# Differential equation

1. Apr 12, 2013

### nick.martinez

1. The problem statement, all variables and given/known data
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

Last edited by a moderator: Apr 13, 2013
2. Apr 12, 2013

### nick.martinez

what am i doing wrong?

3. Apr 13, 2013

### Saitama

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

4. Apr 13, 2013

### nick.martinez

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: Apr 13, 2013
5. Apr 13, 2013

### Saitama

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

6. Apr 13, 2013

### nick.martinez

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. Apr 13, 2013

### nick.martinez

im stuck at -7.7=e^(-1-k)

8. Apr 13, 2013

### Saitama

You solved the differential equation correctly.

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

9. Apr 13, 2013

### nick.martinez

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. Apr 13, 2013

### Saitama

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:
$$-ln|8-x|=y+k$$
Find out k from here.