- #1

asdf1

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another O.D.E. question:

xyy`= 2y^2 +4x^2

also unseparable...

xyy`= 2y^2 +4x^2

also unseparable...

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- Thread starter asdf1
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- #1

asdf1

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another O.D.E. question:

xyy`= 2y^2 +4x^2

also unseparable...

xyy`= 2y^2 +4x^2

also unseparable...

- #2

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.

[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.

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- #3

GCT

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it's separable, use v=y/x

- #4

asdf1

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why'd you think of v=y/x?

- #5

TD

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- #6

asdf1

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=> if not, then check if it's homogeneous, and use that substitution...

thanks! :)

- #7

TD

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The substitution used here a very common one.

- #8

asdf1

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what are the most common substitutions used?

- #9

Benny

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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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- #10

asdf1

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ok~ thanks! :)

- #11

asdf1

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(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'.

- #12

Galileo

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Whoopsie, you've lost an x.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

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)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?

A linear DE of, say, second order:btw, what is meant by the different homogeneous definitions and "zoom"?

[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].

- #13

asdf1

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thank you very much! :)

i'll try and be more careful on my calculations~

i'll try and be more careful on my calculations~

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