# Showing that a function satisfies an ODE

• Feb 22nd 2011, 12:49 PM
mbmstudent
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.
• Feb 22nd 2011, 03:05 PM
zzzoak
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

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