Let [tex]A(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex]. We're given that the tank initially holds [tex]A(0)=100[/tex] lbs of salt.
The rate at which salt flows in and out of the tank is given by the relation
[tex]\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}[/tex]
[tex]\implies A'(t)+\dfrac{11}{200+33t}A(t)=484[/tex]
Find the integrating factor:
[tex]\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}[/tex]
Distribute [tex]\mu(t)[/tex] along both sides of the ODE:
[tex](200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}[/tex]
[tex]\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}[/tex]
[tex]A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt[/tex]
[tex]A(t)=22(200+33t)^{2/3}+C[/tex]
Since [tex]A(0)=100[/tex], we get
[tex]100=22(200)^{2/3}+C\implies C\approx-652.39[/tex]
so that the particular solution for [tex]A(t)[/tex] is
[tex]A(t)=22(200+33t)^{2/3}-652.39[/tex]
The tank becomes full when the volume of solution in the tank at time [tex]t[/tex] is the same as the total volume of the tank:
[tex]200+(44-11)t=500\implies 33t=300\implies t\approx9.09[/tex]
at which point the amount of salt in the solution would be
[tex]A(9.09)\approx733.47\text{ lb}[/tex]
