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A simple differential equation

  1. May 15, 2007 #1
    A body is falling in the earth's atmosphere. The forces on it are the earth's gravitational attraction and air resistance, which is proportional to its speed v. Its acceleration a is given by a=g-kv. Expressing a in the form dv/dt, obtain an expression for v. Show that whatever its initial speed, as t becomes very large the final speed of the body tends to g/k. I seem to go wrong somewhere in this equation. And what meaning has the air resistance to this question?
    [tex] \frac{1}{g-kv}\frac{dv}{dt}=1 \\[/tex] Therefore [tex] \int \frac{1}{g-kv} dv = \int dt \\ [/tex] Let u = g-k*v => du/dt=-k therefore dv=-du/k. Therefore the integral = [tex] \frac{-1}{k}\int \frac{1}{u}du [/tex] = -ln(u)/k, which gives [tex] \frac{-\ln(g-kv)}{k}= t+c \\ [/tex] Therefore [tex] \ln{(g-kv)} = -kt+kc [/tex] Therefore [tex] -kv=-g + \exp^{-kt+kc}[/tex] The answer is actually: [tex] v=\frac{g}{k}+c\exp^{-kt} [/tex]
     
    Last edited: May 15, 2007
  2. jcsd
  3. May 15, 2007 #2

    Dick

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    Divide both sides of your solution by -k. Write exp(-kt+kc)/(-k)=exp(-kt)*exp(kc)/(-k). Now define a new constant C=exp(kc)/(-k). Voila. After all, exp(kc)/(-k) is just 'some constant'.
     
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