Given that $z = e^{2x}\frac{dy}{dx}$, express $\frac{dz}{dx}$ in terms of $x$, $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$. Hence show that the differential equation $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = e^{-1-4x}$ can be reduced to $\frac{dz}{dx} = e^{-1-2x}.$ Solve this differential equation, expressing $y$ in terms of $x$. An aviary keeper started an insect breeding programme with an initial number of 5000 insects in a controlled environment to feed the birds in the aviary. The rate of birth of the insects is $\tfrac{1}{25}$ of the number $I$ (in thousands) of insects, at time $t$ days after the start of the breeding programme. The insects are also being fed to the birds at a constant rate of 0.25 (in thousands) per day. Show that the population of insects can be modelled with the differential equation $\frac{dI}{dt} = 0.04(I - 6.25).$ Hence solve the differential equation to obtain the particular solution of $I$ in terms of $t$. By sketching the solution curve, show that it is possible for the insect breeding programme to be depleted of insects after some time. After how many complete days will this happen?

1. Find an expression for $\frac{dz}{dx}$ by differentiating the given substitution with respect to $x$.

$z = e^{2x}\frac{dy}{dx}$
Using the product rule,
$\frac{dz}{dx} = e^{2x} \frac{d}{dx}\left(\frac{dy}{dx}\right) + \frac{dy}{dx} \frac{d}{dx}\left(e^{2x}\right)$
$\frac{dz}{dx} = e^{2x}\frac{d^2y}{dx^2} + 2e^{2x}\frac{dy}{dx}$

2. Show that the differential equation reduces to the required form.

Given the differential equation:
$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = e^{-1-4x}$
Multiply both sides by $e^{2x}$:
$e^{2x}\left(\frac{d^2y}{dx^2} + 2\frac{dy}{dx}\right) = e^{2x}(e^{-1-4x})$
$e^{2x}\frac{d^2y}{dx^2} + 2e^{2x}\frac{dy}{dx} = e^{-1-2x}$
Since the left-hand side is exactly $\frac{dz}{dx}$ from Step 1:
$\frac{dz}{dx} = e^{-1-2x} \quad \text{(Shown)}$

3. Solve for $z$ by integrating.

$\frac{dz}{dx} = e^{-1-2x}$
Integrating both sides with respect to $x$:
$z = \int e^{-1-2x} \, dx$
$z = -\frac{1}{2}e^{-1-2x} + A$
where $A$ is an arbitrary constant.

4. Substitute $z$ back to express $\frac{dy}{dx}$ in terms of $x$.

Substitute $z = e^{2x}\frac{dy}{dx}$ back into the equation:
$e^{2x}\frac{dy}{dx} = -\frac{1}{2}e^{-1-2x} + A$
Divide both sides by $e^{2x}$ (or multiply by $e^{-2x}$):
$\frac{dy}{dx} = -\frac{1}{2}e^{-1-4x} + Ae^{-2x}$

5. Solve for $y$ by integrating $\frac{dy}{dx}$.

$y = \int \left( -\frac{1}{2}e^{-1-4x} + Ae^{-2x} \right) \, dx$
$y = -\frac{1}{2}\left(\frac{e^{-1-4x}}{-4}\right) + A\left(\frac{e^{-2x}}{-2}\right) + B$
$y = \frac{1}{8}e^{-1-4x} - \frac{A}{2}e^{-2x} + B$
Letting $C_1 = -\frac{A}{2}$ and $C_2 = B$, we obtain the general solution:
$y = \frac{1}{8}e^{-1-4x} + C_1e^{-2x} + C_2$

1. Express the net rate of change of the insect population as the birth rate minus the consumption rate.

$\frac{dI}{dt} = \text{Rate of birth} - \text{Rate of feeding}$
$\frac{dI}{dt} = \frac{1}{25}I - 0.25$

2. Convert the fractional rate to a decimal and factorise to obtain the required form.

$\frac{dI}{dt} = 0.04I - 0.25$
$\frac{dI}{dt} = 0.04\left(I - \frac{0.25}{0.04}\right)$
$\frac{dI}{dt} = 0.04(I - 6.25) \quad \text{(Shown)}$

1. Separate the variables and integrate both sides.

$\int \frac{1}{I - 6.25} \, dI = \int 0.04 \, dt$
$\ln |I - 6.25| = 0.04t + c$

2. Rewrite the equation in exponential form to remove the logarithm.

$I - 6.25 = \pm e^{0.04t + c}$
$I - 6.25 = A e^{0.04t}$ (where $A = \pm e^c$)

3. Use the initial condition to determine the arbitrary constant $A$.

The initial number of insects is $5000$, so $I = 5$ when $t = 0$.
$5 - 6.25 = A e^{0.04(0)}$
$A = -1.25$

4. Express $I$ as the subject of the formula.

$I - 6.25 = -1.25e^{0.04t}$
$I = 6.25 - 1.25e^{0.04t}$

1. State the particular solution from (b)(ii) and describe its sketch to show depletion.

From (b)(ii), the particular solution is $I = 6.25 - 1.25e^{0.04t}$.

A sketch of $I$ against $t$ for $t \ge 0$ would show the $I$-intercept at $(0, 5)$. Since $e^{0.04t}$ increases as $t$ increases, the curve is strictly decreasing. The curve will eventually intersect the $t$-axis ($I=0$) and become negative. This shows that the insect population will drop to $0$ and be depleted after some time.

2. Set $I = 0$ to find the exact time of depletion and deduce the number of complete days.

To find when the insects are depleted, set $I = 0$:
$6.25 - 1.25e^{0.04t} = 0$
$1.25e^{0.04t} = 6.25$
$e^{0.04t} = 5$
$0.04t = \ln 5$
$t = 25\ln 5 \approx 40.236 \text{ days}$

Since $I > 0$ at $t = 40$ and $I < 0$ at $t = 41$, the population will be fully depleted after $41$ complete days.