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Solving a differential equation

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve (xy+y2+x2) dx -( x2 )dy = 0

    2. Relevant equations
    to verify if exact [PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngM/[B][PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngy [Broken] = [PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngN/[B][PLAIN]http://upload.wikimedia.org/math/5/2/c/52cc749bb1c32abf1dccf613bd847a6e.pngx[/B][/B][/B][/B] [Broken]

    3. The attempt at a solution
    So I can see it isn't separable and linear, so I thought of solving it through substitution

    i did y=ux and dy= u dx + x du
    and I substituted them in my first equation giving me
    (xu+u2+x^2) dx - x2(udx + xdu) = 0]

    I'm stuck here... the answer key factors out x^2 from the first equation leaving it with x^2(u+u^2+1), which doesn't make sense to me.. because if I factored out x^2 I would be left with (x-1u+x-2u2+1)
    So any help on how I am seeing this wrong?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 2, 2015 #2

    ShayanJ

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    ## \partial_y M=\partial_y(xy+y^2+x^2)=x+2y ##
    ## \partial_x N=\partial_x(-x^2)=-2x ##
    ## \partial_y M \neq \partial_x N!!! ##
     
  4. Feb 2, 2015 #3
    I know how to verify if exact, hence why I chose substitution, but I am kinda stuck at how to factor my xα. Unless I didn't understand your answer correctly, to me it seems as if you're solving to see whether or not if exact.
     
  5. Feb 2, 2015 #4

    ShayanJ

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    That means your differential isn't exact which means it can't be integrated in this form, so you should turn it into an exact differential. There is such a method in your toolbox which you can't integrate the differential without it. You remember it?
     
  6. Feb 2, 2015 #5
    Well I know I can use substitution to reach a separable equation. If not, I know I can use υ(x,y) and multiply it to my equation. My question was more towards the substitution method because I don't completely grasp the subject.
     
  7. Feb 2, 2015 #6

    ShayanJ

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    By some manipulation, you can get ## y'=\frac{y}{x}+\frac{y^2}{x^2}+1 ##. Now try your substitution!
     
  8. Feb 2, 2015 #7
    I'm sorry, I'm not understanding. why you took the derivative of y. I was taught ( and my book explains) a different way. For instance, we just have to find a coeffiecient of xα to then create a separable DE. anyway, I'll check again later, maybe this will help getting a clearer answer? :/
     
  9. Feb 2, 2015 #8

    ShayanJ

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    ## M dx+N dy=0 \Rightarrow N dy=-M dx \Rightarrow \frac{dy}{dx}=- \frac{M}{N}##
     
  10. Feb 2, 2015 #9

    Ray Vickson

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    If you substitute ##y = x u## into the DE
    [tex] x^2 \frac{dy}{dx} = x^2 + xy + y^2 \; \Rightarrow \frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2[/tex]
    the resulting DE for ##u## is very simple and is straightforward to solve.
     
    Last edited by a moderator: May 7, 2017
  11. Feb 2, 2015 #10
    Yeah it was a stupid mistake, I was going fast and stressing out over an exam. I should've noticed that replacing y = ux into the equation can allow me to factor out x^2 because the function is homogeneous.
     
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