- #1

Augustine Duran

- 39

- 1

## Homework Statement

Solve: ##\frac {dx} {dt}## ##\text{= 8-3x , x(0)=4}##

## The Attempt at a Solution

Step 1:

##\int \frac 1 {8-3x} \, dx## = ##\int \, dt##

Step 2:

- ##\frac 1 3## ##\text{ln|8-3x| = t+c}##

From here I am going to try to get it into explicit form

Step 3:

##\text{ln|8-3x| = -3t-3c}##

Step 4:

##e^{ln|8-3x|}## = ##e^{-3t-3c}##

Step 5:

##\text{|8-3x|}## = ##e^{-3t-3c}##

im not sure what to do about the absolute values. I was thinking of cheating and just dropping them and then just solving for x which would give me

Step 6:

##\text{x= }## ##\frac 1 3####\text{(8-}####e^{-3t-3c}####\text{)}##

Plug in initial conditions x(0) = 4

Step 7:

##\text{4= }## ##\frac 1 3####\text{(8-}####e^{-3(0)-3c}####\text{)}##

Step 8: Simplify

##\text{4}##=-##e^{-3c}##

The problem here is that if i take the ln of both sides to solve for C it will be undefined because of the negative sign. Probably dropping the absolute signs in step 6 without a reason is a no no but I am not sure what to do.

Any help would be appreciated.