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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

I should specify p=p(t)

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

This is the step before integration

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

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

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

:(

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!

loved it

I aim to please ;)

linear operators at the rescue

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