Please show solution and working out for dp/dt=-2p @ p(0)=20. And please list steps to always take with differential equations in order. I g

Does the following help?

Going through Kahn acadamy vids now. 

Alan's understanding of calculus is much better than mine.  I am sure his method is the best one, however

I am having trouble following his logic.    I will come back to that.

This is how I would do it:

$$\begin{array}{rlll}
\frac{dp}{dt}&=&-2p\qquad \qquad &p(0)=20\\\\
\frac{dt}{dp}&=&\frac{1}{-2p}\qquad \qquad &p(0)=20\\\\
\int{\frac{dt}{dp}}dp&=&\int \frac{1}{-2p}dp\\\\
t&=&\frac{1}{-2}ln(p)+c &when\; t=0\;p=20\;\;substitute\; to\; find\; c
\\\\
0&=&\frac{1}{-2}ln(20)+c\\\\
c&=&\frac{ln(20)}{2}\\\\
t&=&\frac{-ln(p)}{2}+\frac{ln(20)}{2}\\\\
2t&=&-ln(p)+ln(20)\\\\
2t&=&ln(\frac{20}{p})\\\\
e^{2t}&=&e^{ln(\frac{20}{p})}\\\\
p&=&\frac{20}{e^{2t}}\\\\

\end{array}$$

Ok Alan, I just worked out your answer.   

I didn't know I could do this

$$\begin{array}{rll}
\frac{}{}dp&=&-2dt\\\\
\int \frac{1}{p}dp&=&\int-2 dt
\end{array}$$

You are integrating both sides but one is being integrated with respect to p and the other with respect to t.

I didn't know you could do this.  Can you try to think of a simple explanation of why this is allowed?

Thank you.  

Thanks Alan but now I cannot understand why 

$$\dfrac{1}{p}\; \dfrac{d}{dt}\;p=\dfrac{d}{dt}ln(p)$$

Use the chain rule to get:

$$\frac{d\ln{p}}{dt}= \frac{d\ln{p}}{dp}\frac{dp}{dt}=\frac{1}{p}\frac{dp}{dt}$$

Thanks Alan. 

Thanks. That and kahn acadamy helped a bit. It is integration always to get a function right? 

Melody the reason to integrate is 2 fold as far as i saw on kahn acadamy. First we want an equation of a function as a rezult zo have to undo the differentiation that has occured. Sec9ndly to apply integration is applying small change in dp and dt resprctively as inthe above example and alans explaination. Thanks guys.