Algebra help, logarithms, calculus

Question

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!

Answer 1

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...

Answer 2

Hint: Integrate both sides; the left with respect to $p$, the right with respect to $h$.

Answer 3

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

$$\log(p)= \frac{g}{\lambda R}\log(T_0-\lambda h) + C$$