2)(C) A garden centre advertises young plants to be used as hedging. After planting, the growth $$ G $$ metres (ie the increase in height) after $$ t $$ years is modelled by the differential equation $$ \frac{d G}{d t}=\frac{25 k-G}{25} $$ where $$ k $$ is a constant and $$ G=0 $$ when $$ t=0 $$. (a) Express $$ G $$ in terms of $$ t $$ and $$ k $$. (b) Given that a plant grows $$ 0 \cdot 6 $$ metres by the end of 5 years, find the value of (c) On the plant labels it states that the expected growth after 10 years is approximately 1 metre. Is this claim justified? (d) Given that the initial height of the plants was 0.3 m , what is the likely long-term height of the plants?

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To solve the problem, we will follow the steps outlined in the prompt.

### Part (a): Express $$ G $$ in terms of $$ t $$ and $$ k $$

We start with the differential equation:

$$
\frac{dG}{dt} = \frac{25k - G}{25}
$$

This is a first-order linear differential equation. We can rearrange it to isolate $$ G $$:

$$
\frac{dG}{dt} + \frac{1}{25}G = k
$$

Next, we will use an integrating factor to solve this equation. The integrating factor $$ \mu(t) $$ is given by:

$$
\mu(t) = e^{\int \frac{1}{25} dt} = e^{\frac{t}{25}}
$$

Now, we multiply the entire differential equation by the integrating factor:

$$
e^{\frac{t}{25}} \frac{dG}{dt} + \frac{1}{25} e^{\frac{t}{25}} G = k e^{\frac{t}{25}}
$$

The left-hand side can be rewritten as the derivative of a product:

$$
\frac{d}{dt} \left( e^{\frac{t}{25}} G \right) = k e^{\frac{t}{25}}
$$

Now, we integrate both sides with respect to $$ t $$:

$$
e^{\frac{t}{25}} G = \int k e^{\frac{t}{25}} dt
$$

The integral on the right side is:

$$
\int k e^{\frac{t}{25}} dt = 25k e^{\frac{t}{25}} + C
$$

where $$ C $$ is the constant of integration. Thus, we have:

$$
e^{\frac{t}{25}} G = 25k e^{\frac{t}{25}} + C
$$

Now, we can solve for $$ G $$:

$$
G = 25k + Ce^{-\frac{t}{25}}
$$

To find $$ C $$, we use the initial condition $$ G(0) = 0 $$:

$$
0 = 25k + C \implies C = -25k
$$

Substituting $$ C $$ back into the equation for $$ G $$:

$$
G = 25k - 25k e^{-\frac{t}{25}} = 25k(1 - e^{-\frac{t}{25}})
$$

Thus, the expression for $$ G $$ in terms of $$ t $$ and $$ k $$ is:

$$
G(t) = 25k(1 - e^{-\frac{t}{25}})
$$

### Part (b): Find the value of $$ k $$

We are given that the plant grows $$ 0.6 $$ metres by the end of $$ 5 $$ years:

$$
G(5) = 0.6
$$

Substituting $$ t = 5 $$ into the equation for $$ G $$:

$$
0.6 = 25k(1 - e^{-\frac{5}{25}})
$$

Calculating $$ e^{-\frac{5}{25}} = e^{-0.2} $$:

Now we can solve for $$ k $$:

$$
0.6 = 25k(1 - e^{-0.2})
$$

Calculating $$ 1 - e^{-0.2} $$:

$$
1 - e^{-0.2} \approx 1 - 0.8187 \approx 0.1813
$$

Now substituting this value back into the equation:

$$
0.6 = 25k(0.1813)
$$

Solving for $$ k $$:

$$
k = \frac{0.6}{25 \times 0.1813}
$$

Calculating this value:

$$
k \approx \frac{0.6}{4.5325} \approx 0.132
$$

### Part (c): Is the claim justified?

We need to check if the expected growth after $$ 10 $$ years is approximately $$ 1 $$ metre:

Substituting $$ t = 10 $$ into the equation for $$ G $$:

$$
G(10) = 25k(1 - e^{-\frac{10}{25}})
$$

Using $$ k \approx 0.132 $$:

Calculating $$ e^{-\frac{10}{25}} = e^{-0.4} $$:

$$
G(10) = 25 \times 0.132(1 - e^{-0.4})
$$

Calculating $$ 1 - e^{-0.4} $$:

$$
1 - e^{-0.4} \approx 1 - 0.6703 \approx 0.3297
$$

Now substituting back:

$$
G(10) \approx 25 \times 0.132 \times 0.3297
$$

Calculating this value:

$$
G(10) \approx 1.089
$$

Since $$ G(10) \approx 1.089 $$ metres, the claim that the expected growth after $$ 10 $$ years is approximately $$ 1 $$ metre is justified.

### Part (d): Long-term height of the plants

The long-term height of the plants can be found by taking the limit as $$ t $$ approaches infinity:

$$
\lim_{t \to \infty} G(t) = 25k(1 - e^{-\frac{t}{25}}) \to 25k
$$

Substituting $$ k \approx 0.132 $$:

$$
\text{Long-term height} = 25 \times 0.132 \approx 3.3 \text{ metres}
$$

Thus, the likely long-term height of the plants is approximately $$ 3.3 $$ metres.
(a) $$ G(t) = 25k(1 - e^{-\frac{t}{25}}) $$ (b) $$ k \approx 0.132 $$ (c) Yes, the claim is justified. (d) The long-term height is approximately 3.3 metres.

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