Solving a Differential Equation by Separation of Variables

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Homework Help Overview

The discussion revolves around solving a differential equation using the method of separation of variables. The original poster presents a specific equation and describes their initial attempts at manipulation, including long division and factoring.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for approaching the problem, including factoring the right-hand side and the implications of the original poster's long division. There is a focus on the steps taken to separate variables and integrate.

Discussion Status

Some participants have offered guidance on factoring and integrating, while others have reflected on their own experiences with the problem. There is acknowledgment of the original poster's progress, but no explicit consensus on the final solution has been reached.

Contextual Notes

Participants mention the use of a specific textbook for further study and express varying opinions on its effectiveness, indicating a broader context of learning resources in differential equations.

MidgetDwarf
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Solve the given differential equation by separation of variables.

(dy/dx)= (xy+3x-y-3)/(xy-2x+4y-8)

First, I noticed when i divided both sides by the left hand side and multiplied both sides by dx, nothing canceled or seemed to work.

I got to thinking.

on the right hand side I preformed long division.

i divided xy+3x-y-3 by xy-2x+4y-8.

I get 1 + (5x-5y+5)/(xy-2x+4y-8)

(dy/dx)= 1 + (5x-5y+5)/(xy-2x+4y-8)

I am stuck here. Any help is welcomed and appreciated.
 
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I think a better approach is beginning instead by factoring the RHS. From there it should be clear how to solve via separation of variables.
 
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Tallus Bryne said:
I think a better approach is beginning instead by factoring the RHS. From there it should be clear how to solve via separation of variables.

wow, i over thought this problem. thanks a lot.

factoring the left hand side.

(y+3)(x-1)/(y-2)(x+4)

Left hand side= (y-2)/(y+3)dy

right hand side=(x-1)/(x+4)dx

then I integrate. the process is kind of lengthy, requiring trivial integration.

the answer is (x+4)^5=c(e^x)(e^-y)(y+3)^5.

I can't take you enough.
 
MidgetDwarf said:
the answer is (x+4)^5=c(e^x)(e^-y)(y+3)^5.

That looks like a correct implicit soln; you can also tidy up the RHS of your answer a little by applying some rules of exponents:
\begin{equation} e^{x}e^{-y} = e^{x-y} \end{equation}
 
Tallus Bryne said:
That looks like a correct implicit soln; you can also tidy up the RHS of your answer a little by applying some rules of exponents:
\begin{equation} e^{x}e^{-y} = e^{x-y} \end{equation}

yes, you are correct. thanks a lot.

Do you recommended a an intro ode book?

we are using zill in our class, and it is a bit to chatty. The graphics make the layout of the book a little hard to read in my opinion and he is too loose ( doesn't really use mathematical language) in his explanations.
 
MidgetDwarf said:
Do you recommended a an intro ode book?

I wouldn't be able to help you there. I also used a text co-authored by Zill when I took differential ('Differential Equations with Boundary-Value Problems' by Zill and Cullen 7ed.) when I took differential.
I saw a text by Ross recommended in the thread "How to self-[URL='https://www.physicsforums.com/insights/self-study-basic-high-school-mathematics/']study mathematics?[/URL]" However, I haven't ever had the opportunity to see what it's like myself.
 

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