Unraveling an ODE: Solving a First Order Equation with Separation of Variables

[Augustine Duran]

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.

 

[ehild]

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

What does the absolute value mean? What is |8-3x| if 3x<8 and when 3x>8? Note that you have different c-s in both cases.
You are given the initial condition, which means that 3x>8.

 

[Augustine Duran]

x(t) =
\begin{cases}
3x-8 & \text{if } x > \frac 8 3 \\
8-3x & \text{if } x \leq \frac 8 3
\end{cases}

okay so seeing that my initial conditions give me x=4 i will be using 3x-8 since 4 is larger than 8/3?

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

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

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

$\text{12 = 8+}$$e^{-3c}$

simplify

-$\frac {ln4} 3$=$\text{C}$

is this correct?

 

[ehild]

x(t) =
\begin{cases}
3x-8 & \text{if } x > \frac 8 3 \\
8-3x & \text{if } x \leq \frac 8 3
\end{cases}

okay so seeing that my initial conditions give me x=4 i will be using 3x-8 since 4 is larger than 8/3?

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

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

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

$\text{12 = 8+}$$e^{-3c}$

simplify

-$\frac {ln4} 3$=$\text{C}$

is this correct?

Correct so far; go ahead. What is x(t)?

 

[Augustine Duran]

plugging in C and simplifying i get

$\text{x=}$$\frac 8 3$$\text{+}$$\frac 4 3$$e^{-3t}$

 

[LCKurtz]

Of course, if you aren't required to use separation of variables, it is much easier to write it as $x' + 3x = 8$ and use the linear equation method of integrating factor. That avoids all the fuss with the absolute value signs and logarithms.