\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\begin{document}

\parindent=0pt

QUESTION

Find the general solution of the differential equation
$\ds\frac{dx}{dt}=\frac{1}{xt^3}$.



ANSWER

$\ds\frac{dx}{dt}=\frac{1}{xt^3},$ therefore $\ds \int
x\,dx=\int\frac{1}{t^3}\,dt=\int t^{-3}\,dt$

Therefore $\ds\frac{x^2}{2}=\frac{t^{-2}}{-2}+c,\
x^2=2c-\frac{1}{t^2}$

 Therefore $\ds
x=\pm\left(2c-\frac{1}{t^2}\right)^\frac{1}{2}$




\end{document}
