Separable differential equation

In summary, the conversation discusses how to approach the equation (x + 2y) dy/dx = 1, y(0) = 1 in the form of g(y/x). The suggested approach is to start with u = y/x and find dy/dx from there. However, there is some confusion about whether the equation is in the correct form for this approach to work. The conversation ends with the suggestion to try a linear trial solution.
  • #1
ch2kb0x
31
0

Homework Statement


(x + 2y) dy/dx = 1, y(0) = 1


Homework Equations





The Attempt at a Solution



Problem is, I can't separate it. This might be a homogenous type? If so, how would I make it into the g(y/x) form.

Thank you.
 
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  • #2
Start with u = y/x, or equivalently, y = ux. From this, find dy/dx.
 
  • #3
Okay, so since you said y = ux, I am thinking that this is a homogenous equation...

However, if it is a homogeneous equation, before we can plug in y = ux, aren't we suppose to first have the equation in the form of dy/dx = f(x,y), where there exists a function such that f(x,y) is expressed g(y/x).

Then, AFTEr we can do the y=ux thing. correct me if I am wrong.
 
  • #4
ch2kb0x said:
Problem is, I can't separate it. This might be a homogenous type? If so, how would I make it into the g(y/x) form.

Thank you.

Well, that is because it isn't a separateable equation... It isn't even an ODE. Sure it's right?
 
  • #5
Yeah, copied exactly from textbook.
 
  • #6
ch2kb0x said:
Okay, so since you said y = ux, I am thinking that this is a homogenous equation...

However, if it is a homogeneous equation, before we can plug in y = ux, aren't we suppose to first have the equation in the form of dy/dx = f(x,y), where there exists a function such that f(x,y) is expressed g(y/x).

Then, AFTEr we can do the y=ux thing. correct me if I am wrong.
You can write the equation as dy/dx = 1/(x + 2y), where the right side is a function of x and y. I'm just offering a suggested approach based on your first post. It may or may not work.
 
  • #7
Yeah, it is a function of x and y, but I don't think it's in the g(y/x) form.
 
  • #8
Mark44 said:
You can write the equation as dy/dx = 1/(x + 2y), where the right side is a function of x and y. I'm just offering a suggested approach based on your first post. It may or may not work.

To me it seems that you are suggesting a linear trial solution?
 
  • #9
The OP said he wanted to put this into g(y/x) form, so that made me think of the substitution I suggested. I don't have access to my DE textbooks right now, so I don't have any more ideas on solving this one.
 

Related to Separable differential equation

1. What is a separable differential equation?

A separable differential equation is a type of differential equation where the dependent variable and independent variable can be separated on opposite sides of the equation. This allows for the equation to be solved by integrating both sides separately.

2. How do you solve a separable differential equation?

To solve a separable differential equation, you first need to separate the variables on opposite sides of the equation. Then you can integrate both sides separately and solve for the constant of integration. Finally, you can solve for the dependent variable to get the final solution.

3. What are the applications of separable differential equations?

Separable differential equations are commonly used in physics, engineering, and other sciences to model real-world phenomena. They can also be used in economics and finance to model growth and decay processes.

4. What is the difference between a separable differential equation and a non-separable differential equation?

The main difference between a separable and non-separable differential equation is that the dependent and independent variables cannot be separated in a non-separable equation. This means that the equation cannot be solved by simply integrating both sides, and other techniques must be used.

5. Are there any limitations to using separable differential equations?

While separable differential equations are useful in many applications, they do have some limitations. They can only be used to solve first-order differential equations, and they may not always provide an exact solution. In some cases, numerical methods may need to be used instead.

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