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Have I solved this ODE correctly?

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    y' + siny + xcosy + x = 0

    3. The attempt at a solution
    well, I've used Weierstrauss substitutions: siny = 2t/(1+t^2) , cosy = (1-t^2)/(1+t^2) , dy = 2dt/(1+t^2) where t=tan(x/2).

    2/(1+t^2)dt + 2t/(1+t^2)dx + x(1-t^2)/(1+t^2)dx + xdx = 0
    2dt + 2tdx + x(1-t^2)dx + x(1+t^2)dx = 0
    2dt + (2t + x(1-t^2) + x(1+t^2))dx = 0
    2dt + (2t + x - xt^2 + x + xt^2)dx = 0
    2dt + (2t + 2x)dx = 0
    dt + (t+x)dx = 0

    well, if I multiply e^x to both sides, it'll be an exact differential and after integrating I'll have:
    te^x + xe^x - e^x + C. after substituting t= tan(x/2) we obtain:
    e^x(tan(y/2) + x - 1) + C.

    well, this is a problem that our professor gave in the last lecture for extra score in the final exam. so I wanna be sure that I've solved it and my method is correct.
    Is it correct if I substitute t=tan(x/2)? because the original problem is defined everywhere but after I use Weierstrauss substitutions then I should restrict x to the x's that are defined. (x=pi and -pi are obviously not defined now). Wouldn't that cause problems?
     
  2. jcsd
  3. Oct 24, 2011 #2

    Mark44

    Staff: Mentor

    You should have an equation. Is it te^x + xe^x - e^x + C = 0?
    You can check your solution yourself. Differentiate your final equation implicitly, and substitute into your differential equation.
     
  4. Oct 24, 2011 #3
    Yea. It's te^x + xe^x - e^x + C = 0. (I had written only the answer of the integration).

    Of course, but that would be terribly tedious.
     
  5. Oct 24, 2011 #4

    Mark44

    Staff: Mentor

    So it's OK for us to do the tedious work of checking, but not you?
     
  6. Oct 24, 2011 #5
    lol. I'm asking you to check if I've done it correctly. that means I'm asking others to see whether what I've done makes sense to them or not.
     
  7. Oct 24, 2011 #6

    Mark44

    Staff: Mentor

    Who are you going to ask for a problem on a test? Checking your solution is a good habit to get into when you're solving differential equations.
     
  8. Oct 24, 2011 #7
    well, that's a good advice. I checked my solution and it worked. but there is one question remained to be answered. There are points like x=pi where tan(x/2) is not defined. those points certainly need to be taken care of I think. How should I treat those points?
     
  9. Oct 25, 2011 #8
    well, I realized that I'll have to treat the points of the form y=2kπ + π separately (because for such points tan(y/2) is not defined and I'm missing some solutions of the ODE for that reason). if I plug y=2kπ+π into the equation I'll obtain:

    y' + sin(2kπ+π) + xcos(2kπ+π) + x = 0
    y' + 0 + x(-1) + x = 0 -> y' = 0 which means y = C.

    Do I need to find what C is?
     
  10. Oct 26, 2011 #9
    Any ideas?
     
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