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

Substituting the initial condition $y = 35$ when $x = 4$:
$35 = \frac{10}{3} (4)^{\frac{3}{2}} + \frac{2}{5} (4)^{\frac{5}{2}} + C.$
Calculating $(4)^{\frac{3}{2}} = 8$ and $(4)^{\frac{5}{2}} = 32$:
$35 = \frac{10}{3} \cdot 8 + \frac{2}{5} \cdot 32 + C.$
This simplifies to:
$35 = \frac{80}{3} + \frac{64}{5} + C.$
To combine rac{80}{3} and rac{64}{5}, find a common denominator (15):
$\frac{80}{3} = \frac{400}{15} \quad \text{and} \quad \frac{64}{5} = \frac{192}{15},$
so
$35 = \frac{400 + 192}{15} + C.$
Calculating gives:
$35 = \frac{592}{15} + C \Rightarrow C = 35 - \frac{592}{15} = \frac{525 - 592}{15} = \frac{-67}{15}.$

Substituting $C$ back into the general solution:
$$y = \frac{10}{3} x^{\frac{3}{2}} + \frac{2}{5} x^{\frac{5}{2}} - \frac{67}{15}.$This is the expression for$y$ in its simplest form.