Answer:
Step-by-step explanation:
From the information given:
Tank's Capacity = 600 gal
Original content of water = 200 gal
Salt solution = 100 lb
In the tank:
Suppose x(t) = amount of salt at time (t); &
V(t) = volume of water
Then,
V(t) = 200 + t
and [tex]x'(t) = 3 - \dfrac{2x(t)}{200+t}[/tex]
From the above linear equation, the integrating factor can be computed as:
[tex]x(t) = \Bigg(e^{\int \dfrac{2}{200+t}} \Bigg)^{dt}[/tex]
[tex]= e^{2 \ log (200+t)}[/tex]
= (200 + t)²
The general solution can now be expressed as:
[tex]x(t) = \dfrac{1}{(200+t)^2}\Bigg( \int 3(200+t)^2 \ dt +C\Bigg)[/tex]
We know that C = integrating factor, thus taking the integral:
[tex]x(t) = \dfrac{(200+t)^3 +C}{(200+t)^2}[/tex]
At the initial condition, x(0) = 100
∴
[tex]100= \dfrac{(200)^3 +C}{(200)^2}[/tex]
C = ((200)²×100) - 200³
C = -4 × 10⁶
Hence, at any time t, the amount of salt is:
[tex]x(t) = \dfrac{(200+t^2) - 4\times 10^6}{(200+t)^2}[/tex]
