The variables $w$, $x$ and $y$ are connected by the following differential equations: $e^{2x}\frac{dw}{dx}=w^{2}$ and $\frac{dy}{dx}=w\,.$ Verify that $w=\dfrac{2}{e^{-2x}+A}$ is the general solution of $e^{2x}\dfrac{dw}{dx}=w^{2}\,,\text{ where A is an arbitrary constant. }$ Hence find $y$ in terms of $x$. A tank initially contains 50 grams of salt dissolved in 100 litres of water. Brine that contains 2 grams of salt per litre of brine flows into the tank at a rate of 5 litres per minute. The solution is kept thoroughly mixed and flows out from the tank at a rate of 5 litres per minute. Given that the amount of salt in the tank at time $t$ minutes is given by $S$, show that $\dfrac{dS}{dt}=\dfrac{200-S}{20}\,.$ Hence find the time, in minutes, at which the concentration of salt in the tank reaches 1 gram per litre.

1. Separate the variables $w$ and $x$, and integrate both sides.

$e^{2x} \frac{dw}{dx} = w^2$
$\frac{1}{w^2} \,dw = e^{-2x} \,dx$
$\int w^{-2} \,dw = \int e^{-2x} \,dx$
$-\frac{1}{w} = -\frac{1}{2}e^{-2x} + C$

2. Rearrange the integrated equation to make $w$ the subject and deduce the form of the general solution.

Multiply the equation by $-2$:
$\frac{2}{w} = e^{-2x} - 2C$
$w = \frac{2}{e^{-2x} - 2C}$
Since $C$ is an arbitrary constant of integration, let $A = -2C$, which is also an arbitrary constant.
$w = \frac{2}{e^{-2x} + A} \quad \text{(Verified)}$

1. Substitute the verified expression for $w$ into the differential equation for $y$ and set up the integral.

$\frac{dy}{dx} = w = \frac{2}{e^{-2x}+A}$
$y = \int \frac{2}{e^{-2x}+A} \, dx$

2. Multiply the numerator and denominator of the integrand by $e^{2x}$ to simplify the expression into a recognizable derivative form.

$y = \int \frac{2e^{2x}}{e^{2x}(e^{-2x}+A)} \, dx$
$y = \int \frac{2e^{2x}}{1+Ae^{2x}} \, dx$

3. Integrate using the standard form $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c$.

$y = \frac{1}{A} \int \frac{2Ae^{2x}}{1+Ae^{2x}} \, dx$
$y = \frac{1}{A} \ln|1+Ae^{2x}| + C$
(where $C$ is an arbitrary constant)

1. Separate variables and integrate the differential equation.

$\frac{dS}{dt} = \frac{200 - S}{20}$
$\int \frac{1}{200 - S} \, dS = \int \frac{1}{20} \, dt$
$-\ln|200 - S| = \frac{1}{20}t + C$

2. Express $S$ explicitly in terms of $t$.

$\ln|200 - S| = -\frac{1}{20}t - C$
$200 - S = A e^{-\frac{1}{20}t}$
$S = 200 - A e^{-\frac{1}{20}t}$
(where $A = \pm e^{-C}$)

3. Use the initial condition to find the constant $A$.

Initially, the tank contains $50$ grams of salt. At $t = 0$, $S = 50$:
$50 = 200 - A e^0$
$A = 150$
So the particular solution is:
$S = 200 - 150 e^{-\frac{1}{20}t}$

4. Determine the target amount of salt $S$ for the given concentration.

Since the inflow and outflow rates are both $5 \text{ litres/min}$, the volume of the mixture remains constant at $100 \text{ litres}$.
For a target concentration of $1 \text{ g/litre}$:
$\text{Target } S = 1 \text{ g/litre} \times 100 \text{ litres} = 100 \text{ g}$

5. Substitute the target $S$ to form an equation in $t$.

$100 = 200 - 150 e^{-\frac{1}{20}t}$
$150 e^{-\frac{1}{20}t} = 100$
$e^{-\frac{1}{20}t} = \frac{2}{3}$

6. Solve the exponential equation for $t$.

$-\frac{1}{20}t = \ln\left(\frac{2}{3}\right)$
$t = -20 \ln\left(\frac{2}{3}\right) = 20 \ln 1.5$
$t \approx 8.11 \text{ minutes (to 3 s.f.)}$