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Find the general solution of the differential equation in explicit form,
$\dfrac{dr}{dt} =\dfrac{ k}{r^2}, (t>0, 0 < r < 10)$

I've split it into two so I have

$r^2 dr = k dx$
so when integrated,

$\frac13 r^3 = \frac12 k^2 + c$

Is this correct?

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    $k$ is not a function of $x. \frac 13 r^3 = kx + c$ – 2017-02-22
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    No, because $k$ is a constant. – 2017-02-22

2 Answers 2

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you have $$\frac{dr}{dt}=\frac{K}{r^2}$$ multiplying by $r^2$ and $dt$ we get $$r^2dr=Kdt$$ and integrating with respect to $t$ $$\frac{r^3}{3}=Kt+C$$ from here we get $$r=\sqrt[3]{3Kt+C}$$

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$2$ mistakes I need to point:

  1. In the first line, one of the integration variables, that is, the variable with respect to which the integration is to be carried out, or, the independent variable was $t$ while the other (dependent) variable was $r$. In the second step of integration the variable $t$ got transformed to the variable $x$, apparently for no reason.
  2. Then, again in the last line of your calculation whereas the integration variable in the previous line was $x$, it now got changed to $k$ and the result was accordingly obtained.

Hope this helps you.