Question

Obtain the general solution of the differential equation $$ \frac{\mathrm{d} T}{\mathrm{~d} \theta}-\mu T=-\mu K $$ where \(\mu\) and \(K\) are constants.

Short answer

Based on the given step-by-step solution, answer the following question: **Question:** Find the general solution of the first-order linear differential equation \(\frac{\mathrm{d} T}{\mathrm{~d} \theta}-\mu T=-\mu K\), where \(\mu\) and \(K\) are constants. **Answer:** The general solution for the given differential equation is \(T(\theta) = -K + Ce^{\mu\theta}\), where \(C\) is an arbitrary constant.

Step-by-step solution

Identify the linear differential equation

The given differential equation is $$ \frac{\mathrm{d} T}{\mathrm{~d} \theta}-\mu T=-\mu K $$ where \(\mu\) and \(K\) are constants. The dependent variable is \(T\), and the independent variable is \(\theta\). It is a first-order linear differential equation.

Find the integrating factor

To find the integrating factor, we need to find the function \(I(\theta)\) such that $$ I(\theta) = e^{\int -\mu ~d\theta} $$ Calculate the integral: $$ \int -\mu ~d\theta = -\mu\theta + C $$ Since we only need the integrating factor, we can ignore the constant \(C\). Then, $$ I(\theta) = e^{-\mu\theta} $$

Multiply the differential equation by the integrating factor

To find the general solution, multiply both sides of the differential equation by the integrating factor \(I(\theta)\): $$ e^{-\mu\theta}\left(\frac{\mathrm{d} T}{\mathrm{~d} \theta}-\mu T\right) = e^{-\mu\theta}(-\mu K) $$

Simplify and rewrite the equation as the product rule

Simplifying the equation, we get: $$ \frac{\mathrm{d}}{\mathrm{d}\theta}(Te^{-\mu\theta}) = -\mu Ke^{-\mu\theta} $$ Now, the left side of the equation is the product rule for the derivative of \(T(\theta)e^{-\mu\theta}\).

Integrate both sides

Integrate both sides of the equation with respect to \(\theta\): $$ \int \frac{\mathrm{d}}{\mathrm{d}\theta}(Te^{-\mu\theta}) ~d\theta = \int -\mu Ke^{-\mu\theta} ~d\theta $$ On the left side, we have the integral of a derivative, which gives back the original function: $$ Te^{-\mu\theta} = \int -\mu Ke^{-\mu\theta} ~d\theta + C $$ The right side can be integrated to find the general solution.

Integrate the right side and find the general solution

When integrating the right side of the equation, we get: $$ -\mu K\int e^{-\mu\theta} ~d\theta = -K e^{-\mu\theta} + C $$ Now equating both sides of the equation, we get: $$ Te^{-\mu\theta} = -K e^{-\mu\theta} + C $$

Solve for T

Multiply both sides by \(e^{\mu\theta}\) to isolate \(T\): $$ T(\theta) = -K + Ce^{\mu\theta} $$ This is the general solution for the given differential equation.

Key concepts

first-order linear differential equation

Differential equations are equations that involve a function and its derivatives. In this context, a **first-order linear differential equation** involves the first derivative of the function. These equations are significant because they appear frequently in various scientific fields to describe changing systems. The general form of a first-order linear differential equation is \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \). In the given problem, the differential equation \( \frac{dT}{d\theta} - \mu T = -\mu K \) fits this form, where:\

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• \( T \) is the dependent variable.
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• \( \theta \) is the independent variable.
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• \( -\mu \) represents the function \( P(x) \).
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• \( -\mu K \) is analogous to \( Q(x) \).
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\Understanding and solving first-order linear differential equations are critical skills for modeling real-world phenomena where there's growth or decay, such as temperature variation, population dynamics, or electrical circuits.

integrating factor

The **integrating factor** is an essential tool for solving first-order linear differential equations. Its function is to simplify the equation into a form that can be easily integrated. To find the integrating factor, you need to compute \( I(x) = e^{\int P(x) \, dx} \), where \( P(x) \) is the coefficient of \( y \) in the differential equation. For the given equation \( \frac{dT}{d\theta} - \mu T = -\mu K \), \( P(x) \) is \( -\mu \). Thus, the integrating factor \( I(\theta) \) becomes \

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• \( I(\theta) = e^{\int -\mu \, d\theta} = e^{-\mu \theta} \).
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\Using the integrating factor simplifies the differential equation to a form where the derivative \( \frac{d}{d\theta}(T e^{-\mu \theta}) \) becomes apparent, enabling straightforward integration to solve for \( T \). This step is like a magic trick in differential equations, making complex problems simpler.

general solution

Once the differential equation is transformed with the integrating factor, finding the **general solution** becomes an exercise in integration. After applying the integrating factor to \( \frac{dT}{d\theta} - \mu T = -\mu K \), the equation becomes:\

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• \( \frac{d}{d\theta}(T e^{-\mu \theta}) = -\mu K e^{-\mu \theta} \)
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\Integrating both sides gives us the integrated form of the function. On the left, integration simply reverses the derivative, returning \( T e^{-\mu \theta} \). On the right, integrating \( -\mu K e^{-\mu \theta} \) results in \( -K e^{-\mu \theta} \), with an added constant of integration \( C \), leading to:

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• \( T e^{-\mu \theta} = -K e^{-\mu \theta} + C \)
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To solve for \( T \), we multiply through by \( e^{\mu \theta} \), observing that it cancels the exponent terms, simplifying to:

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• \( T(\theta) = -K + Ce^{\mu \theta} \)
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This expression represents the general solution to the differential equation, capturing all possible behaviors of the system, with \( C \) embodying any initial conditions related to the problem.