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Differential equation 5y4y' = x2y' + 2xy

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of:

    5y4y' = x2y' + 2xy


    2. The attempt at a solution

    Well I've so far tried to simplify by making the equation really:

    (5y4-x2)y' - 2xy = 0

    Now this will let us use exact equations such as:

    N(x,y)= 5y4-x2
    and
    M(x,y)= -2xy

    since ∂N=∂M

    After this though im not sure what to do.
     
  2. jcsd
  3. Feb 19, 2013 #2

    vela

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    Re: differential equation

    If you have ##\Phi(x,y)=0##, when you differentiate it, you get
    $$\frac{\partial\Phi}{\partial x}dx + \frac{\partial\Phi}{\partial y}dy = 0.$$ Compare that to what you have
    $$(-2xy)\,dx + (5y^4-x^2)\,dy = 0.$$ How would you (partially) recover ##\Phi## from M(x,y) or N(x,y)?
     
  4. Feb 19, 2013 #3
    Re: differential equation

    I then took the integral of M(x,y) thus giving me f-x2y

    Correct or no?
     
  5. Feb 19, 2013 #4

    vela

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    Re: differential equation

    That's good, so you have ##\Phi=-x^2y + f(y)##. Now differentiate that with respect to y and compare the result to N(x,y).
     
  6. Feb 19, 2013 #5
    Re: differential equation

    finding the solution to the f(y) gives us y5 then plugging this into the integral part we get a solution of:

    F(x,y) = y5-x2y

    Is that it?
     
  7. Feb 19, 2013 #6
    Could you help with another? The differential equation is:

    y' = 2(xy' + y)y3
     
  8. Feb 19, 2013 #7

    vela

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    Almost. You just need to include the constant of integration.
     
  9. Feb 19, 2013 #8
    F(x,y) = y5-x2y + C ?
     
  10. Feb 19, 2013 #9

    vela

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    Yup, that's right, and F(x,y) should be set to 0. Remember you're trying to find y(x), not F(x,y). F(x,y)=0 specifies y implicitly as a function of x.
     
  11. Feb 19, 2013 #10
    Thanks! Could you help with the other equation I posted? This one I don't even have a clue how to start
     
  12. Feb 19, 2013 #11

    vela

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    You could start by checking if it's exact.
     
  13. Feb 19, 2013 #12
    No it's not, I got

    N=1-2xy3
    and
    M=-2y4

    which their partials don't equal eachother
     
  14. Feb 19, 2013 #13

    vela

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    In that case, try finding an integrating factor that'll make it exact.
     
  15. Feb 19, 2013 #14

    SammyS

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    It would have been better to start a new thread for this new problem.

    (xy' + y) is d(xy)/dx .

    See if you can use that.
     
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