Answer:
[tex]\mathbf{A(t) = 180 - 130e ^{-\dfrac{t}{30}}}[/tex]
Step-by-step explanation:
Given that:
A tank contains 180 liters of fluid in which 50 grams dissolved inside.
Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min
The salt pumped out[tex]= \dfrac{6 L}{180 L} = \dfrac{1}{30}[/tex] of initial amount added salt
At (t = 0) = 50
To determine the number A (t)
[tex]\dfrac{dA}{dt}=Rate_{in} - Rate _{out}[/tex]
[tex]A' = 6 - \dfrac{1}{30}A[/tex]
[tex]A' + \dfrac{1}{30}A = 6[/tex]
Integrating factor [tex]y = e^{\int\limits pdt[/tex]
[tex]y = e^{\int\limits \dfrac{1}{30}dt}[/tex]
[tex]y = e^{\dfrac{t}{30}}[/tex]
[tex](e^{ \frac{t}{30}}A)' =4 e ^{\dfrac{t}{30}}+c[/tex]
Taking integral on the both sides;
[tex]Ae ^{\dfrac{t}{30}}= 6 * 30 e^{\dfrac{t}{30}} + c[/tex]
[tex]A = 180+ ce^ {-\dfrac{t}{30}}[/tex]
At A(t = 0) = 50
50 = 180 + C (assuming C = [tex]ce ^{-\dfrac{t}{30}}[/tex])
C = 50 - 180
C = 130
[tex]\mathbf{A(t) = 180 - 130e ^{-\dfrac{t}{30}}}[/tex]
