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Differential Equation problem

  1. Jun 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Substituition y=vx , differential equation x2dy/dx = y2-2x2 can be show in the form x dv/dx = (v-2)(v+1)

    Hence , solve the differential equation x dv/dx = (v-2)(v+1) ,expressing answer in the form of y as a function of x in the case where y > 2x > 0 .


    3. The attempt at a solution

    I can only show the equation ,but can't solve the equation as the answer is y =x(2+Ax2) / (1-Ax2)

    TQ.
     
  2. jcsd
  3. Jun 19, 2011 #2

    vela

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    Show us what you tried so far.
     
  4. Jun 19, 2011 #3
    I start after from my showing.
    x dv/dx = (v-2)(v+1)
    dv/dx = (v-2)(v+1) / x
    dx/dv = x / (v-x)(v+1)
    1/x dx = 1 / (v-2)(v+1) dv

    Then I start to integrate , I am still new , so I don't know how to use to type the symbol.Sorry for the inconvenience.

    1/x dx = 1/3 ( (1 / (v-2) )- ( 1/ (v+1) ) -----(1/3 from partial fraction )
    ln x + c = 1/3 ln (v-2) - ln (v+1)
    ln x + c = 1/3 ln (v-2)/(v+1)

    Then I forgot know how to continue as i left my exercise book at home. I can only done so far.Then how could I continue ?
     
  5. Jun 19, 2011 #4

    HallsofIvy

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    This is wrong. If y= xv then dy/dx= x dv/dx+ v, not just x dv/dx.

     
  6. Jun 20, 2011 #5
    no , I have done using dy/dx= x dv/dx+ v, from the equation x2dy/dx = y2-2x2 .
     
  7. Jun 20, 2011 #6

    vela

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    Use the properties of logarithms:
    [tex]\begin{align*}
    \log ab &= \log a + \log b \\
    b \log a &= \log a^b
    \end{align*}[/tex]
    and exponentiate to get rid of the logs.
     
  8. Jun 20, 2011 #7
    Applying the properties of logarithm, i get
    y= x(2+Ax^3)/1-Ax^3
    which is supposed to be 'Ax^2?' HmMm.
     
  9. Jun 21, 2011 #8
    Same answer as mine.I got y= x(2+Ax^3)/1-Ax^3 .But the textbook's anwser is saying that 'Ax^2.The textbook probably wrong?
     
  10. Jun 21, 2011 #9

    vela

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    It's straightforward enough to check. Just plug your answer back into the original differential equation and see if it works.
     
  11. Jun 22, 2011 #10
    I will try , thank you so much for your help ^^
     
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