Solve Diff Eq: Is it Homogenous?

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In summary, the student attempted to solve a differential equation but made a mistake. They replaced each variable with its corresponding y-value and simplified the equation to x*(dv/dx) = (1/v) + v. However, when solving for v, they got y^2=kx^4-x^2 which is not the correct equation. Can someone help them please?
  • #1
S.Amstein
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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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  • #2
Try reading this.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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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  • #6
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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  • #7
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?
 

Related to Solve Diff Eq: Is it Homogenous?

1. What is a Homogeneous Differential Equation?

A homogeneous differential equation is a type of differential equation where all the terms can be written in terms of the dependent variable and its derivatives. In other words, the equation is "homogeneous" because all the terms have the same degree.

2. How can I recognize if a Differential Equation is Homogeneous?

To recognize if a differential equation is homogeneous, you can check if all the terms can be written in terms of the dependent variable and its derivatives. If this is the case, then the equation is homogeneous.

3. What is the general solution to a Homogeneous Differential Equation?

The general solution to a homogeneous differential equation is a solution that satisfies the equation for all possible values of the independent variable. It is usually expressed in terms of a constant value, known as the arbitrary constant.

4. How is a Homogeneous Differential Equation solved?

A homogeneous differential equation can be solved using various methods such as separation of variables, substitution, or integrating factors. The method used depends on the specific equation and its characteristics.

5. Are there any real-life applications of Homogeneous Differential Equations?

Yes, homogeneous differential equations have many real-life applications in fields such as physics, engineering, and economics. For example, they can be used to model the growth of populations, chemical reactions, and electrical circuits.

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