Problem 17
Question
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. $$ \cos x \frac{d y}{d x}+(\sin x) y=1 $$
Step-by-Step Solution
Verified Answer
The general solution is \( y = \sin x + C \cos x \), valid on \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), with no transient terms.
1Step 1: Identification of Equation Type
The given differential equation is \( \cos x \frac{d y}{d x} + (\sin x) y = 1 \). It has the form of a first-order linear differential equation, which can be written as \( \frac{dy}{dx} + P(x) y = Q(x) \). In this case, \( P(x) = \frac{\sin x}{\cos x} = \tan x \) and \( Q(x) = \frac{1}{\cos x} = \sec x \).
2Step 2: Find the Integrating Factor
To solve the linear differential equation, we use an integrating factor. The integrating factor is calculated as \( \mu(x) = e^{\int P(x) \, dx} = e^{\int \tan x \, dx} \). The integral of \( \tan x \) is \( \ln |\sec x| \), so \( \mu(x) = e^{\ln |\sec x|} = \sec x \).
3Step 3: Multiply Through by the Integrating Factor
Multiply the entire differential equation by the integrating factor \( \sec x \): \( \sec x[\cos x \frac{dy}{dx} + (\sin x)y] = \sec x \). This simplifies to \( \frac{d}{dx}[y \sec x] = \sec^2 x \).
4Step 4: Integrate Both Sides
Integrate both sides with respect to \( x \). The left side becomes \( y \sec x \), and the integral of \( \sec^2 x \) with respect to \( x \) is \( \tan x \). So we have \( y \sec x = \tan x + C \), where \( C \) is the constant of integration.
5Step 5: Solve for \( y \)
Divide both sides of the equation by \( \sec x \) to solve for \( y \). This gives \( y = \sin x + C \cos x \).
6Step 6: Determine the Interval of Validity
The interval over which \( \sec x \) is defined is \( x eq \frac{\pi}{2} + k\pi \), where \( k \) is any integer. The largest interval for the solution without discontinuities is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
7Step 7: Identify Transient Terms
A transient term is one which disappears as \( x \to \infty \). In the solution \( y = \sin x + C \cos x \), neither \( \sin x \) nor \( C \cos x \) disappears as \( x \to \infty \) within their defined intervals. Therefore, there are no transient terms.
Key Concepts
Integrating FactorTransient TermsInterval of Validity
Integrating Factor
The concept of an integrating factor is a key tool in solving first-order linear differential equations. Essentially, an integrating factor turns a non-exact differential equation into an exact one, making it easier to solve. It is particularly useful when the equation takes the form \( \frac{dy}{dx} + P(x)y = Q(x) \). In this form, the function \( P(x) \) can often be simplified, as seen in our exercise where \( P(x) = \tan x \).
The integrating factor, \( \mu(x) \), is calculated as \( e^{\int P(x) \, dx} \). In this exercise, since \( P(x) = \tan x \), we compute the integral \( \int \tan x \, dx \), which equals \( \ln |\sec x| \). Applying the rule \( e^{\ln |a|} = |a| \), the integrating factor \( \mu(x) \) is \( \sec x \).
Once you have the integrating factor, you multiply the entire differential equation by it. This turns the left side of the equation into the derivative of a product of functions, greatly simplifying the integration process on both sides.
The integrating factor, \( \mu(x) \), is calculated as \( e^{\int P(x) \, dx} \). In this exercise, since \( P(x) = \tan x \), we compute the integral \( \int \tan x \, dx \), which equals \( \ln |\sec x| \). Applying the rule \( e^{\ln |a|} = |a| \), the integrating factor \( \mu(x) \) is \( \sec x \).
Once you have the integrating factor, you multiply the entire differential equation by it. This turns the left side of the equation into the derivative of a product of functions, greatly simplifying the integration process on both sides.
Transient Terms
Transient terms in mathematical solutions usually relate to terms that diminish to zero as the variable approaches a certain limit, such as infinity. However, not all differential equations have transient terms. They often become negligible over time, contributing temporarily to the solution's behavior.
For our differential equation solution \( y = \sin x + C \cos x \), neither \( \sin x \) nor \( C \cos x \) disappears as \( x \to \infty \) when considered across their respective defined intervals. This means that both terms are persistent and continue to affect the solution indefinitely, especially because trigonometric functions like sine and cosine oscillate forever, never reaching a stable state.
In this particular exercise, you conclude that no transient terms exist since the behavior of \( \sin x \) and \( C \cos x \) does not stabilize or vanish.
For our differential equation solution \( y = \sin x + C \cos x \), neither \( \sin x \) nor \( C \cos x \) disappears as \( x \to \infty \) when considered across their respective defined intervals. This means that both terms are persistent and continue to affect the solution indefinitely, especially because trigonometric functions like sine and cosine oscillate forever, never reaching a stable state.
In this particular exercise, you conclude that no transient terms exist since the behavior of \( \sin x \) and \( C \cos x \) does not stabilize or vanish.
Interval of Validity
The interval of validity provides the scope over which the solution to a differential equation remains valid without discontinuities or undefined points. For the solution of our differential equation, the function \( \sec x \) was both our integrating factor and a component of the solution.
The function \( \sec x \) is undefined when its denominator, \( \cos x \), equals zero. Specifically, this occurs at \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer. Therefore, the largest interval of validity for the solution without crossing these points of discontinuity is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
Choosing the correct interval ensures the solution remains meaningful and defines the domain in which you can expect the solution to behave correctly within the equation's context.
The function \( \sec x \) is undefined when its denominator, \( \cos x \), equals zero. Specifically, this occurs at \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer. Therefore, the largest interval of validity for the solution without crossing these points of discontinuity is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
Choosing the correct interval ensures the solution remains meaningful and defines the domain in which you can expect the solution to behave correctly within the equation's context.
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