A large reservoir initially contains pure water. Water containing salt begins to flow into the reservoir. The solution is kept uniform by stirring and leaves the reservoir through an outlet at its base. Let $x$ grams represent the amount of salt in the reservoir and let $t$ minutes represent the time since the salt water began flowing into the reservoir.

• Attempting to multiply through by $(t + 1)$ and rearrange M1

• $(t + 1)\frac{dx}{dt} + x = 10(t + 1)e^{-\frac{t}{4}}$ A1

• $\frac{d}{dt}\left(x(t + 1)\right) = 10(t + 1)e^{-\frac{t}{4}}$

• $x(t + 1) = \int{10(t + 1)e^{-\frac{t}{4}}}dt$ A1

• Attempting to integrate the RHS by parts M1

• $= -40(t + 1)e^{-\frac{t}{4}} + 40\int{e^{-\frac{t}{4}}}dt$

• $= -40(t + 1)e^{-\frac{t}{4}} - 160e^{-\frac{t}{4}} + C$ A1

Condone the absence of $C$.

Method #1

• Substituting $t = 0, x = 0 \Rightarrow C = 200$ M1

• $x = \frac{-40(t + 1)e^{-\frac{t}{4}} - 160e^{-\frac{t}{4}} + 200}{t + 1}$ A1

• Using $-40e^{-\frac{t}{4}}$ as the highest common factor of $-40(t + 1)e^{-\frac{t}{4}}$ and $-160e^{-\frac{t}{4}}$ M1

Method #2

• Using $-40e^{-\frac{t}{4}}$ as the highest common factor of $-40(t + 1)e^{-\frac{t}{4}}$ and $-160e^{-\frac{t}{4}}$ giving

  $x(t + 1) = -40e^{-\frac{t}{4}}(t + 5) + C$ M1A1

• Substituting $t = 0, x = 0 \Rightarrow C = 200$ M1

Final step for both methods

• $x(t) = \frac{200 - 40e^{-\frac{t}{4}}(t + 5)}{t + 1}$ (Answer Given)

8 marks total

Method #1

• Integrating factor formula: $I(t) = e^{\int P(t) dt}$ M1

• $e^{\int \frac{1}{t + 1} dt}$

• $= e^{\ln(t + 1)}$ A1

• $= t + 1$

Method #2

• Attempt product rule differentiation on $\frac{d}{dt}(x(t + 1))$ M1

• $\frac{d}{dt}(x(t + 1)) = \frac{dx}{dt}(t + 1) + x$

• $= (t + 1)(\frac{dx}{dt} + \frac{x}{t + 1})$ A1

• Therefore, $t + 1$ is an integrating factor for this differential equation

2 marks total

• Recognize that we need to find the minimum value of $\frac{dx}{dt}$ M1

• Set up the equation for the second derivative:

  $\frac{d^2x}{dt^2} = -\frac{5}{2}e^{-\frac{t}{4}} + \frac{x}{(t+1)^2} - \frac{1}{t+1}\frac{dx}{dt}$

• Set $\frac{d^2x}{dt^2} = 0$ and solve for $t$

• Or, graph $\frac{dx}{dt}$ and find the minimum point

• The amount of salt is decreasing most rapidly at $t = 12.9$ minutes A1

2 marks total