202210201648 Solution to 1972-CE-AMATH-II-XX

It is given that $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=4x-\frac{3}{2}x^2}$ and that $y=10$ when $x=2$ . Find $y$ in terms of $x$ .

Roughwork.

$\begin{aligned} y & = \int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int\bigg( 4x-\frac{3}{2}x^2\bigg)\,\mathrm{d}x \\ & = \frac{4x^{(1)+1}}{(1)+1} -\frac{3}{2}\bigg(\frac{x^{(2)+1}}{(2)+1}\bigg) + C\textrm{ for some constant} \\ & = 2x^2+\frac{1}{2}x^3 + C \\ \end{aligned}$

$\begin{aligned} 10 & = 2(2)^2+\frac{1}{2}(2)^3+C \\ C & = -2 \\ \end{aligned}$

In conclusion,

$y(x)=2x^2+\frac{1}{2}x^3-2$ .

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