Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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


Then integration yields

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

and exponentiating gives


Then integrating


leads to

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

share|cite|improve this answer
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


share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.