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Solve the Differential Equation y"+16y=4cos(x)? watch

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    I must find the general solution of the following differential equation:
    y"+16y=4cos(x)

    This is what I have:

    Complimentary solution yc(x)
    Solve for r:
    r^2+16 = 0
    r = ±4i
    Plug into formula.
    yc(x) = C1sin(4x)+C2cos(4x)

    Particular test solution yt(x)
    yt(x) = Acos(x)+Bsin(x)
    y't(x) = Bcos(x)-A*sin(x)
    y''t(x) = -Acos(x)-Bsin(x)
    Set equal to RHS of original equation and solve for A and B:
    -Acos(x)-Bsin(x) = 4cos(x)
    A = -4, B = 0
    Particular solution:
    yp(x) = -4cos(x)

    Final solution:
    y(x) = yc(x) + yp(x)
    y(x) = C1sin(4x)+C2cos(4x)-4cos(x)

    I know the answer is supposed to be:
    y(x) = C1sin(4x)+C2cos(4x)+(4/15)*cos(x)

    So where did I go wrong with the particular solution? I think my particular test solution may be wrong, so how do I find a particular test solution?
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    -Acos(x)-Bsin(x) + 16(Acos(x)+Bsin(x))= 4cos(x)...do you mean?
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    I think that's a different path to solving it... I'm supposed to come up with a complimentary and particular solution and add those to get the total general solution.

    In my notes there is an example where we solved:
    y''+9y=2cos(3x)+3sin(3x)
    The complimentary solution was
    yc(x)=C1cos(3x)+C2sin(3x)
    and the particular test solution was
    yt(x)=x(Acos(3x)+Bsin(3x))
    (Particular test solution multiplied by x due to similarity to complimentary solution.)
    We then found the second derivative of that, set it equal to the original equation's RHS, and solved for A and B, then substituted those values in for the A and B in the original particular test solution to get the particular solution, which was added to the complimentary solution to get the final solution of:
    y(x) = C1cos(3x)+C2sin(3x) - (1/2)xcos(3x)+(1/3)xsin(3x)
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    (Original post by Josh_Dey)
    0.o
    ...?
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    (Original post by Zach827)
    Set equal to RHS of original equation and solve for A and B:
    -Acos(x)-Bsin(x) = 4cos(x)
    Not sure you did this bit right. I put it in and get A = 4/15 as your answer suggests.
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    (Original post by Sasukekun)
    Not sure you did this bit right. I put it in and get A = 4/15 as your answer suggests.
    Okay, but how?
    -Acos(x)-Bsin(x) = 4cos(x)
    B = 0
    -Acos(x) = 4cos(x)
    If A=-4
    -(-4)cos(x) = 4cos(x)
    4cos(x) = 4cos(x)


    What did you do to get to 4/15?
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    -Acos(x)-Bsin(x) = 4cos(x)

    You didn't add 16y.
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    (Original post by Zach827)
    Okay, but how?
    -Acos(x)-Bsin(x) = 4cos(x)
    B = 0
    -Acos(x) = 4cos(x)
    If A=-4
    -(-4)cos(x) = 4cos(x)
    4cos(x) = 4cos(x)


    What did you do to get to 4/15?
    You seem to be missing the 16y.
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    Oh! Okay, thanks. I'll try redoing it to see if I get the right answer.
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    Sorry is this C4?
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    Answer :-

    y = a cos(4x) + b sin(4x) + (4/15) cos(x)



    y"+16y=4cos(x)


    y'' + 16 y = 0

    y_p = A cos(x) + B sin(x), and seek the coefficients A and B'

    y = y_homogeneous + y_particular.

    y = a cos(4x) + b sin(4x) + (4/15) cos(x) (???) )) Tis been a very long time since' hehehe

    +NNM+
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    (Original post by Nash24)
    Sorry is this C4?
    No, you'd usually get this sort of question in 1st year uni maths.
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    (Original post by ttoby)
    No, you'd usually get this sort of question in 1st year uni maths.
    Oh thank god..
    I have a C4 mock in a few days and most of this was not making any sense to me.
    Thanks.
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    It's for a Differential Equations class (book involves Linear Algebra).
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    The solution is:

    

y(x) = c_1 sin(4 x)+c_2 cos(4 x)+\frac{4 cos(x)}{15}

    I just used a computer algebra to solve it. (i.e. I cheated)
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    (Original post by noobynoo)
    The solution is:

    

y(x) = c_1 sin(4 x)+c_2 cos(4 x)+\frac{4 cos(x)}{15}

    I just used a computer algebra to solve it. (i.e. I cheated)
    Lol, I know the solution >.< I was having trouble getting to it.... TI-89 calculators are amazing.
 
 
 
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