problem when integrating an equation

hyourinn

New member
Joined
Oct 2, 2018
Messages
11
acceleraion of certain particle is described by a = 16-4v where v is velocity, if when t =0 v is equal to 5 determine the standart equation of v by integrating a

the books shows answer like this
a=dv/dt
16-4v=dv/dt
4(4-v)=dv/dt
-4dt=dv/(v-4) <- integrating both
-4t + C = ln v-4 then determine C by given condition
(t=0 v=5)
C = ln 1 = 0
then v = 4 + e^-4t is the general equation ###book ans###

and my attempt is not simplify the 16-4v to 4(4-v)
but just leave it as 16-4v but it don't give same result
here my attempt

a=dv/dt
16-4v=dv/dt
-dt=dt/(4v-16) then integrating both side
-t + C= ln (4v -16) determine value of C
(when t=0 v=5)
C = ln 4 then back to
-t + ln 4 = ln (4v - 16) or i can write it as
-t = ln(4v - 16) - ln 4 and by logaritm properties is turned to
-t = ln ((4v-16)/4)
4e^-t = 4v -16 simplyfied
4 + e^-t = v ###mine###

There is significant difference between both result, so i hope anyone can show me where did i go wrong
 

Harry_the_cat

Full Member
Joined
Mar 16, 2016
Messages
943
acceleraion of certain particle is described by a = 16-4v where v is velocity, if when t =0 v is equal to 5 determine the standart equation of v by integrating a

the books shows answer like this
a=dv/dt
16-4v=dv/dt
4(4-v)=dv/dt
-4dt=dv/(v-4) <- integrating both
-4t + C = ln v-4 then determine C by given condition
(t=0 v=5)
C = ln 1 = 0
then v = 4 + e^-4t is the general equation ###book ans###

and my attempt is not simplify the 16-4v to 4(4-v)
but just leave it as 16-4v but it don't give same result
here my attempt

a=dv/dt
16-4v=dv/dt
-dt=dt/(4v-16) then integrating both side
-t + C= ln (4v -16) determine value of C The RHS here should have a factor of 1/4 as well. Fix this up and it gives the same as the book.
(when t=0 v=5)
C = ln 4 then back to
-t + ln 4 = ln (4v - 16) or i can write it as
-t = ln(4v - 16) - ln 4 and by logaritm properties is turned to
-t = ln ((4v-16)/4)
4e^-t = 4v -16 simplyfied
4 + e^-t = v ###mine###

There is significant difference between both result, so i hope anyone can show me where did i go wrong
see comment in red
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,792
To integrate \(\displaystyle \int \frac{dv}{4v- 16}\), let \(\displaystyle u= 4v- 16\). Then \(\displaystyle du= 4 dv\) so that \(\displaystyle \frac{1}{4}du= dv\). So \(\displaystyle \int\frac{dv}{4v- 16}= \frac{1}{4}\int\frac{du}{u}= \frac{1}{4} ln(|u|)+ C= \frac{1}{4}ln(|4v- 16|)+ C\). That's where the "1/4" Harry_the_Cat refers to comes from.
 
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hyourinn

New member
Joined
Oct 2, 2018
Messages
11
thanks everyone
 
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