Solve 1-Param DE: dr/dβ cot β - r = 2

[Nano-Passion]

Homework Statement

Find 1-parameter family of solutions for the folloing differential equation:

dr/dβ cot (β) - r = 2

The Attempt at a Solution

dr/dβ cot β - r = 2
dr/β cot β = 2 + r
dr = (2+r) dβ/cot β
integral dr/2+r = integral tan β dβ
ln (2+r) = - ln (cos β) + c

I take everything to the e power

2+r = - cos β + e^c
r = c cos - 2

While the book has r = c sec - 2. I suspect this might be because of taking e^- ln, if it is, please explain why that happens.

 

[HallsofIvy]

You have errors when taking e to the power of each side. (NOT "to the e power"!)
$e^{-ln(cos(\beta))+ C}$ is NOT "$-cos(\beta)+ e^c$" for two reasons. First, $e^{-ln(A)}= e^{ln(1/A)}= 1/A$, not -A. Second $e^{A+ B}= e^Ae^B$, not $e^A+ e^B$.
$$e^{-ln(cos(\beta))+ c}= e^{ln(1/cos(\beta)}e^c= C/cos(\beta)= C sec(\beta)$$
where $C= e^c$.

 

[Nano-Passion]

You have errors when taking e to the power of each side. (NOT "to the e power"!)

Sorry, I was sloppy with my wording hehe.

$e^{-ln(cos(\beta))+ C}$ is NOT "$-cos(\beta)+ e^c$" for two reasons. First, $e^{-ln(A)}= e^{ln(1/A)}= 1/A$, not -A. Second $e^{A+ B}= e^Ae^B$, not $e^A+ e^B$.

What i did was $e^A+e^B$, then I incorporated the constant when I wrote $c cos$

$$e^{-ln(cos(\beta))+ c}= e^{ln(1/cos(\beta)}e^c= C/cos(\beta)= C cos (\beta)$$
where $C= e^c$.

Many thanks!