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Homework Help: Plz help in the result of: sin3xdx+2y(cos3x)^3dy =0

  1. Nov 4, 2008 #1
    1. Solve by Seperable Variable: sin3xdx + 2ycos33xdy = 0

    2. We need write it in the form: dy/dx = g(x).h(y)

    3. The attempt at a solution:

    a few algebraic operations lead to: dy/dx = (-sin3x/cos33x).(1/2y) with h(y) = 1/2y
    now we have (y2)' = 2ydy = -sin3xdx/cos33x
    integrating we get: y2 = [tex]\int[/tex]-sin3xdx/cos33x =G(x)

    thus y2 = (1/3).[tex]\int[/tex]d(cos3x)/cos33x = (1/3).(-1/2).(1/cos23x)
    so y2 = -1/6cos23x. :confused:

    here i reached a dead end. is there anything wrong in the attempt or is it right as i think & the given d.e. is wrong ? note that i tried to solve it more then 10 times, i also tried the tan3x & its derivative but the -ve sign persists.

    waiting for your help. thx in advance
     
  2. jcsd
  3. Nov 4, 2008 #2

    gabbagabbahey

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    Assuming that you mean : [tex]y^2=\frac{-1}{6cos^2(3x)}[/tex] , then your solution is correct.

    When you differentiate it you get:

    [tex]2ydy=(-2)(3) \left( \frac{sin^3(3x)}{6cos^3(3x)} \right)=\frac{-sin^3(3x)}{cos^3(3x)}[/tex]

    which gives you back your original ODE...why were you thinking it was incorrect?
     
  4. Nov 4, 2008 #3

    HallsofIvy

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    That is a solution to the differential equation. It is not the general solution.

    What happened to the constant of integration in [itex]y^2 = -\int sin3xdx/cos33x[/itex]
     
  5. Nov 4, 2008 #4
    nop but my doctor gave it besides others as ana assignment & i tried it many times - at first yeh i though it was wrong - so i got the same answer over & over again.
    So there's some problem with the D.E. given. I use the book: A first Course in Differential equations by Dennis Zill 8th ed.
     
  6. Nov 4, 2008 #5
    ofcorse its a solution. there's many sol.s , but eventhough the answer y2 is -ve which is impossible.
     
  7. Nov 4, 2008 #6

    tiny-tim

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    Welcome to PF!

    Hi bobmerhebi! Welcome to PF! :smile:
    hmm … that took a long time …

    it's quicker to start by separating them into the form h(y)dy = g(x)dx …

    the clue's in the word "separable" :wink:
    Yes, you're wondering how the LHS can be positive and the RHS negative …

    that'll teach you not to leave out the constant of integration in future, won't it? :rolleyes:
     
  8. Nov 4, 2008 #7
    Re: Welcome to PF!

     
  9. Nov 4, 2008 #8

    tiny-tim

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    Sorry, bobmerhebi, but that makes no sense at all …

    a constant of integration should just be added … at the end. :smile:
     
  10. Nov 4, 2008 #9
    so it should look like this: y2 = -1/6cos23x+ c? im not getting it! It seems to be the same but a famile of solutions which i think are still invalid. Am I missing something?
     
  11. Nov 4, 2008 #10

    tiny-tim

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    Yes! :biggrin:

    (except personally I'd write it y2 = C - 1/6cos23x)
     
  12. Nov 4, 2008 #11

    gabbagabbahey

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    What is wrong with taking the square root of a negative number? You just get an imaginary value of y, no big deal....

    If you only want real-valued solutions, then you have to restrict your Domain.
     
  13. Nov 4, 2008 #12
    in this case the constant should be greater or equal than 1/6cos23x. so here there should be a condition right?

    thx 4 the help
     
  14. Nov 4, 2008 #13

    gabbagabbahey

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    Your constant is a constant...how can it always be larger than [itex]\frac{1}{6cos^2(3x)}[/itex]? ??

    Instead, if you want only the real-valued solutions, you have to restrict your x-values (your Domain): only values of x for which [itex]\frac{-1}{6cos^2(3x)}+C \geq 0[/itex] give real solutions for y.

    Your constant [itex]C[/itex] is determined by the initial conditions of the system. For example; if you are told that y(0)=1/sqrt(6), then C will be 1/3.
     
    Last edited: Nov 4, 2008
  15. Nov 4, 2008 #14

    tiny-tim

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    No, the constant should be ≥ 0. :smile:

    (and as gabbagabbahey says, x is limited)
     
  16. Nov 4, 2008 #15
    aha ok then the independent variable should be over a restricted domain to get real valued solutions otherwise the solution is in the complex system. thank you all 4 ur help
     
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