Differential equation

  • #1

Homework Statement


differential equation
(dr/dy)+r=8 ; r(1)=0.3

The Attempt at a Solution


(dr/dy)=(8-r)

∫[dr/(8-r)]=∫dy

ln|8-r|+c1=y+c2 ; k=c2-c1

ln|8-r|=y+k ; r(1)=0.3
then take of e^ both sides which gives
8-r=e^(y+k)
8-e^(y+k)=r and we know r(1)=0.3 which means 0.3=8-e^(1+k); then solve for k
k=ln|7.7|-1 then when plug this in to r(y)= 8-e^(y+k) for some reason webassign says
r(y)= 8-7.7*e^(1-y)

but i get r(y)=8-7.7*e^(y-1) are they equivalent
 
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Answers and Replies

  • #3
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92
nick.martinez said:
1. Homework Statement
differential equation
(dr/dy)+r=8 ; r(1)=0.3
3. The Attempt at a Solution
(dr/dy)=(8-r)

∫[dr/(8-r)]=∫dy

ln|8-r|+c1=y+c2 ; k=c2-c1

ln|8-r|=y+k ; r(1)=0.3
then take of e^ both sides which gives
8-r=e^(y+k)
8-e^(y+k)=r and we know r(1)=0.3 which means 0.3=8-e^(1+k); then solve for k
k=ln|7.7|-1 then when plug this in to r(y)= 8-e^(y+k) for some reason webassign says
r(y)= 8-7.7*e^(1-y)

but i get r(y)=8-7.7*e^(y-1) are they equivalent
what am i doing wrong?

You have done almost right but check the integral of dr/(8-r).
 
  • #4
You have done almost right but check the integral of dr/(8-r).

i think when i was taking the intgral of dr/(8-r) i was forgetting the absolute value sign for the natural log which is -ln|8-r|+c1=y+c2 ;c2-c1=k

then giving me: -ln|8-r|=y+k then becomes ln|8-r|=-y-k ; then take e^ of both sides giving me

e^(ln|8-r|)=e^(-y-k)

r-8=e^(-y-k) this then becomes

r=e^(-y-k)+1 we know r(1)=0.3

0.3=e^(-1-k)+8
-7.7=e^(-1-k)
im stuck here because now i cant take the log of both sides becuase of the negative symbol on left hand side.
 
Last edited:
  • #5
3,816
92
i think when i was taking the intgral of dr/(8-r) i was forgetting the absolute value sign for the natural log which is ln|1-r|+c1=y+c2 ;c2-c1=k

then giving me: -ln|1-r|=y+k then becomes ln|1-r|=-y-k ; then take e^ of both sides giving me

e^(ln|1-r|)=e^(-y-k)

r-1=e^(-y-k) this then becomes

r=e^(-y-k)+1 we know r(1)=0.3

0.3=e^(-1-k)+1
-0.7=e^(-1-k)+1
im stuck here because now i cant take the log of both sides becuase of the negative symbol on left hand side.

What's the integral of dr/(8+ax) ?
 
  • #6
What's the integral of dr/(8+ax) ?

i did a quick edit check it out i made the -8 in the ln|8-R| into ones now its right but im still stuck
 
  • #7
i did a quick edit check it out i made the -8 in the ln|8-R| into ones now its right but im still stuck


im stuck at -7.7=e^(-1-k)
 
  • #8
3,816
92
You solved the differential equation correctly.

nick said:
e^(ln|8-r|)=e^(-y-k)

r-8=e^(-y-k)

Why did you change |8-r| to r-8? :confused:
 
  • #9
You solved the differential equation correctly.



Why did you change |8-r| to r-8? :confused:

becuase when you take the the e of the absolute value of e^(ln|8-x|) i get |x-8|. dont the absolute values do this?
 
  • #10
3,816
92
becuase when you take the the e of the absolute value of e^(ln|8-x|) i get |x-8|. dont the absolute values do this?

No, I don't think it changes to |x-8|. Instead of this, you can simply put y=1 and r=0.3 in the equation:
[tex]-ln|8-x|=y+k[/tex]
Find out k from here.
 

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