Question

Err...what?

Answer 1

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 \[\frac{ d }{ dt }\ln|p-900|=1/2\]

Answer 2

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

Answer 3

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

Answer 4

I should specify p=p(t)

Answer 5

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

Answer 6

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

Answer 7

This is the step before integration

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

:(

Answer 17

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

Answer 18

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

Answer 19

Yup

Answer 20

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

Answer 21

Yeah neither do I haha, thanks for checking though!

Answer 22

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

Answer 23

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

Answer 24

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.
\[\mu(x)y=\int g(x)\mu(x)\,dx~~\implies~~y=\frac{1}{\mu(x)}\int g(x)\mu(x)\,dx\]

Answer 25

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

Answer 26

loved it

Answer 27

I aim to please ;)

Answer 28

linear operators at the rescue

Answer 29

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

Answer 30

Thanks @zzr0ck3r @freckles @SithsAndGiggles