Given that $z = e^{2x}\frac{dy}{dx}$, express $\frac{dz}{dx}$ in terms of $x$, $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$. Hence show that the differential equation $\displaystyle \frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} = e^{-1-4x}$ can be reduced to $\displaystyle \frac{dz}{dx} = e^{-1-2x}$. 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 $\displaystyle \frac{dI}{dt} = 0.04(I - 6.25).$ 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. Differentiate $z$ with respect to $x$ using the product rule.

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

2. Show the reduction of the given differential equation.

Given the differential equation:
$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = e^{-1-4x}$
Multiply both sides by $e^{2x}$:
$e^{2x}\left( \frac{d^2y}{dx^2} + 2\frac{dy}{dx} \right) = e^{2x} e^{-1-4x}$
Substitute the expression for $\frac{dz}{dx}$ derived in the previous step:
$\frac{dz}{dx} = e^{-1-2x} \text{ (Shown)}$

3. Integrate to find $z$, remembering the constant of integration.

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

4. Substitute back $z = e^{2x}\frac{dy}{dx}$ and make $\frac{dy}{dx}$ the subject.

$e^{2x}\frac{dy}{dx} = -\frac{1}{2}e^{-1-2x} + C$
Divide by $e^{2x}$ (or multiply by $e^{-2x}$):
$\frac{dy}{dx} = -\frac{1}{2}e^{-1-2x} \cdot e^{-2x} + C e^{-2x}$
$\frac{dy}{dx} = -\frac{1}{2}e^{-1-4x} + Ce^{-2x}$

5. Integrate $\frac{dy}{dx}$ to find the general solution for $y$.

$y = \int \left( -\frac{1}{2}e^{-1-4x} + Ce^{-2x} \right) dx$
$y = -\frac{1}{2} \left( \frac{e^{-1-4x}}{-4} \right) + C \left( \frac{e^{-2x}}{-2} \right) + D$
$y = \frac{1}{8}e^{-1-4x} - \frac{C}{2}e^{-2x} + D$
Letting $A = -\frac{C}{2}$ and $B = D$, we get:
$y = \frac{1}{8}e^{-1-4x} + Ae^{-2x} + B$

1. Separate the variables and integrate both sides.

$\int \frac{1}{I - 6.25} \, \mathrm{d}I = \int 0.04 \, \mathrm{d}t$
$\ln |I - 6.25| = 0.04t + C$

2. Convert the logarithmic equation to exponential form to simplify finding the constant.

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

3. Apply the initial condition to find the constant.

Given the initial number of insects is 5000, and $I$ is in thousands, we have $I = 5$ when $t = 0$.
$5 - 6.25 = A e^{0.04(0)}$
$A = -1.25$

4. Express $I$ explicitly in terms of $t$.

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

1. Identify the particular solution from part (b)(i) and sketch its curve to show the possibility of depletion.

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

A sketch of $I$ against $t$ would show:

• A starting $I$-intercept at $(0, 5)$.
• A strictly decreasing, concave-downwards curve.
• The curve crossing the horizontal $t$-axis at some $t > 0$.

Because the sketch shows the solution curve intersecting the $t$-axis (where $I = 0$), it demonstrates that the insect population will drop to zero, meaning depletion is possible.

2. Calculate the exact time when the population reaches zero.

Set $I = 0$ to find the time of depletion:
$6.25 - 1.25e^{0.04t} = 0$
$1.25e^{0.04t} = 6.25$
$e^{0.04t} = 5$
$0.04t = \ln 5$
$t = 25\ln 5 \approx 40.2359$

3. Determine the number of complete days before depletion happens.

Since depletion occurs at $t \approx 40.2$ days, the population drops to zero during the 41st day.
Therefore, this will happen after 40 complete days.