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Chapter 20: Problem 3

Use a package to find the general solution of \(\frac{\mathrm{d} T}{\mathrm{~d} \theta}=\mu(T-K)\) where \(\mu\) and \(K\) are constants.

Short Answer

Expert verified
Question: Determine the general solution of the given first-order ODE: \(\frac{\mathrm{d} T}{\mathrm{~d}\theta}=\mu(T-K)\), where \(\mu\) and \(K\) are constants. Answer: The general solution is given by \(T(\theta) = K + Ae^{\mu\theta}\), where \(A\) is an arbitrary constant.

Step by step solution

01

Identify the type of ODE

This is a first-order linear ODE since it can be written in the form \(\frac{\mathrm{d} T}{\mathrm{~d}\theta} + p(T) = q(\theta)\), where \(p(T) = -\mu\) and \(q(\theta)= \mu K\) .
02

Separate the variables

Separate the variables by dividing both sides by \((T-K)\) and multiplying both sides by \(\mathrm{d}\theta\) to get: \(\frac{\mathrm{d}T}{T-K} = \mu \mathrm{d}\theta\)
03

Integrate both sides

Now we will integrate both sides with respect to their respective variables: \(\int \frac{\mathrm{d}T}{T-K} = \int \mu \mathrm{d}\theta\)
04

Evaluate the integrals

Evaluate the integrals as follows: \(\ln |T-K| = \mu\theta + C\), where \(C\) is the constant of integration.
05

Solve for T

Now, we must solve for \(T\): \(\Rightarrow T-K = \pm e^{\mu\theta + C}\) Expressing the constant as \(A=\pm e^{C}\) and multiplying both sides by \(e^{\mu\theta}\), we get: \(\Rightarrow T-K = Ae^{\mu\theta}\) Finally, add \(K\) to both sides to obtain the general solution: \(\Rightarrow T(\theta) = K + Ae^{\mu\theta}\), where \(A\) is an arbitrary constant.

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Most popular questions from this chapter

Using software, obtain a symbolic solution of \(i R+L \frac{\mathrm{d} i}{\mathrm{~d} t}=E\) where \(R, L\) and \(E\) are constants subject to the initial condition \(i(0)=0\).

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