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Chapter 20: Problem 10

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

Expert verified
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

01

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.
02

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, $$ I(\theta) = e^{-\mu\theta} $$
03

Multiply the differential equation by the integrating factor

To find the general solution, multiply both sides of the differential equation by the integrating factor \(I(\theta)\): $$ e^{-\mu\theta}\left(\frac{\mathrm{d} T}{\mathrm{~d} \theta}-\mu T\right) = e^{-\mu\theta}(-\mu K) $$
04

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 \(T(\theta)e^{-\mu\theta}\).
05

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.
06

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: $$ Te^{-\mu\theta} = -K e^{-\mu\theta} + C $$
07

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.

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Most popular questions from this chapter

Classify the following equations, specifying the order and type (linear or non-linear): (a) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-\frac{\mathrm{d} y}{\mathrm{~d} x}=x^{2}\) (b) \(\frac{\mathrm{d} y}{\mathrm{~d} t}+\cos y=0\)

By integrating twice find the general solution of \(y^{\prime \prime}=12 x^{2}\)

Integrate twice the differential equation $$ \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{w}{2}\left(l x-x^{2}\right) $$ where \(w\) and \(l\) are constants, to find a general solution for \(y\).

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Identify the dependent and independent variables of the following differential equations. Give the order of the equations and state which are linear. (a) \(\frac{\mathrm{d} y}{\mathrm{~d} x}+9 y=0\) (b) \(\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)\left(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right)+3 \frac{\mathrm{d} y}{\mathrm{~d} x}=0\) (c) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}+5 \frac{\mathrm{d} x}{\mathrm{~d} t}=\sin x\)
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