Answer:
1. [tex]\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}[/tex]
2. [tex]y(40) = 110.873 \ kg[/tex]
Step-by-step explanation:
Given that:
A very large tank initially contains 100 L of pure water.
Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.
. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.
As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L
t = (50 + t ) L
Since it is given that we should consider y(t) to be the amount of salt (in kilograms) in the tank after t minutes.
Then , the differential equation that y satisfies can be computed as follows:
[tex]\dfrac{dy}{dt}=rate_{in} - rate_{out}[/tex]
[tex]\dfrac{dy}{dt}=(0.8)(5) -\dfrac{y(t)}{100+2t} \times3[/tex]
[tex]\dfrac{dy}{dt}=(0.8)(5) -\dfrac{3y}{100+2t}[/tex]
[tex]\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}[/tex]
How much salt is in the tank after 40 minutes?
So,
suppose : [tex]e^{\int \dfrac{3}{100+2t} \ dt} = (t+50)^{3/2}[/tex]
Then ,
[tex]( t + 50)^{3/2} y' + \dfrac{3}{2}(t+50)^{1/2} y = 4(t+50)^{3/2}[/tex]
[tex]( t + 50)^{1.5} y' + \dfrac{3}{2}(t+50)^{0.5} y = 4(t+50)^{3/2}[/tex]
[tex][y\ (t + 50)^{1.5}]' = 4(t+ 50)^{1.5}[/tex]
Taking the integral on both sides; we have:
[tex][y(t + 50)^{1.5}] = 1.6 (t + 50)^{2.5} + C[/tex]
[tex]y = 1.6 (t+50)+C(t+50)^{-1.5}[/tex]
[tex]y(0) = 0 = 1.6(0+50) + C ( 0 + 50)^{-1.5}[/tex]
[tex]0 = 1.6(50) + C ( 50)^{-1.5}[/tex]
[tex]C= -1.6(50)^{2.5}[/tex]
[tex]y(40) = 1.6 (40 + 50)^1 - 1.6 (50)^{2.5}(50+40)^{-1.5}[/tex]
[tex]y(40) = 144 - 1.6 \times 17677.66953 (90)^{-1.5}[/tex]
[tex]y(40) = 144 - 1.6 \times 17677.66953 \times 0.001171213948[/tex]
[tex]y(40) = 144 - 33.12693299[/tex]
[tex]y(40) = 110.873 \ kg[/tex]
