Question

How do you calculate the explicit solution to dP/dt=kP-h?
(k and h are positive constants) How do you calculate the explicit solution to dP/dt=kP-h?
(k and h are positive constants) @Mathematics

Answer 1

this is a first order linear diffy q right?

Answer 2

P' - kP = -h

is the usual set up, but im cant recall to well what an explicit solution refers to

Answer 3

multiply both sides by e^{-kt}

P'e^(-kt) - k e^(-kt)P = -h e^(-kt)

this creates a situation where the left side is the result of a product rule from derivatives

int (P'e^(-kt) - k e^(-kt)P = -h e^(-kt)) dt

e^(-kt)P = -h * int (-h e^(-kt)) dt

Answer 4

this familiar? or do we do separation of variables ?

Answer 5

e^(-kt)P = -h * int (e^(-kt)) dt

e^(-kt)P = -h * (e^(-kt)/-k + C)

e^(-kt)P = -he^(-kt)/-k - hC

P = h/k - hC e^(kt)

is this result a general or explicit? i cant recall

Answer 6

\[dP=(kP-h) dt\]
\[\frac{1}{(kP-h)}dP= dt\]
\[\int\frac{1}{(kP-h)}dP= \int dt\]
\[\frac{1}{k}\int\frac{k}{(kP-h)}dP= t + C\]
\[\frac{1}{k}ln|kP-h|= t + C\]
\[ln|kP-h|= kt + kC\]
\[exp(ln|kP-h|= kt + kC\]
\[kP-h= e^{kt} + C\]
\[kP= e^{kt} + C+h\]
\[P= \frac{e^{kt}}{k} + C\]
maybe

Answer 7

http://www.wolframalpha.com/input/?i=dP%2Fdt%3DkP-h
the wolf likes my first getup

Answer 8

That looks right.
Many thanks!

Answer 9

yep, i get to take diffy qs next term :)

Answer 10

You must be some kind of genius if you can do that and not have diffy qs!

Answer 11

genius maybe :) but i see an algebra mistake on this last bit

\[ln|kP-h|= kt + kC\]
\[e(ln|kP-h|= kt + kC)\]
\[kP-h= e(kt + kC)\]
\[kP-h= e^{kt} * e^{kC}\]
\[kP-h= c_1e^{kt}\]
\[kP= c_1e^{kt}+h\]
\[P= c_1e^{kt}+\frac{h}{k}\]
so it comes out them same; the constant on the "e" tends to absorb any constants that get near it