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.

\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.}