- #1

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for this O.D.E. :

y`= (x^2 + y^2)/xy

it's unseparable, so what other methods can there be taken?

y`= (x^2 + y^2)/xy

it's unseparable, so what other methods can there be taken?

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- Thread starter asdf1
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- #1

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for this O.D.E. :

y`= (x^2 + y^2)/xy

it's unseparable, so what other methods can there be taken?

y`= (x^2 + y^2)/xy

it's unseparable, so what other methods can there be taken?

- #2

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So if you substitute tx and ty for x and y respectively, you'll see that equality holds. Then do following substitution: y = x V(x) into the DE and see what you get with applying dy/dx and some simplifications. It should reduce to separable equation.

- #3

Galileo

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[tex]y'=x/y+y/x[/tex]

Again, with the substitution z=y/x it becomes:

[tex]z+xz'=1/z+z[/tex]

or

[tex]z'=1/(xz)[/tex]

which is separable.

- #4

GCT

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it's homogenous, separable, use v=y/x

- #5

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ok, thanks!

- #6

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y`=x/y +x/y

suppose y=vx

=>v=y/x

=> y`=v+xv`

so v+xv`=1/v+v

=> vdv=dx/x

=> v^2=ln[absolute value(x)]+c`

=> (y^2)/(x^2)=ln[absolute value(x)]+c

=> y^2=2(x^2)ln[absolute value(x)]+cx^2

now if i take the square root on both sides, there should be a positive and negative sign on the right~

the correct answer should only have the positive sign, but how can you be sure that it should be positive?

- #7

Galileo

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Whoopsie, you've forgot a factor 1/2 on the left side.asdf1 said:here's my work:

y`=x/y +x/y

suppose y=vx

=>v=y/x

=> y`=v+xv`

so v+xv`=1/v+v

=> vdv=dx/x

=> v^2=ln[absolute value(x)]+c`

You can use either sign, both will be valid solutions to the ODE. This becomes clear when you plug y back into the ODE to check if it works out. Then, with the benefit of hindsight, you could foresee this, since if y is one solution, then the other is -y and it's derivative is -y'. If you write the ODE as xyy'=x^2+y^2 you can see that if y is a solution, then -y is too.

Ofcourse, if you're given a boundary value or initial value/condition then there will be only one solution. (otherwise the problem is ill-stated).

- #8

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thank you very much! :)

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