Algebra help, logarithms, calculus I'm seriously out of my depth. I have very basic understanding of logarithms and calculus. Please could someone walk me through how to get from:
$$\frac{{d}p}{p} = -\frac{g}{R}\cdot \frac{dh}{T_{0}-\lambda h}$$
To:
$$\log p = \frac{g}{\lambda R}\log(T_{0}-\lambda h) + constant$$
I'd appreciate terminology of any manipulations done so I can Google them for more information. Thank you!
 A: Hint: Integrate both sides; the left with respect to $p$, the right with respect to $h$.
A: This is a first order, non-linear, ordinary differential equation.  (For terminology)  The dependent variable is $p$, and the independent variable is $h$.
The technique used to solve is called "separation of variables." (more terminology)
Starting with your expression:
$$\frac{{d}p}{p} = -\frac{g}{R}\cdot \frac{dh}{T_{0}-\lambda h}$$
$$\int\frac{{d}p}{p} = \int-\frac{g}{R}\cdot \frac{dh}{T_{0}-\lambda h}$$
Recall that $\int\frac{1}{x}dx=\ln |x| + C$.  Thus we have:
$$\ln|p|+C = \int-\frac{g}{R}\cdot \frac{dh}{T_{0}-\lambda h}$$
Now we perform a $u$-substitution on the RHS, letting $u=T_0 -\lambda h$. This implies that $du = -\lambda dh$
$$\ln|p|+C = \int-\frac{g}{R(-\lambda)}\cdot \frac{du}{u}$$
Integrating (using the same rule as above):
$$\ln|p|+C = \frac{g}{R(\lambda)}\cdot \ln|u|$$
Rearranging, and substituting $u$ back in:
$$\ln|p| = \frac{g}{R\lambda}\cdot \ln|T_0 -\lambda h| + C$$
Leave a comment if you need more explanation on any step...
A: Using Clayton's hint (integrating both sides)...
$$\int\frac{dp}{p}= \int\frac{-g}{R}\frac{dh}{T_0-\lambda h}$$ this gives
$$\log(p)+C= -\frac{g}{R(-\lambda)}\int\frac{1}{u}du = \frac{g}{\lambda R}$$  where $u= T_0-\lambda h$, which gives $du = -\lambda dh$ and $C$ is our constant term.
Finally when substituting $u$ back in and combining our constant terms we get 
$$\log(p)= \frac{g}{\lambda R}\log(T_0-\lambda h) + C$$
