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I have a first order differential to solve;

$$\frac{dr}{d\theta}=\cot\theta \cdot r$$

I have solved it to get this in explicit form;

$r = B\sin\theta$ where $B$ is an arbitrary constant.

I just like for someone to tell me whether i am correct or not please.

Michael Hardy
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1 Answers1

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If $f=B\sin\theta$ then $\dfrac{dr}{d\theta}=B\cos\theta$. Is it true that $B\cos\theta = \cot(\theta r)=\cot(\theta B\sin\theta)$? Notice that as $\theta\to0$ we have $\cot(\theta B\sin\theta)\to\infty$ (unless $B=0$) (here I'm not distinguishing between $+\infty$ and $-\infty$ but rather with the $\infty$ that is approached if you go in either of the two directions), and $B\sin\theta\to 0\ne\infty$.

The question has been altered since I wrote what appears above. $$ \frac{dr}{d\theta} = r\cot\theta $$ If the line immediately above this was intended, then I would write $$ \frac{dr}r = \frac{\cos\theta\,d\theta}{\sin\theta} = \frac{du}u $$ so $$ \log_e |r| = \log_e|u| + C = \log_e\sin\theta + C, $$ and hence $$ r = B\sin\theta. $$ This actually leads to $B$ being a nonzero constant, but since we divided by $r$, we must consider separately the case where $r=0$. And checking by substitution shows that $B=0$ also works.

So the solution of the revised question is correct.

Michael Hardy
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  • I'd just like to make sure... your saying my answer is correct? –  Feb 24 '15 at 18:16
  • No; I'm saying it is not. But now I see you've changed the question. I read what I took to be $\cot(\theta r)$, but now it looks as if you meant $(\cot\theta)r$. ${}\qquad{}$ – Michael Hardy Feb 24 '15 at 18:23
  • Sorry, i should have made it clear, the way i originally posted it was misleading. Thanks so much. –  Feb 24 '15 at 18:53