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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?

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2 Answers

up vote 7 down vote accepted

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. –  night owl Jun 1 '11 at 15:33
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$0=v''t+v'=v''t+v'\cdot1=v''t+v't'=(v't)'$.

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There's a prime missing in the first sum. –  joriki Apr 20 '11 at 23:11
    
@joriki, typo corrected. thanks. –  lhf Apr 20 '11 at 23:11
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