Given that $y = 35$ at $x = 4$, find $y$ in terms of $x$, giving each term in its simplest form. - Edexcel - A-Level Maths Pure - Question 6 - 2010 - Paper 2

We know that $y = 35$ when $x = 4$:

$35 = \frac{10}{3} (4)^{\frac{3}{2}} + \frac{2}{5} (4)^{\frac{5}{2}} + C$

Calculating the terms:

1. $4^{\frac{3}{2}} = 8$ thus:

$35 = \frac{10}{3} \cdot 8 + \frac{2}{5} \cdot 32 + C$

2. $\frac{10}{3} \cdot 8 = \frac{80}{3}$ and $\frac{2}{5} \cdot 32 = \frac{64}{5}$

3. Converting to a common denominator of 15 gives:

$\frac{80}{3} = \frac{400}{15}$ $\frac{64}{5} = \frac{192}{15}$

So we have:

$35 = \frac{400}{15} + \frac{192}{15} + C$

Thus:

$C = 35 - \frac{592}{15} = \frac{525 - 592}{15} = \frac{-67}{15}$

Substituting $C$ back into the equation gives:

$y = \frac{10}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} - \frac{67}{15}$

This is the required expression for $y$ in terms of $x$, with each term in its simplest form.