Use an integrating factor to show that the general solution for $\frac{{{\text{d}}x}}{{{\text{d}}t}} - \frac{x}{t} =  - \frac{2}{t},{\text{ }}t > 0$ is $x = 2 + ct$, where $c$ is a constant.

The weight in kilograms of a dog, $t$ weeks after being bought from a pet shop, can be modelled by the following function:

$$w(t) = \left\{ {\begin{array}{*{20}{c}} {2 + ct}&{0 \le t \le 5} \\ {16 - \frac{{35}}{t}}&{t > 5} \end{array}.} \right.$$

Given that $w(t)$ is continuous, find the value of $c$.

Write down

(i)     the weight of the dog when bought from the pet shop;

(ii)     an upper bound for the weight of the dog.

Prove from first principles that $w(t)$ is differentiable at $t = 5$.

integrating factor ${e^{\int { - \frac{1}{2}{\text{d}}t} }} = {e^{ - \ln t}}\left( { = \frac{1}{t}} \right)$     M1A1

$\frac{x}{t} = \int { - \frac{2}{{{t^2}}}{\text{d}}t = \frac{2}{t} + c}$     A1A1

Note:     Award A1 for $\frac{x}{t}$ and A1 for ${\frac{2}{t} + c}$.

$x = 2 + ct$     AG

[4 marks]

given continuity at $x = 5$

$5c + 2 = 16 - \frac{{35}}{5} \Rightarrow c = \frac{7}{5}$     M1A1

[2 marks]

(i)     $2$     A1

(ii)     any value $\ge 16$     A1

Note:     Accept values less than $16$ if fully justified by reference to the maximum age for a dog.

[2 marks]

$\mathop {\lim }\limits_{h \to 0 - } \left( {\frac{{\frac{7}{5}(5 + h) + 2 - \frac{7}{5}(5) - 2}}{h}} \right) = \frac{7}{5}$     M1A1

$\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{16 - \frac{{35}}{{5 + h}} - 16 + \frac{{35}}{5}}}{h}} \right)\;\;\;\left( { = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{\frac{{ - 35}}{{5 + h}} + 7}}{h}} \right)} \right)$     M1

$=$$\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{\frac{{ - 35 + 35 + 7h}}{{(5 + h)}}}}{h}} \right) = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{7}{{5 + h}}} \right) = \frac{7}{5}$     M1A1

both limits equal so differentiable at $t = 5$     R1AG

Note:     The limits $t \to 5$ could also be used.

For each value of $\frac{7}{5}$ obtained by standard differentiation award     A1.

To gain the other 4 marks a rigorous explanation must be given on how you can get from the left and right hand derivatives to the derivative.

Note:     If the candidate works with $t$ and then substitutes $t = 5$ at the end award as follows

First M1 for using formula with $t$ in the linear case, A1 for $\frac{7}{5}$

Award next 2 method marks even if $t = 5$ not substituted, A1 for $\frac{7}{5}$

[6 marks]

Total [14 marks]

This was generally well done. Some candidates did not realize ${e^{ - \ln t}}$ could be simplified to $\frac{1}{t}$.

This part was well done by the majority of candidates.

This part was well done by the majority of candidates.

Some candidates ignored the instruction to prove from first principles and instead used standard differentiation. Some candidates also only found a derivative from one side.