202210131233 Solution to 1969-CE-AMATH-I-XX

Find $y$ in terms of $x$ , if $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=8x-1}$ and $y=-1$ when $x=-1$ .

Roughwork.

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 8x-1 \\ y(x) & = \int \frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (8x-1)\,\mathrm{d}x \\ & = 4x^2-x+C\qquad\textrm{for some constant }C \\ \because\enspace -1 & = y(-1) \\ -1 & = 4(-1)^2-(-1)+C \\ \therefore\enspace C & = -6 \\ \end{aligned}$

In conclusion, $y(x) = 4x^2-x-6$ .

物理後生﹑掣數學之肘.
小子算數﹑指物理之月.

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