Let [tex]q[/tex] be the unknown amount of the chemical originally in the pond, so [tex]Q(0)=q[/tex].
a. The incoming water introduces the chemical at a rate of
[tex]Q'_{\rm in}=\left(0.1\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac25\dfrac{\rm kg}{\rm hr}[/tex]
and the mixture flows out at a rate of
[tex]Q'_{\rm out}=\left(\dfrac Q{2760}\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac Q{690}\dfrac{\rm kg}{\rm hr}[/tex]
so that the net rate of change (in kg/hr) of the chemical in the pond is given by the differential equation,
[tex]\boxed{Q'=\dfrac25-\dfrac Q{690}}[/tex]
b. The ODE is linear; multiplying both sides by [tex]e^{t/690}[/tex] gives
[tex]e^{t/690}Q'+\dfrac{e^{t/690}}{690}Q=\dfrac{2e^{t/690}}5[/tex]
Condense the left side into the derivative of a product:
[tex]\left(e^{t/690}Q\right)'=\dfrac{2e^{t/690}}5[/tex]
Integrate both sides to get
[tex]e^{t/690}Q=276e^{t/690}+C[/tex]
and solve for [tex]Q[/tex] to get
[tex]Q=276+Ce^{-t/690}[/tex]
The pond starts with [tex]q[/tex] kg of the chemical, so when [tex]t=0[/tex] we have
[tex]q=276+C\implies C=q-276[/tex]
so that the amount of chemical in the water at time [tex]t[/tex] is
[tex]Q(t)=276+(q-276)e^{-t/690}[/tex]
As [tex]t\to\infty[/tex], the exponential term will converge to 0, leaving a fixed amount of 276 kg of the chemical in the pond.
