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Differential equations: y’= sin(cx) sin(yx), y(0)=1

  1. Aug 23, 2011 #1
    It’s a non-linear first order ode, the trig form of which is y’= sin(cx) sin(yx), y(0)=1. It’s possible that I’ve forgotten an important detail in the construction of the equation. If there’s someone who has some ideas about solving or approximating this, or who knows how to get mathematica or matlab to do something useful with it, I would be so grateful. Thank you in advance!
  2. jcsd
  3. Aug 24, 2011 #2


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    Here's one way. Since y(0)= 1 and y'(x)= sin(cx)sin(xy), y'(0)= sin(0)sin(0)= 0.
    Differentiating, y''(x)= ccos(cx)sin(xy)+ ysin(cx)cos(xy) and so y''(0)= 0. Differentiating again, y'''(x)= -c^2sin(cx)sin(xy)+ 2cycos(cx)cox(xy)- y^2sin(cx)sin(xy) and y'''(0)= 2cy(0)= 2c. So far, we have, for the Taylor's series about x=0, y(x)= 1+ 0x+ 0x^2+ 2c/3! x^3= 1+ (c/3)x^3. Although, obviously, the calculations are going to get more complicated, continuing in that way, you will have a power series approximatiom for y.
  4. Aug 24, 2011 #3
    This is what I do: give it a few minutes of a little of this, little of that myself. Ok, no go. Then I shamelessly turn to Mathematica's DSolve which attempts to find a symbolic solution. Still no go. I'm feelin' good by now cus' Mathematica can't do it neither. Finally I stop trying to be a purist and just turn to Mathematica's NDSolve which computes a numeric solution of the IVP:

    mysol=NDSolve[{y'[x]==Sin[myc x] Sin[y[x] x],y[0]=0},y,{x,0,1}];

    . . . yeah, I know. I'm not proud. Hall's got a good idea up there. If you're a Mathematica junkie, it wouldn't be hard to get maybe say 20-30 of those derivatives, maybe more, before Mathematica starts choking. Then form the series and compare it to the numerical results. Be good practice for when you really need to do something with the software.
    Last edited: Aug 24, 2011
  5. Aug 25, 2011 #4
    This looks promising! Thankyou! Looking into a Taylor expansion..
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