We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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 got stuck here because I firstly seperate to get 1/dt = -2p/dp. Not sure what to do, and if I intergrate both sides the same way or differently if dp for example is multiplied or devided or other. Thanks.

Stu Aug 9, 2014

#3**+10 **

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}$$

Melody Aug 9, 2014

#4**+5 **

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.

Melody Aug 9, 2014

#6**+5 **

Thanks Alan but now I cannot understand why

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

Melody Aug 9, 2014

#7**+10 **

Use the chain rule to get:

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

.Alan Aug 10, 2014

#9**+10 **

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.

Stu Aug 10, 2014