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 5 - 2010 - Paper 2

Integrating with respect to $x$ gives: $y = \int \left(5x^{-\frac{3}{2}} + x^{\frac{3}{2}}\right) dx$ This results in: $y = 5 \cdot \left(-2x^{-\frac{1}{2}}\right) + \frac{2}{5}x^{\frac{5}{2}} + C$ which simplifies to: $y = -10x^{-\frac{1}{2}} + \frac{2}{5}x^{\frac{5}{2}} + C$

Substituting the values $x = 4$ and $y = 35$ into the equation gives: $35 = -10 \cdot (4^{-\frac{1}{2}}) + \frac{2}{5} \cdot (4^{\frac{5}{2}}) + C$ Calculating $4^{-\frac{1}{2}} = \frac{1}{2}$ and $4^{\frac{5}{2}} = 32$, we simplify: $35 = -10 \cdot \frac{1}{2} + \frac{2}{5} \cdot 32 + C$ This leads to: $35 = -5 + \frac{64}{5} + C$ Finding a common denominator: $35 = \frac{-25 + 64}{5} + C$ So, $C = 35 - \frac{39}{5} = \frac{175 - 39}{5} = \frac{136}{5}$

Substituting $C = \frac{136}{5}$ back into the original expression for $y$ gives: $y = -10x^{-\frac{1}{2}} + \frac{2}{5}x^{\frac{5}{2}} + \frac{136}{5}$ To ensure the answer consists of each term in its simplest form, it can also be rewritten as: $y = -\frac{10}{\sqrt{x}} + \frac{2}{5}x^{\frac{5}{2}} + \frac{136}{5}$