Solving a Differential Equation by Separation of Variables

Click For Summary
The discussion focuses on solving a differential equation using separation of variables. The original equation was manipulated through long division and factoring, leading to the integration of both sides. The final implicit solution is presented as (x+4)^5 = c(e^x)(e^-y)(y+3)^5, with a suggestion to simplify the right-hand side using exponent rules. Participants also discuss the suitability of different introductory ODE textbooks, expressing dissatisfaction with Zill's book for being too informal and visually cluttered. Recommendations for alternative texts include a book by Ross, although personal experience with it is lacking.
MidgetDwarf
Messages
1,588
Reaction score
703
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.
 
Physics news on Phys.org
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.
 
  • Like
Likes MidgetDwarf
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K