Err...what?

- Astrophysics

Err...what?

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- Astrophysics

If we have \[\frac{ dp }{ dt } = \frac{ p-900 }{ 2 } \implies \frac{ dp/dt}{p-900 }= \frac{ 1 }{ 2 }\] so how does this go to \[\frac{ d }{ dt }\ln|p-900|=1/2\]

- Astrophysics

I mean we can do a u - sub, but I find that weird for differentiation

- Astrophysics

I guess it involves the chain rule, but I'm not seeing it xD

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## More answers

- Astrophysics

I should specify p=p(t)

- zzr0ck3r

\(\int\dfrac{1}{p-900}dp=\int\dfrac{1}{2}dt\\\ln(p-900)=\dfrac{1}{2}t+c\)

- Astrophysics

Yeah that's fine I just mean how to go from \[\frac{ dp/dt }{ p-900 }=1/2 \implies \frac{ d }{ dt } \ln|p-900|=1/2\]

- Astrophysics

This is the step before integration

- Astrophysics

Though I would do it your way, I'm just trying to figure out how to use the chain rule here, if that makes sense

- zzr0ck3r

I have no idea... it makes no sense to me
I did
\[\dfrac{dp}{dt}\dfrac{1}{p-900}=\dfrac{1}{2}\\\implies \dfrac{1}{p-900}dp=\dfrac{1}{2}dt\] Then I integrate both sides.

- Astrophysics

Yeah exactly, that's how I'd do it, but in my book it says "By the chain rule the left side" then they show above

- zzr0ck3r

\(\frac{ d }{ dt } \ln|p-900|=1/2 \implies \dfrac{df(p)}{dt}=\dfrac{1}{2}\) which is odd to me.

- zzr0ck3r

the derivative of a function of p with respect to t is a constant 1/2 ?

- Astrophysics

Ok so this is part of a modelling problem it starts off as \[\frac{ dp }{ dt } = \frac{ p-900 }{ 2 } \implies \frac{ dp/dt }{ p-900 } = \frac{ 1 }{ 2 } (eq 6)\] then it says by the chain rule the left side of the equation 6 is the derivative of ln|p-900| with respect to t, so we have \[\frac{ d }{ dt }\ln|p-900|=1/2\] then they integrate

- zzr0ck3r

I don't know man, maybe someone else will help. I suck at calculus...

- Astrophysics

Haha, no worries man, I was like I should know this...but I had no clue, thanks a lot for the help though!

- zzr0ck3r

:(

- Astrophysics

The solution is the same as yours I just don't know why it's written this way

- zzr0ck3r

Do they end with \(p=Ce^{\frac{1}{2}t}+900\)?

- Astrophysics

Yup

- zzr0ck3r

yeah, I dont know what that step is about then.

- Astrophysics

Yeah neither do I haha, thanks for checking though!

- freckles

\[\frac{d}{dt} \ln|p-900| \text{ does equal } (p-900)' \cdot \frac{1}{p-900} =p' \cdot \frac{1}{p-900}=\frac{dp}{dt} \cdot \frac{1}{p-900}\]
but yeah I would have never written it like the way they did

- Astrophysics

Yeah so they sort of our just going backwards? Just a weird way to introduce dif equations I thought haha, thanks @freckles

- anonymous

You do stuff like that plenty of time when solving linear ODEs. Without going into too much detail, consider the general case:
\[\frac{dy}{dx}+f(x)y=g(x)\]
You find some function \(\mu(x)\) which you distribute onto both sides, giving
\[\mu(x)\frac{dy}{dx}+f(x)\mu(x)y=g(x)\mu(x)\]
with the property that
\[\frac{d}{dx}\left[\mu(x)y\right]=\mu(x)\frac{dy}{dx}+f(x)\mu(x)y\]
This means you can solve what remains, i.e.
\[\frac{d}{dx}\left[\mu(x)y\right]=g(x)\mu(x)\]
by simply integrating, i.e.
\[\mu(x)y=\int g(x)\mu(x)\,dx~~\implies~~y=\frac{1}{\mu(x)}\int g(x)\mu(x)\,dx\]

- anonymous

PS: "Without going into too much detail..." by which I mean I'm assuming you haven't tackled ODEs with methods beyond separating variables.

- zzr0ck3r

loved it

- anonymous

I aim to please ;)

- zzr0ck3r

linear operators at the rescue

- Astrophysics

Ha, nice! Thanks @SithsAndGiggles And yeah I just started ODE xD

- Astrophysics

Thanks @zzr0ck3r @freckles @SithsAndGiggles

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