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guess

find the general solution y``-5y`-6y=cos x

  • one year ago
  • one year ago

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  1. scarydoor
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    this is a 2nd order linear differential equation. There is a really easy method for solving these things. Do you know of it? You should have been told it in class. Otherwise you can search google and it'll tell you the method.

    • one year ago
  2. Aperogalics
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    @guess it is a second order equation it as two parts one general and particular and their sum is the complete solution so for general put y''-5y'-6y=0 and replace y^n by D^n and solve quadratic...............try it:)

    • one year ago
  3. Aperogalics
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    @RolyPoly let him solve:)

    • one year ago
  4. RolyPoly
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    First, find the complementary solution by setting the left side equal to 0. y``-5y`-6y=0 <- homogeneous solution, so, λ^2 - 5λ -6 =0 Solve λ \[y_c = c_1e^{\lambda _1 x} + c_2e^{\lambda _2 x}\] Then, find out the particular solution, try \(y_p = Asinx+Bcosx\) (No guarantee that it works though). Differentiate it and solve the A and B. Then, you get \(y_p\) \[y = y_c +y_p\] @Aperogalics I never intend to give out the answer..

    • one year ago
  5. guess
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    y(h)=c1e^(-x)+c2e^(6x) g(x)=cos x y(P)=a cosx+b sin x >> I stood at the 5A-7B=0 -(5B+7A)=0 Then ??

    • one year ago
  6. guess
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    @RolyPoly @Aperogalics @scarydoor thank you but i need some help

    • one year ago
  7. RolyPoly
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    Hmm... Do you mind showing your work for the particular solution?

    • one year ago
  8. scarydoor
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    It's called 'method of undetermined coefficients'. If you're going to be doing differential equations a fair bit, then I recommend this book: http://www.amazon.com/Ordinary-Differential-Equations-Dover-Mathematics/dp/0486649407/ref=sr_1_1?ie=UTF8&qid=1354192173&sr=8-1&keywords=ordinary+differential+equations It's about the best book you can get for it.

    • one year ago
  9. scarydoor
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    Do you know how it works? The method to find a particular solution? That seems to be what you're stuck on. This is the method to use: http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

    • one year ago
  10. guess
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    Of course @RolyPoly y(P)=a cosx+b sin x y`=-A sin x+b cos x y``=a cos x-b sin x y``=-y -y-5y`(-a sin x+b cos x)-6y =cos x 5asinx-5b cos x-7acosx -7b sin x=cos x (5a-7b)sinx-(5b+7a)cos x=cosx 5A-7B=0 -(5B+7A)=0

    • one year ago
  11. guess
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    @scarydoor thanks

    • one year ago
  12. RolyPoly
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    y`=-A sin x+b cos x Isn't y'' = -Acosx - Bsinx?

    • one year ago
  13. RolyPoly
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    But that doesn't affect your conclusion that y'' = -y y(P)=a cosx+b sin x y`=-A sin x+b cos x y``=-a cos x-b sin x = -y So, -y-5y`-6y =cos x -7 y - 5y' =cos x (5a-7b)sinx-(5b+7a)cos x=cosx 5a-7b=0 -(5b+7a)=0 (it should be -(5b+7a) =1?!)

    • one year ago
  14. RolyPoly
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    When you get the two equations: 5a-7b =0 and -(5b+7a) =1 , you can solve a and b and you'll get the particular solution.

    • one year ago
  15. guess
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    @RolyPoly no maybe -5b-7a

    • one year ago
  16. RolyPoly
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    (5a-7b)sinx-(5b+7a)cos x=cosx Compare the terms on the left and right, there is no sinx term, so the coefficient of sinx is 0 Hence you get the first equation 5a-7b =0 there is a cosx term, with the coefficient 1, so the coefficient of cosx term on the left is equal to 1. Hence we should get -(5b+7a) = 1 as the second equation. Got it?

    • one year ago
  17. guess
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    @RolyPoly yes thank you so we get a=7/74 b=5/74 ?

    • one year ago
  18. RolyPoly
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    Not quite.

    • one year ago
  19. RolyPoly
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    You've missed a negative sign for both a and b...

    • one year ago
  20. guess
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    @RolyPoly thank you

    • one year ago
  21. RolyPoly
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    Welcome :)

    • one year ago
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