- #31
Mark44
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What you have is a differential equation. To find a unique solution of a first-order differential equation, you need an initial condition, say m(a) = ma.
Before continuing I should add that m(x) really represents the mass density of your rod. The actual mass of a very short section of length Δx would be m(x)Δx.
Your problem, with the addition of an initial condition, is
$$\frac{m'}{m} = \frac{1}{x}, m(a) = m_a$$
From this we get ln(m) = ln(x) + C. Exponentiating both sides, we get m = eln(x) + C = eC eln(x) = Kx, where K = eC.
So m(x) = Kx.
From the initial condition, m(a) = ma = Ka ##\Rightarrow## K = ma/a.
The unambiguous solution is m(x) = ma/a * x. You can verify that this function satisfies the differential equation and initial condition.
To find the mass of the rod in the interval [a, b], calculate integral of m(x) between a and b.
Before continuing I should add that m(x) really represents the mass density of your rod. The actual mass of a very short section of length Δx would be m(x)Δx.
Your problem, with the addition of an initial condition, is
$$\frac{m'}{m} = \frac{1}{x}, m(a) = m_a$$
From this we get ln(m) = ln(x) + C. Exponentiating both sides, we get m = eln(x) + C = eC eln(x) = Kx, where K = eC.
So m(x) = Kx.
From the initial condition, m(a) = ma = Ka ##\Rightarrow## K = ma/a.
The unambiguous solution is m(x) = ma/a * x. You can verify that this function satisfies the differential equation and initial condition.
To find the mass of the rod in the interval [a, b], calculate integral of m(x) between a and b.