How to Find the General Solution for dy/dx=3√(xy)

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Discussion Overview

The discussion revolves around finding the general solution for the differential equation dy/dx = 3√(xy). Participants explore methods of solving the equation, including separation of variables and integration techniques, while sharing their approaches and solutions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the equation is separable and provides a solution: y = [9√(x) + C]^(2/3).
  • Another participant questions the validity of this solution, proposing an alternative solution: y = [(x^(3/2) + C)]^2, claiming it is the answer provided in their answer key.
  • Some participants express confusion about the integration process and mention difficulties in solving the equation.
  • A later reply acknowledges a mistake in interpreting the equation and provides a corrected solution, reiterating that the equation is separable.
  • There is a discussion about the learning process in differential equations, with one participant expressing appreciation for the help received.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct general solution, as multiple competing views and solutions are presented. The discussion remains unresolved regarding which solution is correct.

Contextual Notes

Some participants mention confusion over integration techniques and the need for simultaneous integration, indicating potential gaps in understanding the problem-solving process.

Who May Find This Useful

Students learning differential equations, educators seeking examples of student discussions, and individuals interested in mathematical problem-solving techniques.

rapwaydown
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how to solve dy/dx=3√(xy) for a general solution?

PS: √ is square root sign, just in case if it is not clear.

Thank You
 
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That equation is separable. Let me explain.

dy/dx = 3/sqrt(xy)
<=>
dy/dx = 3/sqrt(x)sqrt(y)
<=>
sqrt(y)dy = [3/sqrt(x)]dx
<=>
(2/3)y^(3/2) = 6sqrt(x) + C
<=>
y^(3/2) = 9sqrt(x) + C
<=>
y = [9sqrt(x)+C]^(2/3)

I think that's right. Does that look ok?
 
Last edited:
is your answer the same as y=[(x^(3/2)+c)]^2
 
Nope.
 
csprof2000 said:
Nope.

oh, i guess your answer above is wrong.
because i have the answer key, and it said the answer for this question should be y=[(x^(3/2)+c)]^2

Thank you a lot.
im self learning Differential Equations, i really appreciate your help.
 
you have to integrate both sides simultaneously.
 
K.J.Healey said:
you have to integrate both sides simultaneously.

yea, that is what the book suggest me to do, but i got stuck when trying to integrate both sides
 
I accidentally did

dy/dx = 3 / sqrt(xy).

I'm surprised you didn't catch this. Oh well. Sorry for the confusion. Your problem is actually easier than the one I did.

dy/dx = 3sqrt(xy)
<=>
dy/sqrt(y) = 3sqrt(x)dx
<=>
2sqrt(y) = 2x^(3/2) + C
<=>
y = [x^(3/2)+C]^2

Hopefully that answers your question. It's separable.
 
csprof2000 said:
I accidentally did

dy/dx = 3 / sqrt(xy).

I'm surprised you didn't catch this. Oh well. Sorry for the confusion. Your problem is actually easier than the one I did.

dy/dx = 3sqrt(xy)
<=>
dy/sqrt(y) = 3sqrt(x)dx
<=>
2sqrt(y) = 2x^(3/2) + C
<=>
y = [x^(3/2)+C]^2

Hopefully that answers your question. It's separable.

its ok. thanks a lot csprof2000
im self learning this, so you will be surprise by how simple my questions are.
they are easy for you, but not esay for me.
im doing my best to get an overall knowledge of Differential Equations before i enter college.
thanks again csprof2000.
i have one question though, are you a teacher or professor?
 
  • #10
A professor, but thankfully not of anything as hard as this. They have me doing the introductory sequence, mostly...
 
  • #11
csprof2000 said:
A professor, but thankfully not of anything as hard as this. They have me doing the introductory sequence, mostly...

this explain why i spend like about 2 hours on 3 questions, and still can't figure out any of them, but you spend like a few minutes and sloved one of the questions.
where do you teach?
 
  • #12
I'd rather not say, if that's alright. I like to maintain some anonymity in these forums, and I don't want any of my students figuring out whom I am.

But it's in the USA, in the South. A state university...
 

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