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dh/(2.5-h)=0.016 dt
now integrating, -log(2.5-h)=0.016 t+c

Is this enough, ersp?

i think so my brains kinda died doing this work XD thanks :)

ok :)

actually i dont get where the minus log comes from when integrating

Comes from the -h...

\[\int\limits_{}{}\frac{dh}{2.5-h}=-\int\limits_{}{}\frac{(2.5-h)'dh}{2.5-h}=-\log (2.5-h)+c\]

I got bumped. Is this explanation okay?

i think so cheers

so what method is this?

Method? To solve the differential equation?

The d.e. is separable.

Or, you can just take the reciprocal of both sides and integrate directly, but it's the same thing.

thanks makes a bit more sense to me now, my mind is really failing me today!

np ;)

shouldnt both sides be logged?

lol, it happens :p

Thanks for fanning me, ersp :D

after that much help it seems logical! XD