Salt flows in at a rate of
(2 g/L) * (4 L/min) = 8 g/min
and out at a rate of
(B/(200 + t) g/L) * (3 L/min) = 3B/(200 + t) g/min
where B is the amount of salt in the tank at time t.
Then the net rate at which B changes is governed by the ODE,
[tex]B'=8-\dfrac{3B}{200+t}[/tex]
[tex]B'+\dfrac{3B}{200+t}=8[/tex]
Multipy both sides by [tex](200+t)^3[/tex]:
[tex](200+t)^3B'+3B(200+t)^2=8(200+t)^3[/tex]
[tex]\left(B(200+t)^3\right)'=8(200+t)^3[/tex]
Integrate both sides:
[tex]B(200+t)^3=2(200+t)^4+C[/tex]
[tex]B=2(200+t)+C(200+t)^{-3}=\dfrac{2(200+t)^4+C}{(200+t)^3}[/tex]
The tank starts with 30 g of salt, so B(0) = 30, which gives
[tex]30=2(200) + C(200)^{-3}\implies C=-2,960,000,000[/tex]
