v"t + v' = 0

my professor solved this by guessing, but i was wondering how would I solve it the technical way?

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- Mar 11th 2009, 07:31 PMp00ndawgtechnical solution for this 2nd order de..
v"t + v' = 0

my professor solved this by guessing, but i was wondering how would I solve it the technical way? - Mar 11th 2009, 07:41 PMTheEmptySet
I am guessing that v is a function of t if so

let $\displaystyle u=v' \implies u'=v''$

subbing in we get

$\displaystyle t u'+u=0 \iff t\frac{du}{dt}=-u \iff \frac{du}{u}=-\frac{du}{t}$

integrating both sides gives

$\displaystyle \ln|u|=-\ln|t|+\ln|c| \iff ln|u|=\ln \left( \frac{c}{t}\right)$

$\displaystyle u=\frac{c}{t}$ but u=v' so we get

$\displaystyle \frac{dv}{dt}=\frac{c}{t} \iff dv=\frac{c}{t}dt \implies v=c\ln|t|+d$

I hope this helps.