Solving 2tv' - v = 0: Is it Separable?

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How do you find a solution for:

2tv' - v = 0

The text says it's separable but I'm not seeing it. I'm just learning so extra details are appreciated. Thanks.

(this should have been posted in the homework section - but I can't seem to move it there, sorry)
 
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2t dv/dt = v
==> 2t dv = v dt
Can you continue?
 
well I think I want to switch things around so v and t are separated

1/v dv = 1/2t dt

integrating both sides
ln(v) = 1/2 ln(t)+C

e^x both sides:
v = k*sqrt(t)

Which looks right! Is that how I was meant to do it? If so, thanks - I got stalled before when I got the ln() on both sides.
 
Looking right might not be good enough. You can check by taking the derivative and verifying that tv' - v = 0.
 
Mark44 said:
Looking right might not be good enough. You can check by taking the derivative and verifying that tv' - v = 0.
Verifying that 2tv' - v = 0 :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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