dx}\left(e^{2x}y\right) = 0\). This step involves recognizing that \(e^{2x} y' + 2e^{2x} y\) is the derivative of \(e^{2x} y\).
05

Integrate Both Sides

Integrate both sides with respect to \(x\). Since the right-hand side is zero, integrating yields: \(e^{2x}y = C\), where \(C\) is the constant of integration.
06

Solve for \(y\)

To find \(y\), solve for it by dividing both sides by \(e^{2x}\) to get: \(y = Ce^{-2x}\). This is the general solution of the differential equation.
07

Identify the Transient Term

In the solution \(y = Ce^{-2x}\), the transient term is \(Ce^{-2x}\). It represents the part of the solution that approaches zero as \(x\) increases.

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Key Concepts

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

Integrating Factor
An integrating factor is a function used to simplify the process of solving first-order linear differential equations. Specifically, it is employed to convert a non-precise differential equation into an exact one, which can be readily integrated. The standard form of a first-order linear differential equation is given by
  • \( y' + P(x)y = Q(x) \)
In this equation, \( P(x) \) is a function of \( x \), and \( Q(x) \) is the non-homogeneous part. The integrating factor, usually denoted as \( \mu(x) \), is expressed as:
  • \( \mu(x) = e^{\int P(x) \, dx} \)
This function, when multiplied by the entire differential equation, allows the left side to be written as a derivative of a product. Thereby simplifying it greatly and enabling us to solve the equation more easily. Let鈥檚 say for
  • \( y' + 2y = 0 \)
the integrating factor would be \( e^{\int 2 \, dx} = e^{2x} \), aiding in forming the derivative \( \frac{d}{dx}(e^{2x}y) = 0 \). Integrating factors are crucial in solving linear equations efficiently.
Homogeneous Differential Equation
A homogeneous differential equation is a type where the function is set to zero, meaning every term involves the dependent variable or its derivatives. Thus, it can be written in this simplified form:
  • \( y' + P(x)y = 0 \)
For our specific case, \( y' + 2y = 0 \) clearly fits this pattern, where the right side is zero making it homogeneous. The term homogeneous indicates a scenario where all output terms result directly from input terms and their coefficients. This characteristic often makes them easier to solve since they do not have additional peripherals like \( Q(x) \) adding complexity. The significance of recognizing a differential equation as homogeneous early on includes simplifying the solution-path choice and focusing on solving linear parts with straightforward techniques without worrying about external functions.
General Solution
The general solution of a differential equation encompasses all possible solutions formed by varying constants. For first-order linear differential equations such as \( y' + 2y = 0 \), it provides a way to express the infinite set of solutions in a comprehensive formula. To arrive at this, once the integrating factor technique is applied and simplified to obtain:
  • \( \frac{d}{dx}(e^{2x}y) = 0 \)
Integration produces a constant on one side because the right-hand outcome is zero after integration:
  • \( e^{2x}y = C \)
Here, \( C \) represents the constant of integration that embodies various solution possibilities. Finally, solving for \( y \) gives:
  • \( y = Ce^{-2x} \)
This formula shows that every solution rolls back to an original constant multiplying the exponential function. The part \( Ce^{-2x} \) collapses as \( x \) increases, identifying it as the transient term indicating how responses diminish over time. Therefore, the general solution in these cases is not merely a set result but a framework defining how different conditions manifest across \( x \).

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Most popular questions from this chapter

(a) Separating variables we have \(2 y d y=(2 x+1) d x .\) Integrating gives \(y^{2}=x^{2}+x+c .\) When \(y(-2)=-1\) we find \(c=-1,\) so \(y^{2}=x^{2}+x-1\) and \(y=-\sqrt{x^{2}+x-1} .\) The negative square root is chosen because of the initial condition. (b) From the figure, the largest interval of definition appears to be approximately \((-\infty,-1.65)\). (c) Solving \(x^{2}+x-1=0\) we get \(x=-\frac{1}{2} \pm \frac{1}{2} \sqrt{5},\) so the largest interval of definition is \(\left(-\infty,-\frac{1}{2}-\frac{1}{2} \sqrt{5}\right)\) The right-hand endpoint of the interval is excluded because \(y=-\sqrt{x^{2}+x-1}\) is not differentiable at this point.

For \(\frac{d T}{d t}-k T=-T_{m} k\) an integrating factor is \(e^{\int(-k) d t}=e^{-k t}\) so that \(\frac{d}{d t}\left[e^{-k t} T\right]=-T_{m} k e^{-k t}\) and \(T=T_{m}+c e^{k t}\) for \(-\infty< t<\infty .\) If \(T(0)=T_{0}\) then \(c=T_{0}-T_{m}\) and \(T=T_{m}+\left(T_{0}-T_{m}\right) e^{k t}\).

Write the differential equation as $$\frac{d v}{d x}+\frac{1}{x} v=32 v^{-1},$$ and let \(u=v^{2}\) or \(v=u^{1 / 2} .\) Then $$\frac{d v}{d x}=\frac{1}{2} u^{-1 / 2} \frac{d u}{d x},$$ and substituting into the differential equation, we have $$\frac{1}{2} u^{-1 / 2} \frac{d u}{d x}+\frac{1}{x} u^{1 / 2}=32 u^{-1 / 2} \quad \text { or } \quad \frac{d u}{d x}+\frac{2}{x} u=64.$$ The latter differential equation is linear with integrating factor \(e^{\int(2 / x) d x}=x^{2},\) so $$\frac{d}{d x}\left[x^{2} u\right]=64 x^{2}$$ and $$x^{2} u=\frac{64}{3} x^{3}+c \quad \text { or } \quad v^{2}=\frac{64}{3} x+\frac{c}{x^{2}}.$$

Separating variables, we have \\[\frac{d y}{y^{2}-y}=\frac{d x}{x} \quad \text { or } \quad \int \frac{d y}{y(y-1)}=\ln |x|+c.\\] Using partial fractions, we obtain $$\begin{aligned} \int\left(\frac{1}{y-1}-\frac{1}{y}\right) d y &=\ln |x|+c \\ \ln |y-1|-\ln |y| &=\ln |x|+c \\ \ln \left|\frac{y-1}{x y}\right| &=c \\ \frac{y-1}{x y} &=e^{c}=c_{1}.\end{aligned}$$ Solving for \(y\) we get \(y=1 /\left(1-c_{1} x\right) .\) We note by inspection that \(y=0\) is a singular solution of the differential equation. (a) Setting \(x=0\) and \(y=1\) we have \(1=1 /(1-0),\) which is true for all values of \(c_{1} .\) Thus, solutions passing through (0,1) are \(y=1 /\left(1-c_{1} x\right)\). (b) Setting \(x=0\) and \(y=0\) in \(y=1 /\left(1-c_{1} x\right)\) we get \(0=1 .\) Thus, the only solution passing through (0,0) is \(y=0\). (c) Setting \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\) we have \(\frac{1}{2}=1 /\left(1-\frac{1}{2} c_{1}\right),\) so \(c_{1}=-2\) and \(y=1 /(1+2 x)\). (d) \(\operatorname{Setting} x=2\) and \(y=\frac{1}{4}\) we have \(\frac{1}{4}=1 /\left(1-2 c_{1}\right),\) so \(c_{1}=-\frac{3}{2}\) and \(y=1 /\left(1+\frac{3}{2} x\right)=2 /(2+3 x)\).

For \(\frac{d r}{d \theta}+r \sec \theta=\cos \theta\) an integrating factor is \(e^{\int \sec \theta d \theta}=e^{\ln |\sec x+\tan x|}=\sec \theta+\tan \theta\) so that \(\frac{d}{d \theta}[(\sec \theta+\tan \theta) r]=1+\sin \theta\) and \((\sec \theta+\tan \theta) r=\theta-\cos \theta+c\) for \(-\pi / 2 <\theta < \pi / 2\).
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