Err...what?

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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 \[\frac{ d }{ dt }\ln|p-900|=1/2\]
I mean we can do a u - sub, but I find that weird for differentiation
I guess it involves the chain rule, but I'm not seeing it xD

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I should specify p=p(t)
\(\int\dfrac{1}{p-900}dp=\int\dfrac{1}{2}dt\\\ln(p-900)=\dfrac{1}{2}t+c\)
Yeah that's fine I just mean how to go from \[\frac{ dp/dt }{ p-900 }=1/2 \implies \frac{ d }{ dt } \ln|p-900|=1/2\]
This is the step before integration
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
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.
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
\(\frac{ d }{ dt } \ln|p-900|=1/2 \implies \dfrac{df(p)}{dt}=\dfrac{1}{2}\) which is odd to me.
the derivative of a function of p with respect to t is a constant 1/2 ?
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
I don't know man, maybe someone else will help. I suck at calculus...
Haha, no worries man, I was like I should know this...but I had no clue, thanks a lot for the help though!
:(
The solution is the same as yours I just don't know why it's written this way
Do they end with \(p=Ce^{\frac{1}{2}t}+900\)?
Yup
yeah, I dont know what that step is about then.
Yeah neither do I haha, thanks for checking though!
\[\frac{d}{dt} \ln|p-900| \text{ does equal } (p-900)' \cdot \frac{1}{p-900} =p' \cdot \frac{1}{p-900}=\frac{dp}{dt} \cdot \frac{1}{p-900}\] but yeah I would have never written it like the way they did
Yeah so they sort of our just going backwards? Just a weird way to introduce dif equations I thought haha, thanks @freckles
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. \[\mu(x)y=\int g(x)\mu(x)\,dx~~\implies~~y=\frac{1}{\mu(x)}\int g(x)\mu(x)\,dx\]
PS: "Without going into too much detail..." by which I mean I'm assuming you haven't tackled ODEs with methods beyond separating variables.
loved it
I aim to please ;)
linear operators at the rescue
Ha, nice! Thanks @SithsAndGiggles And yeah I just started ODE xD

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