2
$\begingroup$

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?

$\endgroup$

2 Answers 2

7
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – night owl
    Jun 1, 2011 at 15:33
2
$\begingroup$

$0=v''t+v'=v''t+v'\cdot1=v''t+v't'=(v't)'$.

$\endgroup$
2
  • $\begingroup$ There's a prime missing in the first sum. $\endgroup$
    – joriki
    Apr 20, 2011 at 23:11
  • $\begingroup$ @joriki, typo corrected. thanks. $\endgroup$
    – lhf
    Apr 20, 2011 at 23:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .