Q.Let a tank contain $1000$ litres of brine which contains $250$ g of salt per litre. Brine containing $200$ g of salt per litre flows into the tank at the rate of $25$ litres per minute and the mixture flows out at the same rate. Assume that the mixture is kept uniform all the time by stirring. What would be the amount of salt in the tank at any time $t$?

Let $y(t)$ be the salt (kg) at time $t$, with $y(0) = 250$. Salt enters at $25 \times 0.2 = 5$ kg/min and leaves at $25 \times \dfrac{y}{1000} = \dfrac{y}{40}$ kg/min, so

$\frac{dy}{dt} + \frac{y}{40} = 5.$ …

Step 1 — Identify. Track the total salt in a well-stirred tank whose volume stays constant (equal inflow and outflow).

Step 2 — Set up variables and initial condition. Let $y = y(t)$ denote the amount of salt (in kg) in the tank $t$ minutes after the flow starts; assume $y$ is differentiable. The volume stays at $1000$ litres because inflow and outflow are both $25$ L/min. Initially the brine holds $250$ g/L over $1000$ L, so

Step 3 — Mathematical formulation. Change in $y$ comes from inflow minus outflow.

Inflow: $25\text{ L/min} \times 200\text{ g/L} = 5000\text{ g/min} = 5\text{ kg/min}$.

Outflow: at time $t$ the tank holds $\tfrac{y}{1000}$ kg of salt per litre, so the leaving rate is $25 \times \dfrac{y}{1000} = \dfrac{y}{40}$ kg/min.

Hence

Step 4 — Solve. Equation $(1)$ is linear with integrating factor $e^{\int \frac{1}{40}\,dt} = e^{t/40}$. Then

so

Applying $y(0)=250$: $250 = 200 + C \Rightarrow C = 50$. Therefore

Solving $(3)$ for $t$ gives the time at which the salt equals $y$ kg:

$t = 40\,\log_e\!\left(\frac{50}{\,y - 200\,}\right). \quad(4)$ …

Method: Setting Up and Solving a Rate-In/Rate-Out Linear Differential Equation

This is the standard technique for any "mixing tank" / "inflow-outflow" word problem: a quantity accumulates from an inflow rate and depletes from an outflow rate that depends on the current amount.

Steps

Step 1: Define the unknown function and its initial condition

Let $y(t)$ be the quantity of interest (salt, concentration, population, …) at time $t$. Read the starting amount directly off the problem statement to get $y(0)$.

Step 2: Write inflow rate and outflow rate separately

Inflow rate = (rate of liquid entering) × (concentration entering) — usually a constant. Outflow rate = (rate of liquid leaving) × (current concentration in the tank) = (rate of liquid leaving) × $\dfrac{y}{\text{volume}}$ — this is where $y$ itself enters the rate, making the equation a genuine ODE rather than plain arithmetic.

Step 3: Form the differential equation

and rearrange into the standard linear first-order form

$\frac{dy}{dt} + P(t)\,y = Q(t).$ …

Common Mistakes

Mistake 1: Computing the outflow rate with the incoming concentration instead of the current concentration

Writing the outflow as $25\times200 = 5000$ g/min (using the inflow concentration) instead of $25\times\dfrac{y}{1000}$ (using the current tank concentration). Why it's wrong: the mixture leaving the tank has whatever concentration is currently inside it (since it's kept uniform by stirring), not the concentration of what's flowing in — that's precisely what makes this a genuine differential equation rather than simple arithmetic. Correct approach: outflow rate must always be (rate out) $\times \dfrac{y(t)}{\text{tank volume}}$.

Mistake 2: A sign error in the rate equation …