Question

Verify that this is a solution of the differential equation

Answer 1

\[\frac{dP}{dt}=P(1-P) , P=\frac{c_1e^t}{1+c_1e^t}\]

Answer 2

\[\frac{dP}{dt}=[c_1e^t(1+c_1e^t)]'\]

Answer 3

just take the derivative and show that it is the same as \(P(1-P)\)
that proves that it is a potential solution

Answer 4

i'm not getting the right derivativei don't beilve

Answer 5

\[\frac{dP}{dt}=\frac{(-c_1e^t)^2+c_1e^t+(c_1e^t)^2}{(1+c_1e^t)^2}\]

Answer 6

\[P=\frac{c_1e^t}{1+c_1e^t}\]\[u=c_1e^t\]\[v=1+c_1e^t\]\[P'=\frac{u'v-v'u}{v^2}=c_1\frac{e^t+e^2t-e^2t}{(1+c_1e^t)^2}=\frac{c_1e^t}{(1+c_1e^t)^2}=P(\frac1{1+c_1e^t})\]I guess we can use some algebra to show that last part is (1-P)

Answer 7

thats what i got

Answer 8

\[\frac1{1+c_1e^t}=1-P\]just try it
so you wind up with \[P'=P(1-P)\]which is what we wanted

Answer 9

not seeing it?

Answer 10

hold on let me do therest when i got to plugging in it looked completely off equal

Answer 11

yeah it was right it was not simplifying that caused it to look wrong

Answer 12

you could also use long division\[\frac1{1+c_1e^t}=1-\frac{c_1e^t}{1+c_1e^t}=1-P\]