Question

The bottom of a deposit, of 300l of capacity, is cover by a layer of salt. Assuming that the velocity at which the salt is disolved is proportional to the difference between the concentration at this time and the concentration of the saturated solution ( 1 kg of salt for 3 liters of water), and also take into account that the amount of pure water disolve 1/3 de Kg of salt per minute. Find the ODE giving the amount of salt x(t), as a function of time, integrate it, and determine the amount of salt in the solution after one hour.Find the solutions of the ODE that satisfy the indicated conditions when x → ±[infinity]

Answer 1

Explanation:

Assuming x(t) to be the number of salt

The maximum bottom of deposits =300 Litre

Initial Concentration of the saturated salt solution = 1 kg of salt for 3 liters of water

`x_ic_i= \frac{1}3`

⇒x = 0

The differential equation is :

`(dx(t))/(dt)`
`={x_ic_i}-{x_oc_o}`

`= (1)/(3) - (x(t))/(300)`

`(dx(t))/(dt) + (x(t))/(300) = (1)/(3)`

Using the first order differential equation of the form :

x' + p(t) x = q(t)

where

`p(t) = (1)/(300)` ;
`q(t) = (1)/(3)`

By integrating the factor:

`\mu = e^(\int\limitsP(t)dt ) \\ \\ =e^{\int\limits (1)/(300)dt } \\ \\ = e^{(t)/(300) }`

`x(t) = (1)/(\mu(t) )\int\limits \mu (t) * q(t) dt \\ \\ = e ^(- t/300) \ \ [\int\limits e ^(+ t/300) * (1)/(3)dt ]`

`= e ^(- t/300) \ \ [(1)/(3) * (e^(t/300))/(1/300) + C ]`

`x(t) = 100 + Ce ^{-t/300`

Thus the differential equation solution is :
`x(t) = 100 + Ce ^{-t/300`

However; we have initial concentration x(0) = 0

SO;

`0 = 100 + Ce^(-0/300)`

C = -100

`x(t) = 100 -100e ^{-t/300`

when x → ±[infinity]

`\infty = 100 + Ce^(- \infty /300)`

`x(t) = 100 -100e ^{-t/300`

Now; the amount of salt after an hour is as follows:

1 hour = 60 minutes

t = 60 min

`x(60) = 100 -100e ^{-60/300`

= 100 -100 (0.818)

= 100 - 81.8

= 18.2

x(60) = 18.2 kg