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Math Help - Particular solution to a differential equations

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    Particular solution to a differential equations

    Find the particular solution to y^{(4)}+5y''+4y = \sin x + \cos 2x

    First, I found that the general solution is y_h= C_1 \cos (2x) + C_2 \sin (2x) + C_3 \cos (-2x) +  C_4 \sin (-2x) + C_5 \cos (x) + C_6 \sin (x) + C_7 \cos (-x) + C_8 \sin (-x)

    So the list of terms that I obtain here is:  \{ \cos (2x), \sin (2x),  \cos (-2x) , \sin (-2x) , \cos (x) , \sin (x) , \cos (-x) , \sin (-x) \}

    Now, for my guess of the particular solution, I have list of terms  \{ \sin x , \cos (2x) \}

    Since I have repeats, I would take the derivative, so I have new list  \{ \sin x , \cos x , \cos (2x) , \sin (2x) \}

    So my particular solution is: y_p = A \sin x + B \cos x + C \cos (2x) + D \sin (2x)

    I know I'm missing something, I think I'm suppose to multiply x by one of the list, but I just can't remember what the method was, please help. Thanks.
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    Quote Originally Posted by tttcomrader View Post
    Find the particular solution to y^{(4)}+5y''+4y = \sin x + \cos 2x

    First, I found that the general solution is y_h= C_1 \cos (2x) + C_2 \sin (2x) + C_3 \cos (-2x) +  C_4 \sin (-2x) + C_5 \cos (x) + C_6 \sin (x) + C_7 \cos (-x) + C_8 \sin (-x)

    So the list of terms that I obtain here is:  \{ \cos (2x), \sin (2x), \cos (-2x) , \sin (-2x) , \cos (x) , \sin (x) , \cos (-x) , \sin (-x) \}

    Now, for my guess of the particular solution, I have list of terms  \{ \sin x , \cos (2x) \}

    Since I have repeats, I would take the derivative, so I have new list  \{ \sin x , \cos x , \cos (2x) , \sin (2x) \}

    So my particular solution is: y_p = A \sin x + B \cos x + C \cos (2x) + D \sin (2x)

    I know I'm missing something, I think I'm suppose to multiply x by one of the list, but I just can't remember what the method was, please help. Thanks.
    First off, you have some redundancy. The complimentary solution is only

    y = c_1 \sin x + c_2 \cos x + c_3 \sin 2x + c_4 \cos 2x

    Second, since the homogeneous terms appears as part of your complimentary solution you would need to seek a particluar solution of the form

    y = c_1 x \sin x + c_2 x \cos x + c_3 x \sin 2x + c_4 x \cos 2x
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    So if the same terms appear, I just need to multiply everything by x?
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    Quote Originally Posted by tttcomrader View Post
    So if the same terms appear, I just need to multiply everything by x?
    Yes or higher. For example

    y'' - 2y' + y = e^x

    For the particular solution you'll need y_p = Ax^2 e^x as e^x\;\text{and}\; x e^x are a part of the complimentary solution.
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    So for y''+2y'-3y=1-xe^x

    I have the complimentary solution y = C_1e^{-3x}+C_2e^x

    So my terms are  \{ 1 , xe^x \}

    But this time there are no repeats, and I tried with y=A+Bxe^x , but things went wrong, everything cancel, what should I do next?
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    Quote Originally Posted by tttcomrader View Post
    So for y''+2y'-3y=1-xe^x

    I have the complimentary solution y = C_1e^{-3x}+C_2e^x

    So my terms are  \{ 1 , xe^x \}

    But this time there are no repeats, and I tried with y=A+Bxe^x , but things went wrong, everything cancel, what should I do next?
    Because e^x is a part of your complimentary solution and x e^x is a part of your homogeneous term, then you would seek a particular solution of the form

    y_p = A x e^x + B x^2 e^x.
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    Okay, I tried with this particular solution, but weird things happened:

    After plugging in everything to the original equations, I have the following:

    y=Axe^x+Bx^2e^x
    y'=Ae^x+(A+2B)xe^x+Bx^2e^x
    y''=(A+2B)e^x+(A+2B)xe^x+Bx^2e^x

    So I have

    (B+2b-3B)x^2e^x+(A+2B+2A+4B-3A)xe^x+(A+2B+2A)e^x=1+xe^x
    (3A+2B)e^x+(6B)xe^x=1+(1)xe^x

    But I don't have anything on the left to equal to the 1 on the right, what am I doing wrong? I really feel stupid now because I'm actually tutoring this class, I took DEQ like 3 years ago, I JUST CAN'T UNDERSTAND WHY I AM NOT GETTING THIS RIGHT AS A GRADUATE STUDENT!!!!

    Perhaps I need y_p=A+Bxe^x+Cx^2e^x because of the constant term in the RHS?
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    Quote Originally Posted by tttcomrader View Post
    So for y''+2y'-3y=1-xe^x
    I have the complimentary solution y = C_1e^{-3x}+C_2e^x

    So my terms are  \{ 1 , xe^x \}

    But this time there are no repeats, and I tried with y=A+Bxe^x , but things went wrong, everything cancel, what should I do next?
    Quote Originally Posted by tttcomrader View Post
    Okay, I tried with this particular solution, but weird things happened:

    After plugging in everything to the original equations, I have the following:

    y=Axe^x+Bx^2e^x
    y'=Ae^x+(A+2B)xe^x+Bx^2e^x
    y''=(A+2B)e^x+(A+2B)xe^x+Bx^2e^x

    So I have

    (B+2b-3B)x^2e^x+(A+2B+2A+4B-3A)xe^x+(A+2B+2A)e^x=1+xe^x
    (3A+2B)e^x+(6B)xe^x=1+(1)xe^x

    But I don't have anything on the left to equal to the 1 on the right, what am I doing wrong? I really feel stupid now because I'm actually tutoring this class, I took DEQ like 3 years ago, I JUST CAN'T UNDERSTAND WHY I AM NOT GETTING THIS RIGHT AS A GRADUATE STUDENT!!!!

    Perhaps I need y_p=A+Bxe^x+Cx^2e^x because of the constant term in the RHS?
    Quote Originally Posted by danny arrigo View Post
    Because e^x is a part of your complimentary solution and x e^x is a part of your homogeneous term, then you would seek a particular solution of the form

    y_p = A x e^x + B x^2 e^x.
    This is how to find the particular solution corresponding to the -x e^x term on the RHS of the DE.

    To find the particular solution corresponding to the constant term on the RHS of the DE you try one of the form y = a and get a = -1/3.

    Then you add these two particular solutions together to get the overall particular solution.
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