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Astrophysics
 one year ago
Err...what?
Astrophysics
 one year ago
Err...what?

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0If we have \[\frac{ dp }{ dt } = \frac{ p900 }{ 2 } \implies \frac{ dp/dt}{p900 }= \frac{ 1 }{ 2 }\] so how does this go to \[\frac{ d }{ dt }\lnp900=1/2\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I mean we can do a u  sub, but I find that weird for differentiation

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I guess it involves the chain rule, but I'm not seeing it xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I should specify p=p(t)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(\int\dfrac{1}{p900}dp=\int\dfrac{1}{2}dt\\\ln(p900)=\dfrac{1}{2}t+c\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah that's fine I just mean how to go from \[\frac{ dp/dt }{ p900 }=1/2 \implies \frac{ d }{ dt } \lnp900=1/2\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0This is the step before integration

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Though 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
 one year ago
Best ResponseYou've already chosen the best response.1I have no idea... it makes no sense to me I did \[\dfrac{dp}{dt}\dfrac{1}{p900}=\dfrac{1}{2}\\\implies \dfrac{1}{p900}dp=\dfrac{1}{2}dt\] Then I integrate both sides.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.1\(\frac{ d }{ dt } \lnp900=1/2 \implies \dfrac{df(p)}{dt}=\dfrac{1}{2}\) which is odd to me.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1the derivative of a function of p with respect to t is a constant 1/2 ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ok so this is part of a modelling problem it starts off as \[\frac{ dp }{ dt } = \frac{ p900 }{ 2 } \implies \frac{ dp/dt }{ p900 } = \frac{ 1 }{ 2 } (eq 6)\] then it says by the chain rule the left side of the equation 6 is the derivative of lnp900 with respect to t, so we have \[\frac{ d }{ dt }\lnp900=1/2\] then they integrate

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I don't know man, maybe someone else will help. I suck at calculus...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, no worries man, I was like I should know this...but I had no clue, thanks a lot for the help though!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0The solution is the same as yours I just don't know why it's written this way

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Do they end with \(p=Ce^{\frac{1}{2}t}+900\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1yeah, I dont know what that step is about then.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah neither do I haha, thanks for checking though!

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dt} \lnp900 \text{ does equal } (p900)' \cdot \frac{1}{p900} =p' \cdot \frac{1}{p900}=\frac{dp}{dt} \cdot \frac{1}{p900}\] but yeah I would have never written it like the way they did

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah so they sort of our just going backwards? Just a weird way to introduce dif equations I thought haha, thanks @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0PS: "Without going into too much detail..." by which I mean I'm assuming you haven't tackled ODEs with methods beyond separating variables.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1linear operators at the rescue

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ha, nice! Thanks @SithsAndGiggles And yeah I just started ODE xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @zzr0ck3r @freckles @SithsAndGiggles
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