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Unseparable O.D.E.

  • Thread starter asdf1
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734
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another O.D.E. question:
xyy`= 2y^2 +4x^2
also unseparable...
 

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Galileo
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I`m a bit rusty on D.E.'s, but you can write it as:

[tex]y'=2(y/x)+4(x/y)[/tex]

IIRC, whenever you have y'=F(y/x), the equation also called homogeneous (different from the homogeneous linear ODE definition), this one is invariant under 'zoom'.

If you make the subsitution z=y/x, the equation becomes:
[tex]z+xz'=2z+4/z[/tex]
or
[tex]xz'=z+4/z[/tex]
which is separable.
 
Last edited:
GCT
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it's separable, use v=y/x
 
734
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why'd you think of v=y/x?
 
TD
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Galileo explained. It's because the DE is homogeneous (not with the 0 constant, but with the sum of all exponents of all terms is constant (here 2). You can then divide so the DE becomes a function of (y/x). Susbtitution will make it separable.
 
734
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so the first step when you see an O.D.E you have to check if it's separable
=> if not, then check if it's homogeneous, and use that substitution...
thanks! :)
 
TD
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Checking if it's separable should indeed always be the first you do and when it's not, try to see if you can make it separable with a substitution.
The substitution used here a very common one.
 
734
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what are the most common substitutions used?
 
584
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Well y = v(x)x is a common one and if I recall correctly, for second order isobaric equations you can make the substitution y = vx^m if there is a single value of m which makes the equation dimensionally consistent (you assign weights to y, dy, x and dx).
 
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734
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ok~ thanks!!! :)
 
734
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hmmm... here's my work~
(i forgot to include that the original condition is y(2)=4, sorry about that!)
y`= 2y/x+4x/y
suppose y=vx
=>v=y/x
so y`=v+v`x
=> v+xv`=2v +4/v
=> v`x=v+4/v
=> v`x=(v^2 +4)/v
=> v/(v^2 +4)dv=dx
=> 1/2[dv^2/(v^2 +4)] =dx
=> 1/2ln[absolute value (v^2 +4)] =x+ c`
=> ln[absolute value (y^2/(x^2) +4)]=2x+c`
=> y^2/(x^2) +4 =e^(2x +c`)
=> y^2 +4x^2 =c(x^2)*[e^(2x)]
=> y^2=c(x^2)*[e^(2x)]-4x^2
now my problem is that when you take the square root on both sides of the equations, the right side should have a positive and negative root, but the correct answer should only have a positive square root~
why?
btw, what is meant by the different homogeneous definitions and "zoom"?

Galileo said:
whenever you have y'=F(y/x), the equation also called homogeneous (different from the homogeneous linear ODE definition), this one is invariant under 'zoom'.
 
Galileo
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asdf1 said:
hmmm... here's my work~
(i forgot to include that the original condition is y(2)=4, sorry about that!)
y`= 2y/x+4x/y
suppose y=vx
=>v=y/x
so y`=v+v`x
=> v+xv`=2v +4/v
=> v`x=v+4/v
=> v`x=(v^2 +4)/v
=> v/(v^2 +4)dv=dx
Whoopsie, you've lost an x.

now my problem is that when you take the square root on both sides of the equations, the right side should have a positive and negative root, but the correct answer should only have a positive square root~
why?
The actual answer also needs a square root since you'll get a relation between y^2 and x. Now if this equation is true both the positive and negative square roots must be valid solutions to the ODE, but a little bit of thought shows that no more than one can fit the same boundary condition (y(2)=4)
btw, what is meant by the different homogeneous definitions and "zoom"?
A linear DE of, say, second order:
[tex]y''+A(x)y'+B(x)y=r(x)[/tex]
is called homogeneous if r(x)=0. Your ODE is a also called homogeneous, but it's in a different context, so it naturally hasn't the same meaning.

Since the ODE can be written in terms of y/x. If we make the change of variables [itex]\hat y=ay[/itex] and [itex]\hat x=ax[/itex] for some positive a. So that we're scaling the y and x-axes by the same amount (this is what I meant by zoom), the ODE looks the same, since [itex]\hat y/\hat x=y/x[/itex].
 
734
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thank you very much!!! :)
i'll try and be more careful on my calculations~
 

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