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QUESTION

Find the general solution of the differential equation
$\ds\frac{dx}{dt}=\frac{x^3}{t^2}$.

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ANSWER

$\ds\frac{dx}{dt}=\frac{x^3}{t^2},$ therefore $\ds\int
x^{-3}\,dx=\int t^{-2}\,dt\\
\frac{x^{-2}}{-2}=\frac{t^{-1}}{-1}+c,$ therefore
$\ds-\frac{1}{2x^2}=-\frac{1}{t}+c$




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