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Math Help - Showing that a function satisfies an ODE

  1. #1
    Newbie
    Joined
    Jan 2011
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    Showing that a function satisfies an ODE

    Hi, I have the following problem and I am stuck... Thanks for any help you can give me!

    The problem says:

    The atmospheric pressure p>0 decreases with height x according to the equation

    dp/dx=-gk

    where k is the air density, and g is the (constant) acceleration due to gravity. The density k is related to the pressure and temperature T by the ideal-gas law

    p=c*k*T

    where c>0 is a constant. This assignment examines the pressure distribution in an atmosphere under the assumption that the temperature decreases with height according to

    T=T_0+T_1e^{-x/H}

    where T_0, T_1 and H are positive constants.

    Q1: Show that p satisfies the ODE

    dp/dx=-(L*p)/(T_0+T_1e^{-x/H})

    where L is a constant which you will relate to g and c.
    __________________________


    So here's what I understand and what I have done so far.

    dp/dk=cT

    and dp/dx=-gk, so I need to find a way to transform dp/dk to dp/dx? What am I doing wrong?

    I have seen other examples, but they do it in a different way. For example, they have

    dp/dx=-gk, p=kbT/m (where b is a positive constant and m is the mass of a molecule of air). In one of the examples it says: if T is constant, show that

    dk/dx=-((mg)/(bT))k

    This makes more sense when it comes to transforming the dk/dx to dp/dx, since

    p=(kbT)/m, dp/dk=bT/m, and

    dk/dx=(dp/dx)/(dp/dk)=-((mkg)/(bT))

    My teacher usually mistypes things, but no one else has complained about it so I don't know if he's mistyped it or if I don't know how to do it.
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  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    We have

    <br />
\displaystyle<br />
\frac{dp}{dx}=-gk \; \; (1)<br />

    From

    <br />
\displaystyle<br />
p=ckT<br />

    we get

    <br />
\displaystyle<br />
k=\frac{p}{cT}<br />

    and inserting to (1) we have

    <br />
\displaystyle<br />
p=-g\frac{p}{cT}. \; \; (2)<br />

    Inserting temperature

    <br />
\displaystyle<br />
T=T_0+T_1 \; e^{-x/H} <br />

    into (2) we get

    <br />
\displaystyle<br />
p=-g\frac{p}{c(T_0+T_1 \; e^{-x/H})}. <br />
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