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Question on ODEs

  1. Nov 1, 2011 #1
    I've been asked to consider differential equations of the form y=xp+f(p), where p=dy/dx, and to show that the general solution is y=cx+f(c).

    Substituting in p, the original equation is y=x(dy/dx)+f(dy/dx), and by differentiating I get:

    dy/dx=x(d2y/dx2) + dy/dx + d2y/fx2 * f`(dy/dx).

    Substracting dy/dx from both sides:

    0 = x(d2y/dx2) + d2y/fx2 * f`(dy/dx).

    Factorising d2y/dx2:

    0=d2y/dx2(x+f`(dy/dx))

    Thus in one case d2y/dx2=0 => dy/dx=c => p=c. Substituting into the original equation we get y=cx+f(c) so that part is done.

    It then asks me to further show that there is a second solution obeying the differentiatial equation:

    d(f(p))/dp + x = 0

    Now I suspected you could solve this by looking at the second case above where x+f`(dy/dx)=0 but I'm not sure how. I noticed you get this new differential equation by simply differentiating the original equation by p. However surely I can't ignore the case of x+f`(dy/dx)=0, it must be involved somehow?

    Thanks for any help
     
  2. jcsd
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