Answer :
The quantity of salt in the tank at period t equivalents A(t) = 200 - 130 e(-t/40).
Initially, the tank has A(0) = 50 g of salt.
The rate of salt entryway is (1 g/L) × (5 L/min) = 5 g/min
And a rate of (A(t) ÷ 200 g/L) × (5 L/min) = (A(t) ÷ 40) g/min,
As an outcome, the quantity of salt in the tank modifications as
A'(t) = (5 - A(t)) ÷ 40.
Solve ODE for A(t):
A'(t) + (A(t) ÷ 40) = 4
[tex]e^{\frac{(t)}{40} }[/tex] A'(t) + ([tex]e^{\frac{(t)}{40} }[/tex] ÷ 40 A(t)) = 4[tex]e^{\frac{(t)}{40} }[/tex]
([tex]e^{\frac{(t)}{40} }[/tex] A(t))' = 4[tex]e^{\frac{(t)}{40} }[/tex]
[tex]e^{\frac{(t)}{40} }[/tex] A(t) = 160[tex]e^{\frac{(t)}{40} }[/tex] + C
A(t) = 160 + C[tex]e^{\frac{(t)}{40} }[/tex]
Given A(0) = 30, conclude that
30 = 160 + C
C = -130,
This denotes that the quantity of salt in the tank at period t equivalents
A(t) = 200 - 130 e(-t ÷ 40).
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