Solve by separation of variables

In summary, to solve the given differential equation by separation of variables, you need to factor the numerator and denominator of the right side, then separate the variables and integrate. You can also use polynomial division to see how many times the denominator goes into the numerator and use that to simplify the integration process.
  • #1
find_the_fun
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Solve given differential equation by separation of variables

\(\displaystyle \frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}\)

So separate x and y terms

\(\displaystyle (xy-2x+4y-8) dy = (xy+3x-y-3)\) ugh I'm stuck:(
 
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  • #2
Re: solve by separation of variables

You want to factor the numerator and denominator of the right side, then you may separate variables.
 
  • #3
Re: solve by separation of variables

I can factor to \(\displaystyle \frac{dy}{dx}=\frac{(x-1)(y+3)}{(x+4)(y-2)}\) and rewriting gives \(\displaystyle \frac{(y-2)}{(y+3)} dy = \frac{(x-1)}{(x+4)} dx\). Am I on the right track? I don't know how to integrate this.
 
  • #4
Re: solve by separation of variables

Yes, you are correct. For the left side, consider:

\(\displaystyle \frac{y-2}{y+3}=\frac{y+3-5}{y+3}=1-\frac{5}{y+3}\)

Do the same kind of thing on the right side, and you should be able to integrate now.
 
  • #5
Re: solve by separation of variables

MarkFL said:
Yes, you are correct. For the left side, consider:

\(\displaystyle \frac{y-2}{y+3}=\frac{y+3-5}{y+3}=1-\frac{5}{y+3}\)

Do the same kind of thing on the right side, and you should be able to integrate now.

Is an alternative to \(\displaystyle \frac{y+3-5}{y+3}\) doing polynomial division and seeing y+3 goes into y-2 once with a remainder of 5? I'm not super clear on the thought process of getting \(\displaystyle 1-\frac{5}{y+3}\).
 
  • #6
Re: solve by separation of variables

find_the_fun said:
Is an alternative to \(\displaystyle \frac{y+3-5}{y+3}\) doing polynomial division and seeing y+3 goes into y-2 once with a remainder of 5? I'm not super clear on the thought process of getting \(\displaystyle 1-\frac{5}{y+3}\).

Yes, although the remainder is actually -5, but then you get the same result. I just find it simpler to do as I did above. To make what I did more clear, consider:

\(\displaystyle \frac{y-2}{y+3}=\frac{(y+3)+(-2-3)}{y+3}=\frac{y+3}{y+3}-\frac{5}{y+3}=1-\frac{5}{y+3}\)
 

FAQ: Solve by separation of variables

What is the concept of "solve by separation of variables"?

"Solve by separation of variables" is a mathematical technique used to solve differential equations by separating the variables in the equation into different parts and finding the solution for each part separately. This method is often used in physics and engineering to solve problems involving changing rates or quantities.

How does "solve by separation of variables" work?

The first step in solving by separation of variables is to rewrite the equation so that one variable is on one side and the other variable is on the other side. Then, the two sides of the equation can be integrated separately, resulting in two separate equations. These equations can then be combined to find the final solution.

When is "solve by separation of variables" typically used?

"Solve by separation of variables" is typically used when dealing with differential equations that are separable, meaning that the variables can be separated into different parts. It is commonly used in physics and engineering problems that involve rates of change or quantities that vary over time.

What are the advantages of using "solve by separation of variables"?

One advantage of using "solve by separation of variables" is that it can simplify complex differential equations and make them easier to solve. It also allows for the use of basic integration techniques, making it a more accessible method for those without advanced mathematical knowledge.

Are there any limitations to using "solve by separation of variables"?

One limitation of using "solve by separation of variables" is that it can only be applied to certain types of differential equations, specifically those that are separable. Additionally, it may not always result in an exact solution and may require additional approximations to find an accurate answer.

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