Solve Diff Eq: Is it Homogenous?

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Homework Help Overview

The discussion revolves around determining whether a given differential equation is homogeneous and, if so, solving it. The specific equation under consideration is dy/dx = (x^2 - 2y^2) / (xy), which falls under the subject area of differential equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the homogeneity of the differential equation and attempt to apply substitutions, such as y = xv, to simplify the equation. There are discussions about potential mistakes in the simplification process and the integration steps involved.

Discussion Status

Some participants have provided guidance on identifying errors in the integration process and the manipulation of logarithmic expressions. There is an ongoing examination of the steps taken, with multiple interpretations of the problem being explored.

Contextual Notes

Participants note the importance of showing work to identify mistakes and clarify assumptions. There is also mention of confusion arising from changes in the problem statement, which may affect the discussion.

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Homework Statement


determine whether the given differential equations are homogenous and, if so, solve them.


Homework Equations



dy/dx = (( x^2 -2y^2)/xy )


The Attempt at a Solution



i assumed it was homoG.

Then i replaced ( x*v) for each Y
and made dy/dx --> v+x*(dv/dx)

i simplify to x*(dv/dx) = (1/v) + v

And i know here that i made a mistake . because the solution is

y^2=kx^4-x^2

Can someone help me please.
 
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S.Amstein said:

Homework Statement


determine whether the given differential equations are homogenous and, if so, solve them.


Homework Equations



dy/dx = (( x^2 -2y^2)/xy )


The Attempt at a Solution



i assumed it was homoG.

Then i replaced ( x*v) for each Y
and made dy/dx --> v+x*(dv/dx)

i simplify to x*(dv/dx) = (1/v) + v

And i know here that i made a mistake . because the solution is

y^2=kx^4-x^2

Can someone help me please.

You need to show your work so we can see where you went wrong, if you did. And since you haven't finished solving for v and hence y, how do you know it is wrong in the first place?
 
Y' = (x^2+2y^2)/xy

make y =xv , y'=v + xv'

simplify to xv'= (1+v^2)/v
rearrange to 1/x dx = (v/(1+v^2))dv

Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C

simplify ----> v = ( x^2 + C) ^ (1/2)

Y = x*( x^2 + C) ^ (1/2)

Y^2= x^4 + (x^2*C)

But the solution in the back is y^2 = kx^4 -x^2

Where did i go wrong?
 
S.Amstein said:
Y' = (x^2+2y^2)/xy

make y =xv , y'=v + xv'

simplify to xv'= (1+v^2)/v
rearrange to 1/x dx = (v/(1+v^2))dv

Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C

simplify ----> v = ( x^2 + C) ^ (1/2)

Y = x*( x^2 + C) ^ (1/2)

Y^2= x^4 + (x^2*C)

But the solution in the back is y^2 = kx^4 -x^2

Where did i go wrong?

It's somewhere between your "Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C" (which looks correct) and the "simplify ----> v = ( x^2 + C) ^ (1/2)" (which looks incorrect). How did you do that? Might also note you've changed your problem statement, which may have caused some confusion.
 
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S.Amstein said:
Y' = (x^2+2y^2)/xy

make y =xv , y'=v + xv'

simplify to xv'= (1+v^2)/v
rearrange to 1/x dx = (v/(1+v^2))dv

Integrate ---> ln ( 1+v^2) ^ (1/2) = ln (x) + C

simplify ----> v = ( x^2 + C) ^ (1/2)
Here is your error. Removing the logarithms gives (1+ v^2)^{1/2}= Cx, NOT what you have.
Squaring, 1+ v^2= 1+ y^2/x^2= Cx^2

Y = x*( x^2 + C) ^ (1/2)
Y^2= x^4 + (x^2*C)

But the solution in the back is y^2 = kx^4 -x^2

Where did i go wrong?
 
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HallsofIvy said:
Here is your error. Removing the logarithms gives (1+ v^2)^{1/2}= Cx, NOT what you have.
Squaring, 1+ v^2= 1+ y^2/x^2= Cx^2

Ahh, there it is. Thank You !

I still don't have a 100% handle on why the removal of logarithms creates a multiplication of the inside terms . is there a theorem or rule for this process?
 

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