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

• Augustine Duran
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.
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.

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.

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?

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)?

plugging in C and simplifying i get

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

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.

## 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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