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

In summary, the student attempted to solve the following equation:##\frac {dx} {dt}##=-3t-3c. However, they ran into trouble when trying to solve for x and ended up with x' = 8.
  • #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.
 
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  • #2
Augustine Duran said:

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.
 
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Likes Augustine Duran
  • #3
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?
 
  • #4
Augustine Duran said:
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)?
 
  • #5
plugging in C and simplifying i get

##\text{x=}####\frac 8 3####\text{+}####\frac 4 3####e^{-3t}##
 
  • #6
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.
 

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

1. What is a first order ODE?

A first order ODE, or ordinary differential equation, is an equation that relates a function with its derivative. It is typically in the form of y' = f(x,y), where y' is the derivative of y with respect to x.

2. How do you solve a first order ODE?

The most common method for solving a first order ODE is by using separation of variables. This involves isolating the y term on one side of the equation and the x term on the other side, then integrating both sides to solve for y. Other methods include using the method of integrating factors or using a substitution to reduce the equation to a simpler form.

3. What are the initial conditions for solving a first order ODE?

Initial conditions are the values of the function and its derivative at a specific point, typically denoted as x = x0. These values are necessary for solving a first order ODE, as they provide a starting point for the solution.

4. Can all first order ODEs be solved analytically?

No, not all first order ODEs can be solved analytically. Some equations are too complex or do not have a closed form solution, meaning they cannot be expressed using standard mathematical functions. In these cases, numerical methods are used to approximate the solution.

5. How are first order ODEs used in real-world applications?

First order ODEs are used to model a wide range of real-world phenomena, such as population growth, chemical reactions, and electrical circuits. By solving these equations, scientists and engineers can make predictions and understand the behavior of these systems.

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