A separable differential equation given this (d)/(dx) = x(1-x) where x(0)=0.1 is this correct:? is this correct:? (dx)/(dt) = x(1-x) t-t_0 = int(dx)/(x(1-x)) Where did the t and t_0 come from?

Davirnoilc

Davirnoilc

Answered question

2022-11-15

a separable differential equation
given this
d d x = x ( 1 x )
where
x ( 0 ) = 0.1
is this correct:?
d x d t = x ( 1 x )
t t 0 = d x x ( 1 x )
Where did the t and t 0 come from?

Answer & Explanation

Geovanni Shelton

Geovanni Shelton

Beginner2022-11-16Added 15 answers

From d x d t = x ( 1 x ) you get d x x ( 1 x ) = d t and then
d x x ( 1 x ) = d t = t t 0 .
d x x ( 1 x ) = ( 1 x + 1 1 x ) d x partial fractions = log x log ( 1 x ) + constant = log x 1 x + constant ,  so  log x 1 x = t t 0  where  t 0  is the “constant''. x 1 x = e t t 0 .  When  t = 0  then  x = 0.1 ,  so you have 0.1 1 0.1 = e 0 t 0 . 1 9 = e t 0 . e t t 0 = e t 9 . x 1 x = e t 9 . 9 x 1 x = e t . 9 x = e t x e t . 9 x x e t = e t . x ( 9 e t ) = e t . x = e t 9 e t = 1 9 e t 1 .
That last form makes it easy to find t as a function of x if you wanted that. Either of the last two forms is an explicit closed-form for x as a function of t, and shows x 1 as t + and x 0  as  t .
Normally I would leave most of this as an exercise, especially since this goes beyond what was asked, but the question seemed like something you wouldn't ask if all this weren't useful.
Fahdvfm

Fahdvfm

Beginner2022-11-17Added 3 answers

Using separation of variables, you get
d t = d x x ( 1 x )
So evaluating the lefthand side, you are left with
t t 0 = d x x ( 1 x )
Where t 0 is the constant of integration. I believe that you are solving a differential equation peratining to motion, so in this case t 0 would be the initial time or starting time.

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