• Support PF! Buy your school textbooks, materials and every day products Here!

Solve this diff eq.

  • Thread starter S.Amstein
  • Start date
  • #1
4
0

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.
 

Answers and Replies

  • #2
phyzguy
Science Advisor
4,496
1,444
Try reading this.
 
  • #3
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751

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
4
0
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
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
Last edited:
  • Like
Likes 1 person
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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?
 
  • Like
Likes 1 person
  • #7
4
0
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 Threads on Solve this diff eq.

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
11
Views
1K
Replies
4
Views
960
  • Last Post
Replies
1
Views
543
  • Last Post
Replies
2
Views
948
  • Last Post
Replies
2
Views
3K
Replies
2
Views
747
Replies
7
Views
513
Replies
4
Views
950
Top