Let [tex]S(t)[/tex] be the amount of sugar in the tank at time [tex]t[/tex].
a. The tank contains only pure water at the start, so [tex]\boxed{S(0)=0}[/tex].
b. The inflow rate of sugar is
[tex]{S_{\rm in}}'=\left(0.05\dfrac{\rm kg}{\rm L}\right)\left(5\dfrac{\rm L}{\rm min}\right)=\dfrac1{40}\dfrac{\rm kg}{\rm min}[/tex]
and the outflow rate is
[tex]{S_{\rm out}}'=\left(\dfrac S{2340}\dfrac{\rm kg}{\rm L}\right)\left(5\dfrac{\rm L}{\rm min}\right)=\dfrac S{468}\dfrac{\rm kg}{\rm min}[/tex]
so the net rate at which [tex]S(t)[/tex] changes over time is governed by
[tex]S'=\dfrac1{40}-\dfrac S{468}\implies S'+\dfrac S{468}=\dfrac1{40}[/tex]
Multiply both sides by [tex]e^{t/468}[/tex],
[tex]e^{t/468}S'+\dfrac{e^{t/468}}{468}S=\dfrac{e^{t/468}}{40}[/tex]
and condense the left side as the derivative of a product,
[tex]\left(e^{t/468}S\right)'=\dfrac{e^{t/468}}{40}[/tex]
Integrate both sides to get
[tex]e^{t/468}S=\dfrac{117e^{t/468}}{10}+C[/tex]
and solve for [tex]S[/tex]:
[tex]S=\dfrac{117}{10}+Ce^{-t/468}[/tex]
With [tex]S(0)=0[/tex], we find [tex]C=-\dfrac{117}{10}=-11.7[/tex], so that
[tex]\boxed{S(t)=11.7-11.7e^{-t/468}}[/tex]
c. As [tex]t\to\infty[/tex], the exponential term will converge to 0, leaving a fixed amount of 11.7 kg of sugar in the solution.
