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

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SUMMARY

The general solution for the differential equation dy/dx = 3√(xy) is y = [9√(x) + C]^(2/3), derived through separation of variables. An alternative solution, y = [(x^(3/2) + C)]^2, was incorrectly suggested by a participant but later clarified. The discussion emphasizes the importance of correctly applying integration techniques and understanding separable equations in differential equations. Participants shared insights on solving the equation and the learning process involved in self-studying this mathematical topic.

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  • Understanding of differential equations, specifically separable equations.
  • Familiarity with integration techniques.
  • Knowledge of square root functions and their properties.
  • Basic algebraic manipulation skills.
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  • Practice integrating functions involving square roots.
  • Explore the implications of initial conditions on general solutions.
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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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