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Differential equations

  1. Oct 1, 2009 #1
    So i've just started Differential equations and already im stuck.

    I have the Question:

    xy' = y² + y It also gives me (u = y/x) and i can see how to get u.


    So now i have y' =(y²/x) + u Now im stuck, i'm not really sure of the exact way of going about this.
     
  2. jcsd
  3. Oct 1, 2009 #2

    Mark44

    Staff: Mentor

    This equation is separable, so that you can get dy and the y terms on one side, and dx and the x term on the other. Then integrate, not forgetting the constant of integration.
     
  4. Oct 1, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you want to treat it as a "homogeneous" equation, which is what the "hint", to take u= y/x implies, then dividing the equation by x gives [itex]y'= y^2/x+ y/x[/itex] as you say and, yes, that is [itex]y'= y^2/x+ u[/itex]. But you can also write [itex]y^2/x= (y/x)y= uy[/itex]. Your equation becomes y'= uy+ u= u(y+1) which is again separable:
    dy/(y+1)= u du.

    I tend to agree with Mark44, that, since it is already separable, doing it as dy/(y^2+ y)= dx/x seems more obvious.
     
  5. Oct 1, 2009 #4
    I would have done it automaticly like that, exept it states * use u=y/x *

    This is what the given solution is:

    PICTAR.jpg

    I think i might stick with my answer and see what he says.
     
  6. Oct 1, 2009 #5

    Mark44

    Staff: Mentor

    Check your solution by calculating dy/dx. It should be that for your solution xdy/dx = y2 + y. If so, you're in business. If not, you need to go back and see where you went wrong.
     
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