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Help with a method please?

  1. Feb 24, 2008 #1
    Hi,

    I am trying to:
    Show that y1(t)=t^2 and y2(t)=t^2+1

    for the differential equation:
    dy/dt=-y^2+y+2t^2y+2t-t^2-t^4

    This is clearly a non-linear, non-seperable equation, and I cannot see how to try the method of exact equations on it. Therefore, I am stuck.

    Anyone have any ideas of what to try?
     
  2. jcsd
  3. Feb 24, 2008 #2
    are you trying to show that y1(t)=t^2 and y2(t)=t^2+1
    are two solutions of that diffl. eq or, just to solve that diff. eq? if the former is the case, all you need to do is find the derivative of y1, y2 and plug in the diff eq and see if the right hand is equal to the left hand. if the latter than ther is more work to do!!
     
  4. Feb 24, 2008 #3
    Trying to show that those are solutions to the diff. eq.

    If I could solve the diff. eq. I guess it would show that those are particular solutions though.
     
  5. Feb 24, 2008 #4
    Show that y1(t)=t^2 and y2(t)=t^2+1

    for the differential equation:
    dy/dt=-y^2+y+2t^2y+2t-t^2-t^4
    y1'=2t, so

    2t=-(t^2)^2+t^2+2t^2*t^2+2t-t^2-t^4=-t^4+2t^4 +2t-t^4=2t, so this is a solution. try the same thing with y2
     
  6. Feb 24, 2008 #5
    Thank you much.
     
  7. Feb 25, 2008 #6
    This equation is a Riccati type equation, i.e. [itex]y'(t)=a(t)\,y(t)^2+b(t)\,y(t)+c(t)[/itex]. The general solution can be found if one knows a particular solution [itex]y_p(t)[/itex] by the transformation [itex]y(t)=y_p(t)+\frac{1}{u(t)}[/itex] which makes it a linear one.
    For the problem at hand if you write [itex]y(t)=t^2+\frac{1}{u(t)}[/itex] then the original ODE reduces to [itex]u'(t)=1-u(t)\Rightarrow u(t)=C\,e^{-t}+1[/itex].
     
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