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Thread: Showing that a function satisfies an ODE

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    16

    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

    $\displaystyle dp/dx=-gk$

    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

    $\displaystyle p=c*k*T$

    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

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

    where $\displaystyle T_0$, $\displaystyle T_1$ and $\displaystyle H$ are positive constants.

    Q1: Show that p satisfies the ODE

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

    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.

    $\displaystyle dp/dk=cT$

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

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

    $\displaystyle dp/dx=-gk$, $\displaystyle p=kbT/m$ (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

    $\displaystyle dk/dx=-((mg)/(bT))k$

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

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

    $\displaystyle dk/dx=(dp/dx)/(dp/dk)=-((mkg)/(bT))$

    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

    $\displaystyle
    \displaystyle
    \frac{dp}{dx}=-gk \; \; (1)
    $

    From

    $\displaystyle
    \displaystyle
    p=ckT
    $

    we get

    $\displaystyle
    \displaystyle
    k=\frac{p}{cT}
    $

    and inserting to (1) we have

    $\displaystyle
    \displaystyle
    p=-g\frac{p}{cT}. \; \; (2)
    $

    Inserting temperature

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

    into (2) we get

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