Answer :
Answer:
387.87 lb
Step-by-step explanation:
Rate of flow of salt solution, [tex]\frac{dy}{dt}[/tex] = (1 lb/gal * 3 gal/min) - ([tex]\frac{y}{300 + (3-1)t}[/tex]lb/gal * 1 gal/min)
[tex]\frac{dy}{dt}[/tex] = 3 - [tex]\frac{y}{300 + 2t}[/tex]
[tex]\frac{dy}{dt}[/tex] + [tex]\frac{y}{300 + 2t}[/tex] = 3
The linear differential equation with Q = [tex]\frac{1}{2t+300}[/tex] , R = 3
I.F. = e[tex]\int\limits {\frac{1}{2t+300} } \, dt[/tex] = [tex](t+150)^{\frac{1}{2} }[/tex]
y * I.F. = [tex]\int\limits {Q *I.F.} \, dt+c[/tex]
y[tex](t+150)^{\frac{1}{2} }[/tex] = [tex]\int\limits {3*(t+150)^{\frac{1}{2} }} \, dt+c[/tex]
At y = 0, t = o, c = [tex]-2(150)^{\frac{3}{2} }[/tex]
y[tex](t+150)^{\frac{1}{2} }[/tex] = [tex]2(t+150)^{\frac{3}{2} }[/tex] [tex]-2(150)^{\frac{3}{2} }[/tex]
y = 2(t + 150) - [tex]\frac{2(150)\frac{3}{2} }{(t+150)\frac{1}{2} }[/tex]
Net rate of solution accumulation = (3 - 1) gal/min = 2 gal/min
300 gallon was already there when t = 0.
Time taken to fill the gallon = 300 gal / 2 min/gal = 150 min
Salt content at t=150 min is given by
y = 2(150+150) - [tex]\frac{2(150)\frac{3}{2} }{(150+150)\frac{1}{2} }[/tex]
y = 387.87 lb