Getting started with differential equations

[DivGradCurl]

Folks, I'm just getting started with differential equations and I need some help.

$$\frac{dp}{dt}=0.5p-450 \qquad (1)$$

$$\frac{dp/dt}{p-900}=\frac{1}{2} \qquad \mbox{if }p \neq 900 \qquad (2)$$

$$\frac{d}{dt}\ln \left| p-900 \right| = \frac{1}{2} \qquad (3)$$

I'm stuck in $$(3)$$. I know

$$\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}$$

but I still don't understand how that is obtained.

Any help is highly appreciated.

 

[arildno]

What do you mean?
$$\frac{d}{dt}ln|x(t)|=\frac{dx}{dt}\frac{1}{x}$$

 

[DivGradCurl]

I've just edited my previous post. There was a typo.

Well, the fact that in general

$$\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}$$

should give some insight into the transition from (2) to (3), but I still don't get it. That's what I mean.

Thanks

 

[arildno]

I've just edited my previous post. There was a typo.

Well, the fact that in general

$$\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}$$

should give some insight into the transition from (2) to (3), but I still don't get it. That's what I mean.

Thanks

First off, this is WRONG!
It should be:
$$\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}\frac{dx}{dt}$$

Do you now see the connection between (2) and (3)?

 

[DivGradCurl]

Ooops. I'm sorry (it was a typo). I mean

$$\frac{d}{dx}\ln \left| x \right|=\frac{1}{x}$$

I still don't get the connection between (2) and (3). Could you please clarify that?

 

[arildno]

Note that, by the chain rule we have :
$$\frac{d}{dt}ln|p(t)-900|=\frac{1}{p-900}\frac{dp}{dt}=\frac{\frac{dp}{dt}}{p-900}$$
Does that help you?

 

[dextercioby]

Have you been taught the METHOD OF SEPARATION OF VARIABLES...?I.e.Working with differentials,just as if they were numbers which could be multiplied & divided with...?

Daniel.

 

[DivGradCurl]

So, my problem is with the chain rule step. Arildno, I don't see why

$$\frac{d}{dt}ln|p(t)-900|=\frac{1}{p-900}\frac{dp}{dt}$$

although I can apply it (for example):

$$\frac{d}{dx}2(4x+1)^2=2\cdot 2 \cdot (4x+1)\cdot 4=64x+16$$

Could please you explain me how that works?

Daniel, I've read about separation of variables. That's going to be useful once I'm done with (3) and move on.

 

[arildno]

Well, the chain rule states, letting x=p-900:
$$\frac{d}{dt}ln|p-900|=\frac{d}{dt}ln|x|=(\frac{d}{dx}ln|x|)\frac{dx}{dp}\frac{dp}{dt},\frac{d}{dx}ln|x|=\frac{1}{x},\frac{dx}{dp}=1$$

were you interested in a more detailed explanation behind the chain rule?

 

[dextercioby]

What is $$\frac{d\ln u(t)}{dt}$$...?You said u knew how to apply the chain rule...


Daniel.

 

[DivGradCurl]

I can now see the way to get there:

$$\frac{dp}{dt}=0.5p-450 \qquad (1)$$

$$\frac{dp/dt}{p-900}=\frac{1}{2} \qquad \mbox{if }p \neq 900 \qquad (2)$$

$$\frac{dp}{dt} \left[ \frac{1}{p-900}\right] =\frac{1}{2}$$

$$\frac{dp}{dt} \left[ \frac{d}{dp} \ln \left| p-900 \right| \right] =\frac{1}{2}$$

$$\frac{d}{dt}\ln \left| p-900 \right| = \frac{1}{2} \qquad (3)$$

Thank you

 

[dextercioby]

Okay,how about posting the sollution to this ODE...?

Daniel.

 

[DivGradCurl]

$$\frac{dp}{dt}=0.5p-450$$

$$\frac{dp/dt}{p-900}=\frac{1}{2} \qquad \mbox{if }p \neq 900$$

$$\frac{dp}{dt} \left[ \frac{1}{p-900}\right] =\frac{1}{2}$$

$$\frac{dp}{dt} \left[ \frac{d}{dp} \ln \left| p-900 \right| \right] =\frac{1}{2}$$

$$\frac{d}{dt}\ln \left| p-900 \right| = \frac{1}{2}$$

$$\int \left[ \frac{d}{dt}\ln \left| p-900 \right| \right] dt = \int \frac{1}{2} dt$$

$$\ln \left| p-900 \right| = \frac{t}{2} + \mathrm{C}$$

$$\left| p-900 \right| = e^{\frac{t}{2} + \mathrm{C}}$$

$$\left| p-900 \right| = e^{\mathrm{C}}e^{\frac{t}{2}}$$

$$p-900 = \pm e^{\mathrm{C}}e^{\frac{t}{2}}$$

$$p(t) = 900 + \mathrm{K}e^{\frac{t}{2}} \quad \mbox{where } \mathrm{K}=\pm e^{\mathrm{C}} \mbox{ is an arbitrary constant.}$$

 

[arildno]

That is correct.

 

[dextercioby]

Look at the bright side.Once u'll learn separation of variables,u'll be able to shortcut many calculations...

Daniel.

 

[DivGradCurl]

Yes, indeed. :tongue2:

$$\frac{dp}{dt}=0.5p-450$$

$$\frac{dp/dt}{p-900}=\frac{1}{2} \qquad \mbox{if }p \neq 900$$

$$\int \frac{dp}{p-900}=\int \frac{dt}{2}$$

$$\ln \left| p-900 \right| = \frac{t}{2} + \mathrm{C}$$

$$\left| p-900 \right| = e^{\frac{t}{2} + \mathrm{C}}$$

$$\left| p-900 \right| = e^{\mathrm{C}}e^{\frac{t}{2}}$$

$$p-900 = \pm e^{\mathrm{C}}e^{\frac{t}{2}}$$

$$p(t) = 900 + \mathrm{K}e^{\frac{t}{2}} \quad \mbox{where } \mathrm{K}=\pm e^{\mathrm{C}} \mbox{ is an arbitrary constant.}$$