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?

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.