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I'm at a loss on ones like this problem. I'm working on a reduction of order problem and have come across the equation v''t+v'=0.

I have the solution manual to the book the problem is from, and it says that solving for v', I should get v'(t)=ct^{-1}, which can be integrated to get $v(t)=c_1*\ln(t)+c_2$.

I cannot for the life of me figure out the method of going from v''t+v'=0 to v'(t)=ct^{-1}. What is the method I should use and what are the steps?

2 Answers 2

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First write w=v' to reduce the order of the equation by one. That leaves w't+w=0. This can be written as

\frac{w'}{w}=-\frac{1}{t}\;.

Then integration yields

$\ln w = -\ln t + \hat{c}_1\;,$

and exponentiating gives

$w=\frac{c_1}{t}\;.$

Then integrating

v'=\frac{c_1}{t}

leads to

$v=c_1\ln t + c_2\;.$

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    I totally agree with the steps shown. Just one slight thing I was unsure about. When integrating $\dfrac{w'}{w}=-\dfrac{1}{t}$, do we not need the absolute value signs because after exponentiating the solution to put the equation in terms of $w$ explicitly, we have have to consider the $|w|$ of $w$ and therefore take $\pm$ to the other side of the equation. Or does this not really matter much here because of some restrictions or the signs will just vanish anyway. Thanks.2011-06-01
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0=v''t+v'=v''t+v'\cdot1=v''t+v't'=(v't)'.

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    @joriki, typo corrected. thanks.2011-04-20