Answer:
The amount of salt in the tank at any time t is [tex]A(t)=120\gamma\gamma(1-e^{-t/60} )[/tex]
Explanation:
By definition, we have that the change rate of salt in the tank is [tex]\frac{dA}{dt}=R_{i}-R_{o}[/tex], where [tex]R_{i}[/tex] is the rate of salt entering and [tex]R_{o}[/tex] is the rate of salt going outside.
Then we have, [tex]R_{i}=2\frac{L}{min}*\gamma\gamma\frac{g}{L}=2 \gamma\gamma\frac{g}{min}[/tex], and
[tex]R_{o}=2\frac{L}{min}*\frac{A}{120} \frac{g}{L}=\frac{A}{60}\frac{g}{min}[/tex]
So we obtain. [tex]\frac{dA}{dt}=2 \gamma\gamma-\frac{A}{60}[/tex], then
[tex]\frac{dA}{dt}+\frac{A}{60}=2 \gamma\gamma[/tex], and using the integrating factor [tex]e^{\int {\frac{1}{60}} \, dt=e^{\frac{t}{60}[/tex], therefore
[tex](\frac{dA }{dt}+\frac{A}{60}}{60}=2 \gamma\gamma)e^{\frac{t}{60}[/tex], we get [tex]\frac{d}{dt}(A*e^{\frac{t}{60}})= 2 \gamma\gamma e^{\frac{t}{60}[/tex], after integrating both sides [tex]A*e^{\frac{t}{60}}= 120 \gamma\gamma e^{\frac{t}{60}}+C[/tex], therefore [tex]A(t)= 120 \gamma\gamma +Ce^{\frac{-t}{60}}[/tex], to find [tex]C[/tex] we know that the tank initially contains 120L of pure water, that means the initial conditions [tex]A(0)=0[/tex], so [tex]0= 120 \gamma\gamma +Ce^{\frac{-0}{60}}[/tex]
[tex]0= 120 \gamma\gamma +C\\C= -120 \gamma\gamma[/tex]
Finally we can write an expression in terms of γγ for the amount of salt in the tank at any time t, it is [tex]A(t)= 120 \gamma\gamma -120 \gamma\gamma e^{\frac{-t}{60}}\\A(t)= 120 \gamma\gamma (1-e^{\frac{-t}{60}})[/tex]
