Separable Equations?

[mystic gyro]

capitalize P, and change the x to h
your function is P(h), not P(x)

then solve for P

lmk what you get then we continue

yep

dP/dh = -kP

nah I said you should solve for P first

you got to
(ln|P| - C) / h = k

then solve for P instead of trying to put k and C in at this step

you can put your C and k in later

then can you solve for ln|P|

ln|P| = kh + C instead

then you can reverse ln by doing e^ to both sides

oh right

after e^ing both sides, what do you get

yep, P = ±e^(-kh+C)

(|| reverses to ±)

from here, its easier to gauge how P should behave based on h

yep

(and the ± disappears since 0.4 = -e^C has no real solutions)

well we originally had ln|P| instead of ln(P)

so if you have ln|y| = x, then thats y = ±e^x

compare to ln(y) = x which would just show y = e^x

[mystic gyro]

the ± would then disappear here from that initial condition

[swift mason]

Solved

Post marked as solved by @umbral marsh.

Use .unsolved if this was a mistake.

[mystic gyro]

np

another thing about what you did originally:

you could find k and C from here, the problem was dividing by h, so once that problem is missing, you can find k and C as usual

this also would avoid the ± problem by never having to reverse ||

in general its best practice to find P if it looks simple to do so, so that you can sort of get any value P(h)