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Math Help - Finding general solution of 2ODE

  1. #1
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    Finding general solution of 2ODE

    The equation which I have is:
    d2y/dx2 - dy/dx - 2y = 2cos(2x).

    I know that I should find the Auxiliary equation, which in this case would be:
    m^2 - m - 2 = 0
    which would generate (m+2)(m-1)=0, giving the equation:
    Ae^1 + Be^-2. The problem I have is finding the complimentary function, as I dont know what to start with. Any help is appreciated, and if possible find general solution. Thanks
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  2. #2
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    try
    y=C \cos 2x + D\sin 2x
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  3. #3
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    Quote Originally Posted by thelostchild View Post
    try
    y=C \cos 2x + D\sin 2x
    Ok I worked it out like that, and I got an answer of C= -30 and D= -10, which sounds a bit iffy to be honest . I dont know the answer for this so I'm not sure if that is the correct answer.
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  4. #4
    Behold, the power of SARDINES!
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    You can ALWAYS check your solution to an ODE by taking the needed derivatives and pluggng them back into the original problem.

    If you just want to check the particular solution only use it(because the complimentry solution should give 0)
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by LooNiE
    http://www.mathhelpforum.com/math-he...tml#post279219

    The way you explained that question I got all dizzy now and I thought I could sort of get it before. How can I 'plug' it in when I am not given any values of dy/dx, t or x?
    y_p=-30\cos(2x)-10\sin(2x)

    Now lets take two derivatives to see what we get

    y_p'=60\sin(2x)-20\cos(2x)

    y_p''=120\sin(2x)+40\sin(2x)

    Now we can plug these back into the original ODE

    y''-y-2y=\cos(2x)

    if after you plug in and get an identity your solution is correct.
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