There's no salt in the tank at the start, so A(0) = 0.
Salt flows in at a rate of
(2 lb/gal) * (5 gal/min) = 10 lb/min
and flows out at a rate of
(A(t)/500 lb/gal) * (5 gal/min) = A(t)/100 lb/min
Then the net flow rate is given by
A'(t) = 10 - A(t)/100
in gal/min.
Solve the ODE:
[tex]A'(t)+\dfrac{A(t)}{100}=10[/tex]
[tex]e^{t/100}A'(t)+e^{t/100}\dfrac{A(t)}{100}=10e^{t/100}[/tex]
[tex]\left(e^{t/100}A(t)\right)'=10e^{t/100}[/tex]
[tex]e^{t/100}A(t)=1000e^{t/100}+C[/tex]
[tex]A(t)=1000+Ce^{-t/100}[/tex]
With A(0) = 0, the above gives us 0 = 1000 + C, so that C = -1000, and the particular solution to this IVP is
[tex]A(t)=1000(1-e^{-t/100})[/tex]
