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First order non linear DE

  1. Sep 20, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Solve the following DE: [itex]2xyy'=4x^2+3y^2[/itex].


    2. Relevant equations
    Bernoulli's DE: [itex]y'+P(x)y=Q(x)y^2[/itex].


    3. The attempt at a solution
    I know that the original DE isn't under Bernoulli's form, but I have thought a lot on the problem and my feeling is that if I could find a change of variable to transform the general DE into a Bernoulli's equation, I'd be done. I have tried [itex]z=4x^2+3y^2[/itex], so [itex]z'=8x+6yy'[/itex] but this leads me nowhere. I am not even sure I can reduce the original DE into a Bernoulli's equation. This is the only way I think I could solve the DE, I don't see any other way.
    I'd love some help.
     
  2. jcsd
  3. Sep 20, 2011 #2
    Divide both sides by x^2

    =)
     
  4. Sep 20, 2011 #3
    Don't you think this should be homogenous ODE instead of Bernoulli? :smile:
     
  5. Sep 20, 2011 #4

    fluidistic

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    Okay thanks guys. I divided by x² but was stuck right away.
    I checked out if it was homogeneous and indeed it was homogeneous of first order. I then opened my cookbook for DE's (Boas, 2nd edition) and followed his advice. I wrote the DE under the form [itex]P(x,y)dx+Q(x,y)dy=0[/itex], with [itex]P(x,y)=-4x^2-3y^2[/itex] and [itex]Q(x,y)=2xy[/itex]. He says I can write the DE under the form [itex]y'=f(y/x)[/itex].
    And thus the change of variable [itex]v=y/x[/itex] is appropriate, according to him. The DE should then be separable. However I tried but got stuck.
    I reach [itex]2x^3vdv+2x^2v^2dx-4x^2-3[x^2(dv)^2+xvdvdx+v^2(dx)^2]=0[/itex]. I don't know how to deal with the squared differentials nor the crossed ones (dxdv).
    Any further help will also be appreciated. :)
     
  6. Sep 20, 2011 #5

    dynamicsolo

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    You should not have squared differentials. If you are using the substitution v = y/x , presumably to replace the y , then you need to differentiate implicitly y = vx to find dy/dx in terms of dv/dx . You should be able to use this to replace y' with v' (and also y/x with v ) to produce a transformed differential equation that is easier to work with. (The result isn't pretty, but it is separable and integrable...)
     
    Last edited: Sep 21, 2011
  7. Sep 21, 2011 #6
    Just to give you a little hint:

    [tex]2xyy'=4x^{2}+3y^{2}[/tex]

    divide by 2xy throughout:

    [tex]y'=2(x/y)+3(y/x)[/tex]

    Let v=y/x ,

    dv/dx = ?
     
  8. Sep 21, 2011 #7

    fluidistic

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    Thanks once again guys!
    [itex]y=xv \Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx}[/itex].
    Thus the original DE transforms into [itex]v+x\frac{dv}{dx}=2v+\frac{3}{2v}\Rightarrow \frac{xdv}{dx}=v+\frac{3}{2v} \Rightarrow \frac{dx}{x}=\left ( v+\frac{3}{2v} \right ) ^{-1}dv[/itex]. Now I must integrate to get v(x). Once I have v(x) I divide it by x in order to get y(x).
    I tried this way too but strangely I don't get "nice" stuff. [itex]dv/dx= \frac{1}{x^2} \left ( \frac{xdy}{dx}-y \right )[/itex]. If I isolate dy/dx, I don't even reach the one I got from the above way. All this, despite that y=xv in the first case and v=y/x in the second case. I don't understand how I can reach a different differential for dy/dx...
     
  9. Sep 21, 2011 #8

    dynamicsolo

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    OK, so far. (And, yes, icystrike has a small typo in the resulting equation. The "3" should be "3/2"...)

    You shouldn't still have that "y" kicking around in there. (I thought you were going from icystrike's equation, so this is a little puzzling. You should again have [itex] v + x\frac{dv}{dx} = 2v + \frac{3}{2v} [/itex].)

    As I said above, the result isn't going to be very pretty. (Solutions of DEs often aren't...) You would reduce your transformed equation to [itex]\frac{dx}{x} = ( \frac{2v}{2v^{2} + 3} ) dv[/itex] and integrate from there.
     
    Last edited: Sep 21, 2011
  10. Sep 21, 2011 #9

    fluidistic

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    Thank you again for your help.
    Ah I see.


    Well I used the derivative of the quotient y(x)/x. It's [y'(x)x-y(x)]/x²
    I see. I think you have a small typo in the right hand side (I think you forgot a ^-1).
     
  11. Sep 21, 2011 #10

    dynamicsolo

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    I already added the ratios and "flipped over" the result from your differential equation result for v : [itex] ( v+\frac{3}{2v} ) ^{-1} = \frac{2v}{2v^{2} + 3} [/itex] .
     
  12. Sep 21, 2011 #11

    fluidistic

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    Ah right I see now!
    Do you have an idea what's my problem using icystrike's suggestion? Why can't I get rid of the y term and I don't reach the same expression for dv/dx.
     
  13. Sep 22, 2011 #12

    dynamicsolo

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    To be frank, I am still trying to figure out how you arrived at your result. The intent of icystrike's suggestion was to use v = y/x to completely eliminate y from the equation; you use Bernoulli's (whichever of the nine of them it was) idea to transform the original differential equation into a new one involving v and x that (hopefully) is easier to solve.

    By making only a partial substitution, you created a situation that is harder to resolve algebraically than it needs to be. (It can probably be solved from there, but less easily. You would get rid of the y term by using y = vx .)
     
  14. Sep 22, 2011 #13

    ehild

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    The equation contains y2 and 2yy'=(y2)', this suggest the substitution z=y2. With this substitution, the equation becomes linear.

    ehild
     
  15. Sep 22, 2011 #14

    fluidistic

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    Icystricke asked me dv/dx rather than dy/dx in function of dv/dx.
    What I did for his way is [itex]v=y/x \Rightarrow \frac{dv}{dx}= \left ( \frac{xdy}{dx} -y \right ) \frac{1}{x^2}[/itex].
    WOW! This work too! After isolating dy/dx and replacing y/x by v, I reach exactly the same expression for dy/dx than you. Nice!
    Thanks guys for all your help. :biggrin:


    Edit: Okay thanks a lot ehild. I'm going to try this tomorrow (it's past 2 am here).
     
  16. Sep 22, 2011 #15

    ehild

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    Then you live far to west from me. :smile:
    Just to be able to check your derivation: it should be very easy, and results in y2=cx3-4x2.

    ehild
     
  17. Sep 22, 2011 #16

    dynamicsolo

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    Thank you -- that's good spotting and the result for y is faster to extract than it is to do so from v . (I assumed from the initial post that it was required to use Bernoulli's method.)
     
  18. Sep 22, 2011 #17
    Yes TS should try this method too. That is a direct consequence of implicit diff.
     
  19. Sep 22, 2011 #18

    fluidistic

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    Thanks. So I reached the same solution as yours. Problem solved. :smile:
     
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